Water collected between towers: Difference between revisions
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Takes the integers as input from command line, prints out usage on incorrect invocation. |
Takes the integers as input from command line, prints out usage on incorrect invocation. |
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<lang C> |
<lang C> |
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/*Abhishek Ghosh, 7th November 2017*/ |
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#include<stdlib.h> |
#include<stdlib.h> |
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#include<stdio.h> |
#include<stdio.h> |
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Revision as of 16:04, 29 July 2018
You are encouraged to solve this task according to the task description, using any language you may know.
- Task
In a two-dimensional world, we begin with any bar-chart (or row of close-packed 'towers', each of unit width), and then it rains, completely filling all convex enclosures in the chart with water.
9 ██ 9 ██ 8 ██ 8 ██ 7 ██ ██ 7 ██≈≈≈≈≈≈≈≈██ 6 ██ ██ ██ 6 ██≈≈██≈≈≈≈██ 5 ██ ██ ██ ████ 5 ██≈≈██≈≈██≈≈████ 4 ██ ██ ████████ 4 ██≈≈██≈≈████████ 3 ██████ ████████ 3 ██████≈≈████████ 2 ████████████████ ██ 2 ████████████████≈≈██ 1 ████████████████████ 1 ████████████████████
In the example above, a bar chart representing the values [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] has filled, collecting 14 units of water.
Write a function, in your language, from a given array of heights, to the number of water units that can be held in this way, by a corresponding bar chart.
Calculate the number of water units that could be collected by bar charts representing each of the following seven series:
[[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]]
See, also:
- Four Solutions to a Trivial Problem – a Google Tech Talk by Guy Steele
- Water collected between towers on Stack Overflow, from which the example above is taken)
- An interesting Haskell solution, using the Tardis monad, by Phil Freeman in a Github gist.
AppleScript
<lang AppleScript>-- WATER COLLECTED BETWEEN TOWERS --------------------------------------------
-- waterCollected :: [Int] -> Int on waterCollected(xs)
set leftWalls to scanl1(my max, xs)
set rightWalls to scanr1(my max, xs)
set waterLevels to zipWith(my min, leftWalls, rightWalls)
-- positive :: Num a => a -> Bool
script positive
on |λ|(x)
x > 0
end |λ|
end script
-- minus :: Num a => a -> a -> a
script minus
on |λ|(a, b)
a - b
end |λ|
end script
sum(filter(positive, zipWith(minus, waterLevels, xs)))
end waterCollected
-- TEST ----------------------------------------------------------------------
on run
map(waterCollected, ¬
[[1, 5, 3, 7, 2], ¬
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2], ¬
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], ¬
[5, 5, 5, 5], ¬
[5, 6, 7, 8], ¬
[8, 7, 7, 6], ¬
[6, 7, 10, 7, 6]])
--> {2, 14, 35, 0, 0, 0, 0}
end run
-- GENERIC FUNCTIONS ---------------------------------------------------------
-- filter :: (a -> Bool) -> [a] -> [a] on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter
-- foldl :: (a -> b -> a) -> a -> [b] -> a on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- init :: [a] -> [a] on init(xs)
if length of xs > 1 then
items 1 thru -2 of xs
else
{}
end if
end init
-- map :: (a -> b) -> [a] -> [b] on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- max :: Ord a => a -> a -> a on max(x, y)
if x > y then
x
else
y
end if
end max
-- min :: Ord a => a -> a -> a on min(x, y)
if y < x then
y
else
x
end if
end min
-- Lift 2nd class handler function into 1st class script wrapper -- mReturn :: Handler -> Script on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- scanl :: (b -> a -> b) -> b -> [a] -> [b] on scanl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
set lst to {startValue}
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
set end of lst to v
end repeat
return lst
end tell
end scanl
-- scanl1 :: (a -> a -> a) -> [a] -> [a] on scanl1(f, xs)
if length of xs > 0 then
scanl(f, item 1 of xs, items 2 thru -1 of xs)
else
{}
end if
end scanl1
-- scanr :: (b -> a -> b) -> b -> [a] -> [b] on scanr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
set lst to {startValue}
repeat with i from lng to 1 by -1
set v to |λ|(v, item i of xs, i, xs)
set end of lst to v
end repeat
return reverse of lst
end tell
end scanr
-- scanr1 :: (a -> a -> a) -> [a] -> [a] on scanr1(f, xs)
if length of xs > 0 then
scanr(f, item -1 of xs, items 1 thru -2 of xs)
else
{}
end if
end scanr1
-- sum :: Num a => [a] -> a on sum(xs)
script add
on |λ|(a, b)
a + b
end |λ|
end script
foldl(add, 0, xs)
end sum
-- tail :: [a] -> [a] on tail(xs)
if length of xs > 1 then
items 2 thru -1 of xs
else
{}
end if
end tail
-- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c] on zipWith(f, xs, ys)
set lng to min(length of xs, length of ys)
set lst to {}
tell mReturn(f)
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, item i of ys)
end repeat
return lst
end tell
end zipWith</lang>
- Output:
<lang AppleScript>{2, 14, 35, 0, 0, 0, 0}</lang>
AWK
<lang AWK>
- syntax: GAWK -f WATER_COLLECTED_BETWEEN_TOWERS.AWK [-v debug={0|1}]
BEGIN {
wcbt("1,5,3,7,2")
wcbt("5,3,7,2,6,4,5,9,1,2")
wcbt("2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1")
wcbt("5,5,5,5")
wcbt("5,6,7,8")
wcbt("8,7,7,6")
wcbt("6,7,10,7,6")
exit(0)
} function wcbt(str, ans,hl,hr,i,n,tower) {
n = split(str,tower,",")
for (i=n; i>=0; i--) { # scan right to left
hr[i] = max(tower[i],(i<n)?hr[i+1]:0)
}
for (i=0; i<=n; i++) { # scan left to right
hl[i] = max(tower[i],(i!=0)?hl[i-1]:0)
ans += min(hl[i],hr[i]) - tower[i]
}
printf("%4d : %s\n",ans,str)
if (debug == 1) {
for (i=1; i<=n; i++) { printf("%-4s",tower[i]) } ; print("tower")
for (i=1; i<=n; i++) { printf("%-4s",hl[i]) } ; print("l-r")
for (i=1; i<=n; i++) { printf("%-4s",hr[i]) } ; print("r-l")
for (i=1; i<=n; i++) { printf("%-4s",min(hl[i],hr[i])) } ; print("min")
for (i=1; i<=n; i++) { printf("%-4s",min(hl[i],hr[i])-tower[i]) } ; print("sum\n")
}
} function max(x,y) { return((x > y) ? x : y) } function min(x,y) { return((x < y) ? x : y) } </lang>
- Output:
2 : 1,5,3,7,2 14 : 5,3,7,2,6,4,5,9,1,2 35 : 2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1 0 : 5,5,5,5 0 : 5,6,7,8 0 : 8,7,7,6 0 : 6,7,10,7,6
C
Takes the integers as input from command line, prints out usage on incorrect invocation. <lang C>
- include<stdlib.h>
- include<stdio.h>
int getWater(int* arr,int start,int end,int cutoff){ int i, sum = 0;
for(i=start;i<=end;i++) sum += ((arr[cutoff] > arr[i])?(arr[cutoff] - arr[i]):0);
return sum; }
int netWater(int* arr,int size){ int i, j, ref1, ref2, marker, markerSet = 0,sum = 0;
if(size<3) return 0;
for(i=0;i<size-1;i++){ start:if(i!=size-2 && arr[i]>arr[i+1]){ ref1 = i;
for(j=ref1+1;j<size;j++){ if(arr[j]>=arr[ref1]){ ref2 = j;
sum += getWater(arr,ref1+1,ref2-1,ref1);
i = ref2;
goto start; }
else if(j!=size-1 && arr[j] < arr[j+1] && (markerSet==0||(arr[j+1]>=arr[marker]))){ marker = j+1; markerSet = 1; } }
if(markerSet==1){ sum += getWater(arr,ref1+1,marker-1,marker);
i = marker;
markerSet = 0;
goto start; } } }
return sum; }
int main(int argC,char* argV[]) { int *arr,i;
if(argC==1) printf("Usage : %s <followed by space separated series of integers>"); else{ arr = (int*)malloc((argC-1)*sizeof(int));
for(i=1;i<argC;i++) arr[i-1] = atoi(argV[i]);
printf("Water collected : %d",netWater(arr,argC-1)); }
return 0; } </lang> Output :
C:\rosettaCode>waterTowers.exe 1 5 3 7 2 Water collected : 2 C:\rosettaCode>waterTowers.exe 5 3 7 2 6 4 5 9 1 2 Water collected : 14 C:\rosettaCode>waterTowers.exe 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 Water collected : 35 C:\rosettaCode>waterTowers.exe 5 5 5 5 Water collected : 0 C:\rosettaCode>waterTowers.exe 8 7 7 6 Water collected : 0 C:\rosettaCode>waterTowers.exe 6 7 10 7 6 Water collected : 0
C++
<lang cpp>
- include <iostream>
- include <vector>
- include <algorithm>
enum { EMPTY, WALL, WATER };
auto fill(const std::vector<int> b) {
auto water = 0;
const auto rows = *std::max_element(std::begin(b), std::end(b));
const auto cols = std::size(b);
std::vector<std::vector<int>> g(rows);
for (auto& r : g) {
for (auto i = 0; i < cols; ++i) {
r.push_back(EMPTY);
}
}
for (auto c = 0; c < cols; ++c) {
for (auto r = rows - 1u, i = 0u; i < b[c]; ++i, --r) {
g[r][c] = WALL;
}
}
for (auto c = 0; c < cols - 1; ++c) {
auto start_row = rows - b[c];
while (start_row < rows) {
if (g[start_row][c] == EMPTY) break;
auto c2 = c + 1;
bool hitWall = false;
while (c2 < cols) {
if (g[start_row][c2] == WALL) {
hitWall = true;
break;
}
++c2;
}
if (hitWall) {
for (auto i = c + 1; i < c2; ++i) {
g[start_row][i] = WATER;
++water;
}
}
++start_row;
}
}
return water;
}
int main() {
std::vector<std::vector<int>> b = {
{ 1, 5, 3, 7, 2 },
{ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 },
{ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 },
{ 5, 5, 5, 5 },
{ 5, 6, 7, 8 },
{ 8, 7, 7, 6 },
{ 6, 7, 10, 7, 6 }
};
for (const auto v : b) {
auto water = fill(v);
std::cout << water << " water drops." << std::endl;
}
std::cin.ignore();
std::cin.get();
return 0;
}</lang>
- Output:
2 water drops. 14 water drops. 35 water drops. 0 water drops. 0 water drops. 0 water drops. 0 water drops.
C#
Version 1
Translation from Visual Basic .NET. See that version 1 entry for code comments and details. <lang Csharp>using System;</lang> <lang Csharp>static void Main(string[] args) {
int[][] wta = {
new int[] {1, 5, 3, 7, 2}, new int[] { 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 },
new int[] { 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 },
new int[] { 5, 5, 5, 5 }, new int[] { 5, 6, 7, 8 },
new int[] { 8, 7, 7, 6 }, new int[] { 6, 7, 10, 7, 6 }};
string blk = "", lf = "\n"; for (int i = 0; i < wta.Length; i++)
{
int bpf; blk = ""; do
{
string floor = ""; bpf = 0; for (int j = 0; j < wta[i].Length; j++)
{
if (wta[i][j] > 0)
{ floor += "██"; wta[i][j] -= 1; bpf += 1; }
else floor += (j > 0 && j < wta[i].Length - 1 ? "≈≈" : " ");
}
if (bpf > 0) blk = floor + lf + blk;
} while (bpf > 0);
while (blk.Contains(" ≈≈")) blk = blk.Replace(" ≈≈", " ");
while (blk.Contains("≈≈ ")) blk = blk.Replace("≈≈ ", " ");
if (args.Length > 0) Console.Write("{0}", blk);
Console.WriteLine("Block {0} retains {1,2} water units.\n", i + 1,
(blk.Length - blk.Replace("≈≈", "").Length) / 2);
}
}
</lang>
- Output:
<lang>Block 1 retains 2 water units.
Block 2 retains 14 water units. Block 3 retains 35 water units. Block 4 retains 0 water units. Block 5 retains 0 water units. Block 6 retains 0 water units. Block 7 retains 0 water units.</lang>
Version 2
Conventional "scanning" algorithm, translated from the second version of Visual Basic.NET, but (intentionally tweaked to be) incapable of verbose output. See that version 2 entry for code comments and details. <lang cSharp>// Variable names key: // i Iterator (of the tower block array). // tba Tower block array. // tea Tower elevation array. // rht Right hand tower column number (position). // wu Water units (count). // bof Blocks on floor (count). // col Column number in elevation array (position).
static void Main(string[] args) {
int i = 1; int[][] tba = {new int[] { 1, 5, 3, 7, 2 },
new int[] { 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 },
new int[] { 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 },
new int[] { 5, 5, 5, 5 }, new int[] { 5, 6, 7, 8 },
new int[] { 8, 7, 7, 6 }, new int[] { 6, 7, 10, 7, 6 }};
foreach (var tea in tba)
{
int rht, wu = 0, bof; do
{
for (rht = tea.Length - 1; rht >= 0; rht--) if (tea[rht] > 0) break;
if (rht < 0) break;
bof = 0; for (int col = 0; col <= rht; col++)
{
if (tea[col] > 0) { tea[col] -= 1; bof += 1; }
else if (bof > 0) wu++;
}
if (bof < 2) break;
} while (true);
System.Console.WriteLine(string.Format("Block {0} {1} water units.", i++,
wu == 0 ? "does not hold any" : "holds " + wu.ToString()));
}
}</lang> Output: <lang>Block 1 holds 2 water units. Block 2 holds 14 water units. Block 3 holds 35 water units. Block 4 does not hold any water units. Block 5 does not hold any water units. Block 6 does not hold any water units. Block 7 does not hold any water units.</lang>
Clojure
Similar two passes algorithm as many solutions here. First traverse left to right to find the highest tower on the left of each position, inclusive of the tower at the current position, than do the same to find the highest tower to the right of each position. Finally, compute the total water units held at any position as the difference of those two heights.
<lang clojure> (defn trapped-water [towers]
(let [maxes #(reductions max %) ; the seq of increasing max values found in the input seq
maxl (maxes towers) ; the seq of max heights to the left of each tower
maxr (reverse (maxes (reverse towers))) ; the seq of max heights to the right of each tower
mins (map min maxl maxr)] ; minimum highest surrounding tower per position
(reduce + (map - mins towers)))) ; sum up the trapped water per position
</lang>
- Output:
<lang clojure>
- in the following, # is a tower block and ~ is trapped water
- 10|
- 9| #
- 8| #
- 7| # ~ ~ ~ ~ #
- 6| # ~ # ~ ~ #
- 5| # ~ # ~ # ~ # #
- 4| # ~ # ~ # # # #
- 3| # # # ~ # # # #
- 2| # # # # # # # # ~ #
- 1| # # # # # # # # # #
- ---+---------------------
- 5 3 7 2 6 4 5 9 1 2
(trapped-water [5 3 7 2 6 4 5 9 1 2]) ;; 14 </lang>
D
<lang D>import std.stdio;
void main() {
int i = 1;
int[][] tba = [
[ 1, 5, 3, 7, 2 ],
[ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 ],
[ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 ],
[ 5, 5, 5, 5 ],
[ 5, 6, 7, 8 ],
[ 8, 7, 7, 6 ],
[ 6, 7, 10, 7, 6 ]
];
foreach (tea; tba) {
int rht, wu, bof;
do {
for (rht = tea.length - 1; rht >= 0; rht--) {
if (tea[rht] > 0) {
break;
}
}
if (rht < 0) {
break;
}
bof = 0;
for (int col = 0; col <= rht; col++) {
if (tea[col] > 0) {
tea[col] -= 1; bof += 1;
} else if (bof > 0) {
wu++;
}
}
if (bof < 2) {
break;
}
} while (true);
write("Block ", i++);
if (wu == 0) {
write(" does not hold any");
} else {
write(" holds ", wu);
}
writeln(" water units.");
}
}</lang>
- Output:
Block 1 holds 2 water units. Block 2 holds 14 water units. Block 3 holds 35 water units. Block 4 does not hold any water units. Block 5 does not hold any water units. Block 6 does not hold any water units. Block 7 does not hold any water units.
Erlang
Implements a version that uses recursion to solve the problem functionally, using two passes without requiring list reversal or modifications. On the list iteration from head to tail, gather the largest element seen so far (being the highest one on the left). Once the list is scanned, each position returns the highest tower to its right as reported by its follower, along with the amount of water seen so far, which can then be used to calculate the value at the current position. Back at the first list element, the final result is gathered.
<lang erlang> -module(watertowers). -export([towers/1, demo/0]).
towers(List) -> element(2, tower(List, 0)).
tower([], _) -> {0,0}; tower([H|T], MaxLPrev) ->
MaxL = max(MaxLPrev, H),
{MaxR, WaterAcc} = tower(T, MaxL),
{max(MaxR,H), WaterAcc+max(0, min(MaxR,MaxL)-H)}.
demo() ->
Cases = [[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]],
[io:format("~p -> ~p~n", [Case, towers(Case)]) || Case <- Cases],
ok.
</lang>
- Output:
1> watertowers:demo(). [1,5,3,7,2] -> 2 [5,3,7,2,6,4,5,9,1,2] -> 14 [2,6,3,5,2,8,1,4,2,2,5,3,5,7,4,1] -> 35 [5,5,5,5] -> 0 [5,6,7,8] -> 0 [8,7,7,6] -> 0 [6,7,10,7,6] -> 0 ok
F#
see http://stackoverflow.com/questions/24414700/water-collected-between-towers/43779936#43779936 for an explanation of this code. It is proportional to the number of towers. Although the examples on stackoverflow claim this, the n they use is actually the distance between the two end towers and not the number of towers. Consider the case of a tower of height 5 at 1, a tower of height 10 at 39, and a tower of height 3 at 101. <lang fsharp> (* A solution I'd show to Euclid !!!!. Nigel Galloway May 4th., 2017
- )
let solve n =
let (n,_)::(i,e)::g = n|>List.sortBy(fun n->(-(snd n))) let rec fn i g e l = match e with | (n,e)::t when n < i -> fn n g t (l+(i-n-1)*e) | (n,e)::t when n > g -> fn i n t (l+(n-g-1)*e) | (n,t)::e -> fn i g e (l-t) | _ -> l fn (min n i) (max n i) g (e*(abs(n-i)-1))
</lang>
- Output:
solve [(1,1);(2,5);(3,3);(4,7);(5,2)] -> 2 solve [(1,5);(2,3);(3,7);(4,2);(5,6);(6,4);(7,5);(8,9);(9,1);(10,2)] -> 14 solve [(1,2);(2,6);(3,3);(4,5);(5,2);(6,8);(7,1);(8,4);(9,2);(10,2);(11,5);(12,3);(13,5);(14,7);(15,4);(16,1)] -> 35 solve [(1,5);(2,5);(3,5);(4,5)] -> 0 solve [(1,5);(2,6);(3,7);(4,8)] -> 0 solve [(1,8);(2,7);(3,7);(4,6)] -> 0 solve [(1,6);(2,7);(3,10);(4,7);(5,6)] -> 0 solve [(1,5);(39,10);(101,3)] -> 368
Factor
<lang factor>USING: formatting kernel math math.order math.statistics sequences ; IN: rosetta-code.water-towers
CONSTANT: test-cases {
{ 1 5 3 7 2 }
{ 5 3 7 2 6 4 5 9 1 2 }
{ 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 }
{ 5 5 5 5 }
{ 5 6 7 8 }
{ 8 7 7 6 }
{ 6 7 10 7 6 }
}
- area ( seq -- n )
dup [ cum-max ] [ <reversed> cum-max reverse ] bi [ min ] 2map swap [ - ] 2map sum ;
test-cases [ dup area "%[%d, %] -> %d\n" printf ] each</lang>
- Output:
{ 1, 5, 3, 7, 2 } -> 2
{ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 } -> 14
{ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 } -> 35
{ 5, 5, 5, 5 } -> 0
{ 5, 6, 7, 8 } -> 0
{ 8, 7, 7, 6 } -> 0
{ 6, 7, 10, 7, 6 } -> 0
Go
<lang go> package main
import "fmt"
func maxl(hm []int ) []int{ res := make([]int,len(hm)) max := 1 for i := 0; i < len(hm);i++{ if(hm[i] > max){ max = hm[i] } res[i] = max; } return res } func maxr(hm []int ) []int{ res := make([]int,len(hm)) max := 1 for i := len(hm) - 1 ; i >= 0;i--{ if(hm[i] > max){ max = hm[i] } res[i] = max; } return res } func min(a,b []int) []int { res := make([]int,len(a)) for i := 0; i < len(a);i++{ if a[i] >= b[i]{ res[i] = b[i] }else { res[i] = a[i] } } return res } func diff(hm, min []int) []int { res := make([]int,len(hm)) for i := 0; i < len(hm);i++{ if min[i] > hm[i]{ res[i] = min[i] - hm[i] } } return res } func sum(a []int) int { res := 0 for i := 0; i < len(a);i++{ res += a[i] } return res }
func waterCollected(hm []int) int { maxr := maxr(hm) maxl := maxl(hm) min := min(maxr,maxl) diff := diff(hm,min) sum := sum(diff) return sum }
func main() {
fmt.Println(waterCollected([]int{1, 5, 3, 7, 2}))
fmt.Println(waterCollected([]int{5, 3, 7, 2, 6, 4, 5, 9, 1, 2}))
fmt.Println(waterCollected([]int{2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1}))
fmt.Println(waterCollected([]int{5, 5, 5, 5}))
fmt.Println(waterCollected([]int{5, 6, 7, 8}))
fmt.Println(waterCollected([]int{8, 7, 7, 6}))
fmt.Println(waterCollected([]int{6, 7, 10, 7, 6}))
}</lang>
- Output:
2 14 35 0 0 0 0
Groovy
<lang Groovy> Integer waterBetweenTowers(List<Integer> towers) {
// iterate over the vertical axis. There the amount of water each row can hold is
// the number of empty spots, minus the empty spots at the beginning and end
return (1..towers.max()).collect { height ->
// create a string representing the row, '#' for tower material and ' ' for air
// use .trim() to remove spaces at beginning and end and then count remaining spaces
towers.collect({ it >= height ? "#" : " " }).join("").trim().count(" ")
}.sum()
}
tasks = [
[1, 5, 3, 7, 2], [5, 3, 7, 2, 6, 4, 5, 9, 1, 2], [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], [5, 5, 5, 5], [5, 6, 7, 8], [8, 7, 7, 6], [6, 7, 10, 7, 6]
]
tasks.each {
println "$it => total water: ${waterBetweenTowers it}"
} </lang>
- Output:
[1, 5, 3, 7, 2] => total water: 2 [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] => total water: 14 [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] => total water: 35 [5, 5, 5, 5] => total water: 0 [5, 6, 7, 8] => total water: 0 [8, 7, 7, 6] => total water: 0 [6, 7, 10, 7, 6] => total water: 0
Haskell
Following the approach of slightly modified cdk's Haskell solution at Stack Overflow. As recommended in Programming as if the Correct Data Structure (and Performance) Mattered it uses Vector instead of Array:
<lang haskell>import Data.Vector.Unboxed (Vector) import qualified Data.Vector.Unboxed as V
waterCollected :: Vector Int -> Int waterCollected =
V.sum . -- Sum of the water depths over each of
V.filter (> 0) . -- the columns that are covered by some water.
(
V.zipWith (-) =<< -- Where coverages are differences between:
(
V.zipWith min . -- the lower water level in each case of:
V.scanl1 max <*> -- highest wall to left, and
V.scanr1 max -- highest wall to right.
)
)
main :: IO () main =
mapM_ (print . waterCollected) [ V.fromList [1, 5, 3, 7, 2] , V.fromList [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] , V.fromList [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] , V.fromList [5, 5, 5, 5] , V.fromList [5, 6, 7, 8] , V.fromList [8, 7, 7, 6] , V.fromList [6, 7, 10, 7, 6] ]
</lang>
- Output:
2 14 35 0 0 0 0
J
Inspired by #Julia.
Solution: <lang j>collectLevels =: >./\ <. >./\. NB. collect levels after filling waterLevels=: collectLevels - ] NB. water levels for each tower collectedWater=: +/@waterLevels NB. sum the units of water collected printTowers =: ' ' , [: |.@|: '#~' #~ ] ,. waterLevels NB. print a nice graph of towers and water</lang>
Examples: <lang j> collectedWater 5 3 7 2 6 4 5 9 1 2 14
printTowers 5 3 7 2 6 4 5 9 1 2
#
#
#~~~~#
#~#~~#
- ~#~#~##
- ~#~####
- ~####
- ~#
- ~#
- ~####
NB. Test cases
TestTowers =: <@".;._2 noun define
1 5 3 7 2 5 3 7 2 6 4 5 9 1 2 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 5 5 5 5 5 6 7 8 8 7 7 6 6 7 10 7 6 )
TestResults =: 2 14 35 0 0 0 0 TestResults -: collectedWater &> TestTowers NB. check tests
1</lang>
Java
<lang Java>public class WaterBetweenTowers {
public static void main(String[] args) {
int i = 1;
int[][] tba = new int[][]{
new int[]{1, 5, 3, 7, 2},
new int[]{5, 3, 7, 2, 6, 4, 5, 9, 1, 2},
new int[]{2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1},
new int[]{5, 5, 5, 5},
new int[]{5, 6, 7, 8},
new int[]{8, 7, 7, 6},
new int[]{6, 7, 10, 7, 6}
};
for (int[] tea : tba) {
int rht, wu = 0, bof;
do {
for (rht = tea.length - 1; rht >= 0; rht--) {
if (tea[rht] > 0) {
break;
}
}
if (rht < 0) {
break;
}
bof = 0;
for (int col = 0; col <= rht; col++) {
if (tea[col] > 0) {
tea[col]--;
bof += 1;
} else if (bof > 0) {
wu++;
}
}
if (bof < 2) {
break;
}
} while (true);
System.out.printf("Block %d", i++);
if (wu == 0) {
System.out.print(" does not hold any");
} else {
System.out.printf(" holds %d", wu);
}
System.out.println(" water units.");
}
}
}</lang>
- Output:
Block 1 holds 2 water units. Block 2 holds 14 water units. Block 3 holds 35 water units. Block 4 does not hold any water units. Block 5 does not hold any water units. Block 6 does not hold any water units. Block 7 does not hold any water units.
JavaScript
ES5
<lang JavaScript>(function () {
'use strict';
// waterCollected :: [Int] -> Int
var waterCollected = function (xs) {
return sum( // water above each bar
zipWith(function (a, b) {
return a - b; // difference between water level and bar
},
zipWith(min, // lower of two flanking walls
scanl1(max, xs), // highest walls to left
scanr1(max, xs) // highest walls to right
),
xs // tops of bars
)
.filter(function (x) {
return x > 0; // only bars with water above them
})
);
};
// GENERIC FUNCTIONS ----------------------------------------
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
var zipWith = function (f, xs, ys) {
var ny = ys.length;
return (xs.length <= ny ? xs : xs.slice(0, ny))
.map(function (x, i) {
return f(x, ys[i]);
});
};
// scanl1 is a variant of scanl that has no starting value argument
// scanl1 :: (a -> a -> a) -> [a] -> [a]
var scanl1 = function (f, xs) {
return xs.length > 0 ? scanl(f, xs[0], xs.slice(1)) : [];
};
// scanr1 is a variant of scanr that has no starting value argument
// scanr1 :: (a -> a -> a) -> [a] -> [a]
var scanr1 = function (f, xs) {
return xs.length > 0 ? scanr(f, xs.slice(-1)[0], xs.slice(0, -1)) : [];
};
// scanl :: (b -> a -> b) -> b -> [a] -> [b]
var scanl = function (f, startValue, xs) {
var lst = [startValue];
return xs.reduce(function (a, x) {
var v = f(a, x);
return lst.push(v), v;
}, startValue), lst;
};
// scanr :: (b -> a -> b) -> b -> [a] -> [b]
var scanr = function (f, startValue, xs) {
var lst = [startValue];
return xs.reduceRight(function (a, x) {
var v = f(a, x);
return lst.push(v), v;
}, startValue), lst.reverse();
};
// sum :: (Num a) => [a] -> a
var sum = function (xs) {
return xs.reduce(function (a, x) {
return a + x;
}, 0);
};
// max :: Ord a => a -> a -> a
var max = function (a, b) {
return a > b ? a : b;
};
// min :: Ord a => a -> a -> a
var min = function (a, b) {
return b < a ? b : a;
};
// TEST ---------------------------------------------------
return [
[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]
].map(waterCollected);
//--> [2, 14, 35, 0, 0, 0, 0]
})();</lang>
- Output:
<lang JavaScript>[2, 14, 35, 0, 0, 0, 0]</lang>
ES6
<lang JavaScript>(() => {
'use strict';
// waterCollected :: [Int] -> Int
const waterCollected = xs => {
const
maxToRight = scanr1(max, xs),
maxToLeft = scanl1(max, xs),
levels = zipWith(min, maxToLeft, maxToRight);
return sum(
zipWith(difference, levels, xs)
.filter(x => x > 0)
);
};
// GENERIC FUNCTIONS -----------------------------------------------------
// difference :: (Num a) => a -> a -> a const difference = (a, b) => a - b;
// max :: Ord a => a -> a -> a const max = (a, b) => a > b ? a : b;
// min :: Ord a => a -> a -> a const min = (a, b) => b < a ? b : a;
// scanl :: (b -> a -> b) -> b -> [a] -> [b]
const scanl = (f, startValue, xs) => {
const lst = [startValue];
return (
xs.reduce((a, x) => {
const v = f(a, x);
return (lst.push(v), v);
}, startValue),
lst
);
};
// scanl1 is a variant of scanl that has no starting value argument
// scanl1 :: (a -> a -> a) -> [a] -> [a]
const scanl1 = (f, xs) =>
xs.length > 0 ? scanl(f, xs[0], xs.slice(1)) : [];
// scanr :: (b -> a -> b) -> b -> [a] -> [b]
const scanr = (f, startValue, xs) => {
const lst = [startValue];
return (
xs.reduceRight((a, x) => {
const v = f(a, x);
return (lst.push(v), v);
}, startValue),
lst.reverse()
);
};
// scanr1 is a variant of scanr that has no starting value argument
// scanr1 :: (a -> a -> a) -> [a] -> [a]
const scanr1 = (f, xs) =>
xs.length > 0 ? scanr(f, xs.slice(-1)[0], xs.slice(0, -1)) : [];
// sum :: (Num a) => [a] -> a const sum = xs => xs.reduce((a, x) => a + x, 0);
// zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
const zipWith = (f, xs, ys) => {
const ny = ys.length;
return (xs.length <= ny ? xs : xs.slice(0, ny))
.map((x, i) => f(x, ys[i]));
};
// TEST ------------------------------------------------------------------
return [
[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1],
[5, 5, 5, 5],
[5, 6, 7, 8],
[8, 7, 7, 6],
[6, 7, 10, 7, 6]
].map(waterCollected);
//--> [2, 14, 35, 0, 0, 0, 0]
})();</lang>
- Output:
<lang JavaScript>[2, 14, 35, 0, 0, 0, 0]</lang>
Julia
Inspired to #Python.
<lang julia>function watercollected(towers::Vector{Int})
high_lft = vcat(0, accumulate(max, towers[1:end-1])) high_rgt = vcat(reverse(accumulate(max, towers[end:-1:2])), 0) waterlvl = max.(min.(high_lft, high_rgt) .- towers, 0) return waterlvl
end
function towerprint(towers, levels)
ctowers = copy(towers)
clevels = copy(levels)
hmax = maximum(towers)
ntow = length(towers)
for h in hmax:-1:1
@printf("%2i |", h)
for j in 1:ntow
if ctowers[j] + clevels[j] ≥ h
if clevels[j] > 0
cell = "≈≈"
clevels[j] -= 1
else
cell = "NN"
ctowers[j] -= 1
end
else
cell = " "
end
print(cell)
end
println("|")
end
println(" " * join(lpad(t, 2) for t in levels) * ": Water lvl")
println(" " * join(lpad(t, 2) for t in towers) * ": Tower lvl")
end
for towers in [[1, 5, 3, 7, 2], [5, 3, 7, 2, 6, 4, 5, 9, 1, 2],
[2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], [5, 5, 5, 5], [5, 6, 7, 8], [8, 7, 7, 6], [6, 7, 10, 7, 6]] towerprint(towers, watercollected(towers)) println()
end</lang>
- Output:
7 | NN |
6 | NN |
5 | NN≈≈NN |
4 | NN≈≈NN |
3 | NNNNNN |
2 | NNNNNNNN|
1 |NNNNNNNNNN|
0 0 2 0 0: Water lvl
1 5 3 7 2: Tower lvl
9 | NN |
8 | NN |
7 | NN≈≈≈≈≈≈≈≈NN |
6 | NN≈≈NN≈≈≈≈NN |
5 |NN≈≈NN≈≈NN≈≈NNNN |
4 |NN≈≈NN≈≈NNNNNNNN |
3 |NNNNNN≈≈NNNNNNNN |
2 |NNNNNNNNNNNNNNNN≈≈NN|
1 |NNNNNNNNNNNNNNNNNNNN|
0 2 0 5 1 3 2 0 1 0: Water lvl
5 3 7 2 6 4 5 9 1 2: Tower lvl
8 | NN |
7 | NN≈≈≈≈≈≈≈≈≈≈≈≈≈≈NN |
6 | NN≈≈≈≈≈≈NN≈≈≈≈≈≈≈≈≈≈≈≈≈≈NN |
5 | NN≈≈NN≈≈NN≈≈≈≈≈≈≈≈NN≈≈NNNN |
4 | NN≈≈NN≈≈NN≈≈NN≈≈≈≈NN≈≈NNNNNN |
3 | NNNNNN≈≈NN≈≈NN≈≈≈≈NNNNNNNNNN |
2 |NNNNNNNNNNNN≈≈NNNNNNNNNNNNNNNN |
1 |NNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNN|
0 0 3 1 4 0 6 3 5 5 2 4 2 0 0 0: Water lvl
2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1: Tower lvl
5 |NNNNNNNN|
4 |NNNNNNNN|
3 |NNNNNNNN|
2 |NNNNNNNN|
1 |NNNNNNNN|
0 0 0 0: Water lvl
5 5 5 5: Tower lvl
8 | NN|
7 | NNNN|
6 | NNNNNN|
5 |NNNNNNNN|
4 |NNNNNNNN|
3 |NNNNNNNN|
2 |NNNNNNNN|
1 |NNNNNNNN|
0 0 0 0: Water lvl
5 6 7 8: Tower lvl
8 |NN |
7 |NNNNNN |
6 |NNNNNNNN|
5 |NNNNNNNN|
4 |NNNNNNNN|
3 |NNNNNNNN|
2 |NNNNNNNN|
1 |NNNNNNNN|
0 0 0 0: Water lvl
8 7 7 6: Tower lvl
10 | NN |
9 | NN |
8 | NN |
7 | NNNNNN |
6 |NNNNNNNNNN|
5 |NNNNNNNNNN|
4 |NNNNNNNNNN|
3 |NNNNNNNNNN|
2 |NNNNNNNNNN|
1 |NNNNNNNNNN|
0 0 0 0 0: Water lvl
6 710 7 6: Tower lvl
Kotlin
<lang scala>// version 1.1.2
fun waterCollected(tower: IntArray): Int {
val n = tower.size
val highLeft = listOf(0) + (1 until n).map { tower.slice(0 until it).max()!! }
val highRight = (1 until n).map { tower.slice(it until n).max()!! } + 0
return (0 until n).map { maxOf(minOf(highLeft[it], highRight[it]) - tower[it], 0) }.sum()
}
fun main(args: Array<String>) {
val towers = listOf(
intArrayOf(1, 5, 3, 7, 2),
intArrayOf(5, 3, 7, 2, 6, 4, 5, 9, 1, 2),
intArrayOf(2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1),
intArrayOf(5, 5, 5, 5),
intArrayOf(5, 6, 7, 8),
intArrayOf(8, 7, 7, 6),
intArrayOf(6, 7, 10, 7, 6)
)
for (tower in towers) {
println("${"%2d".format(waterCollected(tower))} from ${tower.contentToString()}")
}
}</lang>
- Output:
2 from [1, 5, 3, 7, 2] 14 from [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] 35 from [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] 0 from [5, 5, 5, 5] 0 from [5, 6, 7, 8] 0 from [8, 7, 7, 6] 0 from [6, 7, 10, 7, 6]
Perl
<lang perl>use Modern::Perl; use List::Util qw{ min max sum };
sub water_collected {
my @t = map { { TOWER => $_, LEFT => 0, RIGHT => 0, LEVEL => 0 } } @_;
my ( $l, $r ) = ( 0, 0 );
$_->{LEFT} = ( $l = max( $l, $_->{TOWER} ) ) for @t;
$_->{RIGHT} = ( $r = max( $r, $_->{TOWER} ) ) for reverse @t;
$_->{LEVEL} = min( $_->{LEFT}, $_->{RIGHT} ) for @t;
return sum map { $_->{LEVEL} > 0 ? $_->{LEVEL} - $_->{TOWER} : 0 } @t;
}
say join ' ', map { water_collected( @{$_} ) } (
[ 1, 5, 3, 7, 2 ], [ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 ], [ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 ], [ 5, 5, 5, 5 ], [ 5, 6, 7, 8 ], [ 8, 7, 7, 6 ], [ 6, 7, 10, 7, 6 ],
);</lang>
- Output:
2 14 35 0 0 0 0
Perl 6
<lang perl6>sub max_l ( @a ) { [\max] @a } sub max_r ( @a ) { ([\max] @a.reverse).reverse }
sub water_collected ( @towers ) {
return 0 if @towers <= 2;
my @levels = max_l(@towers) »min« max_r(@towers);
return ( @levels »-« @towers ).grep( * > 0 ).sum;
}
say map &water_collected,
[ 1, 5, 3, 7, 2 ], [ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 ], [ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 ], [ 5, 5, 5, 5 ], [ 5, 6, 7, 8 ], [ 8, 7, 7, 6 ], [ 6, 7, 10, 7, 6 ],
- </lang>
- Output:
(2 14 35 0 0 0 0)
PicoLisp
<lang PicoLisp>(de water (Lst)
(sum
'((A)
(cnt
nT
(clip (mapcar '((B) (>= B A)) Lst)) ) )
(range 1 (apply max Lst)) ) )
(println
(mapcar
water
(quote
(1 5 3 7 2)
(5 3 7 2 6 4 5 9 1 2)
(2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1)
(5 5 5 5)
(5 6 7 8)
(8 7 7 6)
(6 7 10 7 6) ) ) )</lang>
- Output:
(2 14 35 0 0 0 0)
Python
Based on the algorithm explained at Stack Overflow:
<lang python>def water_collected(tower):
N = len(tower)
highest_left = [0] + [max(tower[:n]) for n in range(1,N)]
highest_right = [max(tower[n:N]) for n in range(1,N)] + [0]
water_level = [max(min(highest_left[n], highest_right[n]) - tower[n], 0)
for n in range(N)]
print("highest_left: ", highest_left)
print("highest_right: ", highest_right)
print("water_level: ", water_level)
print("tower_level: ", tower)
print("total_water: ", sum(water_level))
print("")
return sum(water_level)
towers = [[1, 5, 3, 7, 2],
[5, 3, 7, 2, 6, 4, 5, 9, 1, 2], [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1], [5, 5, 5, 5], [5, 6, 7, 8], [8, 7, 7, 6], [6, 7, 10, 7, 6]]
[water_collected(tower) for tower in towers]</lang>
- Output:
highest_left: [0, 1, 5, 5, 7] highest_right: [7, 7, 7, 2, 0] water_level: [0, 0, 2, 0, 0] tower_level: [1, 5, 3, 7, 2] total_water: 2 highest_left: [0, 5, 5, 7, 7, 7, 7, 7, 9, 9] highest_right: [9, 9, 9, 9, 9, 9, 9, 2, 2, 0] water_level: [0, 2, 0, 5, 1, 3, 2, 0, 1, 0] tower_level: [5, 3, 7, 2, 6, 4, 5, 9, 1, 2] total_water: 14 highest_left: [0, 2, 6, 6, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8] highest_right: [8, 8, 8, 8, 8, 7, 7, 7, 7, 7, 7, 7, 7, 4, 1, 0] water_level: [0, 0, 3, 1, 4, 0, 6, 3, 5, 5, 2, 4, 2, 0, 0, 0] tower_level: [2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1] total_water: 35 highest_left: [0, 5, 5, 5] highest_right: [5, 5, 5, 0] water_level: [0, 0, 0, 0] tower_level: [5, 5, 5, 5] total_water: 0 highest_left: [0, 5, 6, 7] highest_right: [8, 8, 8, 0] water_level: [0, 0, 0, 0] tower_level: [5, 6, 7, 8] total_water: 0 highest_left: [0, 8, 8, 8] highest_right: [7, 7, 6, 0] water_level: [0, 0, 0, 0] tower_level: [8, 7, 7, 6] total_water: 0 highest_left: [0, 6, 7, 10, 10] highest_right: [10, 10, 7, 6, 0] water_level: [0, 0, 0, 0, 0] tower_level: [6, 7, 10, 7, 6] total_water: 0 [2, 14, 35, 0, 0, 0, 0]
Racket
<lang racket>#lang racket/base (require racket/match)
(define (water-collected-between-towers towers)
(define (build-tallest-left/rev-list t mx/l rv)
(match t
[(list) rv]
[(cons a d)
(define new-mx/l (max a mx/l))
(build-tallest-left/rev-list d new-mx/l (cons mx/l rv))]))
(define (collect-from-right t tallest/l mx/r rv)
(match t
[(list) rv]
[(cons a d)
(define new-mx/r (max a mx/r))
(define new-rv (+ rv (max (- (min new-mx/r (car tallest/l)) a) 0)))
(collect-from-right d (cdr tallest/l) new-mx/r new-rv)]))
(define reversed-left-list (build-tallest-left/rev-list towers 0 null)) (collect-from-right (reverse towers) reversed-left-list 0 0))
(module+ test
(require rackunit)
(check-equal?
(let ((towerss
'[[1 5 3 7 2]
[5 3 7 2 6 4 5 9 1 2]
[2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1]
[5 5 5 5]
[5 6 7 8]
[8 7 7 6]
[6 7 10 7 6]]))
(map water-collected-between-towers towerss))
(list 2 14 35 0 0 0 0)))</lang>
When run produces no output -- meaning that the tests have run successfully.
REXX
version 1
<lang rexx>/* REXX */ Call bars '1 5 3 7 2' Call bars '5 3 7 2 6 4 5 9 1 2' Call bars '2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1' Call bars '5 5 5 5' Call bars '5 6 7 8' Call bars '8 7 7 6' Call bars '6 7 10 7 6' Exit bars: Parse Arg bars bar.0=words(bars) high=0 box.=' ' Do i=1 To words(bars)
bar.i=word(bars,i) high=max(high,bar.i) Do j=1 To bar.i box.i.j='x' End End
m=1 w=0 Do Forever
Do i=m+1 To bar.0
If bar.i>bar.m Then
Leave
End
If i>bar.0 Then Leave
n=i
Do i=m+1 To n-1
w=w+bar.m-bar.i
Do j=bar.i+1 To bar.m
box.i.j='*'
End
End
m=n
End
m=bar.0 Do Forever
Do i=bar.0 To 1 By -1
If bar.i>bar.m Then
Leave
End
If i<1 Then Leave
n=i
Do i=m-1 To n+1 By -1
w=w+bar.m-bar.i
Do j=bar.i+1 To bar.m
box.i.j='*'
End
End
m=n
End
Say bars '->' w Call show Return show: Do j=high To 1 By -1
ol= Do i=1 To bar.0 ol=ol box.i.j End Say ol End
Return</lang>
- Output:
1 5 3 7 2 -> 2
x
x
x * x
x * x
x x x
x x x x
x x x x x
5 3 7 2 6 4 5 9 1 2 -> 14
x
x
x * * * * x
x * x * * x
x * x * x * x x
x * x * x x x x
x x x * x x x x
x x x x x x x x * x
x x x x x x x x x x
2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 -> 35
x
x * * * * * * * x
x * * * x * * * * * * * x
x * x * x * * * * x * x x
x * x * x * x * * x * x x x
x x x * x * x * * x x x x x
x x x x x x * x x x x x x x x
x x x x x x x x x x x x x x x x
5 5 5 5 -> 0
x x x x
x x x x
x x x x
x x x x
x x x x
5 6 7 8 -> 0
x
x x
x x x
x x x x
x x x x
x x x x
x x x x
x x x x
8 7 7 6 -> 0
x
x x x
x x x x
x x x x
x x x x
x x x x
x x x x
x x x x
6 7 10 7 6 -> 0
x
x
x
x x x
x x x x x
x x x x x
x x x x x
x x x x x
x x x x x
x x x x x
version 2, simple numeric list output
<lang rexx>/*REXX program calculates and displays the amount of rainwater collected between towers.*/ call tower 1 5 3 7 2 call tower 5 3 7 2 6 4 5 9 1 2 call tower 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 call tower 5 5 5 5 call tower 5 6 7 8 call tower 8 7 7 6 call tower 6 7 10 7 6 exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ tower: procedure; arg y; #=words(y); t.=0; L.=0 /*the T. array holds the tower heights.*/
do j=1 for #; t.j=word(y, j) /*construct the towers, */
_=j-1; L.j=max(t._, L._) /* " " left-most tallest tower*/
end /*j*/
R.=0
do b=# by -1 for #; _=b+1; R.b=max(t._, R._) /*right-most tallest tower*/
end /*b*/
w.=0 /*rainwater collected.*/
do f=1 for #; if t.f>=L.f | t.f>=R.f then iterate /*rain between towers?*/
w.f=min(L.f, R.f) - t.f; w.00=w.00+w.f /*rainwater collected.*/
end /*f*/
say right(w.00,9) 'units of rainwater collected for: ' y /*display water units.*/
return</lang>
- output:
2 units of rainwater collected for: 1 5 3 7 2
14 units of rainwater collected for: 5 3 7 2 6 4 5 9 1 2
35 units of rainwater collected for: 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1
0 units of rainwater collected for: 5 5 5 5
0 units of rainwater collected for: 5 6 7 8
0 units of rainwater collected for: 8 7 7 6
0 units of rainwater collected for: 6 7 10 7 6
version 3, with ASCII art
This REXX version shows a scale and a representation of the towers and water collected.
It tries to protect the aspect ration by showing the buildings as in this task's preamble. <lang rexx>/*REXX program calculates and displays the amount of rainwater collected between towers.*/
call tower 1 5 3 7 2
call tower 5 3 7 2 6 4 5 9 1 2
call tower 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1
call tower 5 5 5 5
call tower 5 6 7 8
call tower 8 7 7 6
call tower 6 7 10 7 6
exit /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ tower: procedure; arg y; #=words(y); t.=0; L.=0 /*the T. array holds the tower heights.*/
do j=1 for #; t.j=word(y,j); _=j-1 /*construct the towers; max height. */
L.j=max(t._, L._); t.0=max(t.0, t.j) /*left-most tallest tower; build scale.*/
end /*j*/
R.=0
do b=# by -1 for #; _=b+1; R.b=max(t._, R._) /*right-most tallest tower*/
end /*b*/
w.=0 /*rainwater collected.*/
do f=1 for #; if t.f>=L.f | t.f>=R.f then iterate /*rain between towers?*/
w.f=min(L.f, R.f) - t.f; w.00=w.00+w.f /*rainwater collected.*/
end /*f*/
if w.00==0 then w.00='no' /*pretty up wording for "no rainwater".*/
ratio=2 /*used to maintain a good aspect ratio.*/
p.= /*P. stores plot versions of towers. */
do c=0 to #; cc=c*ratio /*construct the plot+scale for display.*/
do h=1 for t.c+w.c; glyph='█' /*maybe show a floor of some tower(s). */
if h>t.c then glyph='≈' /* " " rainwater between towers. */
if c==0 then p.h=overlay(right(h, 9) , p.h, 1 ) /*tower scale*/
else p.h=overlay(copies(glyph,ratio) , p.h, 10+cc) /*build tower*/
end /*h*/
end /*c*/
p.1=overlay(w.00 'units of rainwater collected', p.1, 15*ratio+#) /*append text*/
do z=t.0 by -1 to 0; say p.z /*display various tower floors & water.*/
end /*z*/
return</lang>
- output:
7 ██
6 ██
5 ██≈≈██
4 ██≈≈██
3 ██████
2 ████████
1 ██████████ 2 units of rainwater collected
9 ██
8 ██
7 ██≈≈≈≈≈≈≈≈██
6 ██≈≈██≈≈≈≈██
5 ██≈≈██≈≈██≈≈████
4 ██≈≈██≈≈████████
3 ██████≈≈████████
2 ████████████████≈≈██
1 ████████████████████ 14 units of rainwater collected
8 ██
7 ██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
6 ██≈≈≈≈≈≈██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
5 ██≈≈██≈≈██≈≈≈≈≈≈≈≈██≈≈████
4 ██≈≈██≈≈██≈≈██≈≈≈≈██≈≈██████
3 ██████≈≈██≈≈██≈≈≈≈██████████
2 ████████████≈≈████████████████
1 ████████████████████████████████ 35 units of rainwater collected
5 ████████
4 ████████
3 ████████
2 ████████
1 ████████ no units of rainwater collected
8 ██
7 ████
6 ██████
5 ████████
4 ████████
3 ████████
2 ████████
1 ████████ no units of rainwater collected
8 ██
7 ██████
6 ████████
5 ████████
4 ████████
3 ████████
2 ████████
1 ████████ no units of rainwater collected
10 ██
9 ██
8 ██
7 ██████
6 ██████████
5 ██████████
4 ██████████
3 ██████████
2 ██████████
1 ██████████ no units of rainwater collected
Ruby
<lang ruby> def a(array) n=array.length left={} right={} left[0]=array[0] i=1 loop do
break if i >=n
left[i]=[left[i-1],array[i]].max
i += 1
end right[n-1]=array[n-1] i=n-2 loop do break if i<0
right[i]=[right[i+1],array[i]].max
i-=1 end i=0 water=0 loop do break if i>=n water+=[left[i],right[i]].min-array[i] i+=1 end puts water end
a([ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 ]) a([ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 ]) a([ 5, 5, 5, 5 ]) a([ 5, 6, 7, 8 ]) a([ 8, 7, 7, 6 ]) a([ 6, 7, 10, 7, 6 ]) return</lang> output
14 35 0 0 0 0
Scheme
<lang scheme> (import (scheme base)
(scheme write))
(define (total-collected chart)
(define (highest-left vals curr)
(if (null? vals)
(list curr)
(cons curr
(highest-left (cdr vals) (max (car vals) curr)))))
(define (highest-right vals curr)
(reverse (highest-left (reverse vals) curr)))
;
(if (< (length chart) 3) ; catch the end cases
0
(apply +
(map (lambda (l c r)
(if (or (<= l c)
(<= r c))
0
(- (min l r) c)))
(highest-left chart 0)
chart
(highest-right chart 0)))))
(for-each
(lambda (chart) (display chart) (display " -> ") (display (total-collected chart)) (newline)) '((1 5 3 7 2) (5 3 7 2 6 4 5 9 1 2) (2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1) (5 5 5 5) (5 6 7 8) (8 7 7 6) (6 7 10 7 6)))
</lang>
- Output:
(1 5 3 7 2) -> 2 (5 3 7 2 6 4 5 9 1 2) -> 14 (2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1) -> 35 (5 5 5 5) -> 0 (5 6 7 8) -> 0 (8 7 7 6) -> 0 (6 7 10 7 6) -> 0 (3 1 2) -> 1 (1) -> 0 () -> 0 (1 2) -> 0
Sidef
<lang ruby>func max_l(Array a, m = a[0]) {
gather { a.each {|e| take(m = max(m, e)) } }
}
func max_r(Array a) {
max_l(a.flip).flip
}
func water_collected(Array towers) {
var levels = (max_l(towers) »min« max_r(towers))
(levels »-« towers).grep{ _ > 0 }.sum
}
[
[ 1, 5, 3, 7, 2 ], [ 5, 3, 7, 2, 6, 4, 5, 9, 1, 2 ], [ 2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1 ], [ 5, 5, 5, 5 ], [ 5, 6, 7, 8 ], [ 8, 7, 7, 6 ], [ 6, 7, 10, 7, 6 ],
].map { water_collected(_) }.say</lang>
- Output:
[2, 14, 35, 0, 0, 0, 0]
Tcl
Tcl makes for a surprisingly short and readable implementation, next to some of the more functional-oriented languages. <lang Tcl>namespace path {::tcl::mathfunc ::tcl::mathop}
proc flood {ground} {
set lefts [
set d 0
lmap g $ground {
set d [max $d $g]
}
]
set ground [lreverse $ground]
set rights [
set d 0
lmap g $ground {
set d [max $d $g]
}
]
set rights [lreverse $rights]
set ground [lreverse $ground]
set water [lmap l $lefts r $rights {min $l $r}]
set depths [lmap g $ground w $water {- $w $g}]
+ {*}$depths
}
foreach p {
{5 3 7 2 6 4 5 9 1 2}
{1 5 3 7 2}
{5 3 7 2 6 4 5 9 1 2}
{2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1}
{5 5 5 5}
{5 6 7 8}
{8 7 7 6}
{6 7 10 7 6}
} {
puts [flood $p]:\t$p
}</lang>
- Output:
14: 5 3 7 2 6 4 5 9 1 2 2: 1 5 3 7 2 14: 5 3 7 2 6 4 5 9 1 2 35: 2 6 3 5 2 8 1 4 2 2 5 3 5 7 4 1 0: 5 5 5 5 0: 5 6 7 8 0: 8 7 7 6 0: 6 7 10 7 6
Visual Basic .NET
Version 1
Method: Instead of "scanning" adjoining towers for each column, convert the tower data into a string representation with building blocks, empty spaces, and potential water retention sites. Then "erode" away the water retention sites that are unsupported. This is accomplished with the String Replace() function. The replace operations are unleashed upon the entire "block" of towers, rather than a cell at a time or a line at a time - which perhaps increases the program's execution-time, but reduces program's complexity.
The program can optionally display the interim string representation of each tower block before the final count is completed. I've since modified it to have the same block and wavy characters are the REXX 9.3 output, but used the double-wide columns, as pictured in the task definition area. <lang vbnet>' Convert tower block data into a string representation, then manipulate that. Sub Main()
Dim shoTow As Boolean = Environment.GetCommandLineArgs().Count > 1 ' Show towers.
Dim wta As Integer()() = { ' water tower array (input data).
New Integer() {1, 5, 3, 7, 2}, New Integer() {5, 3, 7, 2, 6, 4, 5, 9, 1, 2},
New Integer() {2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1},
New Integer() {5, 5, 5, 5}, New Integer() {5, 6, 7, 8},
New Integer() {8, 7, 7, 6}, New Integer() {6, 7, 10, 7, 6}}
Dim blk As String, ' String representation of a block of towers.
lf As String = vbCrLf ' Line feed to separate floors in a block of towers.
For i As Integer = 0 To UBound(wta)
Dim bpf As Integer ' Count of tower blocks found per floor.
blk = ""
Do
bpf = 0 : Dim floor As String = "" ' string representation of each floor.
For j As Integer = 0 To UBound(wta(i))
If wta(i)(j) > 0 Then ' Tower block detected, add block to floor,
floor &= "██" : wta(i)(j) -= 1 : bpf += 1 ' reduce tower by one.
Else ' Empty space detected, fill when not first or last column.
' "Almost equal to" characters are possible water retention cells.
floor &= If(j > 0 AndAlso j < UBound(wta(i)), "≈≈", " ")
End If
Next
If bpf > 0 Then blk = floor & lf & blk ' Add floors until blocks are gone.
Loop Until bpf = 0 ' No tower blocks left, so terminate.
' Now erode potential water retention cells from left and right
While blk.Contains(" ≈≈") : blk = Replace(blk, " ≈≈", " ") : End While
While blk.Contains("≈≈ ") : blk = Replace(blk, "≈≈ ", " ") : End While
' Optionally show towers w/ water marks.
If shoTow Then Console.Write("{0}{1}", lf, Wide(blk))
' Now remove all building blocks and whitespace, leaving only water marks.
' Then count the remaining characters with the Len() function.
Console.Write("Block {0} retains {1,2} water units.{2}", i + 1,
(Len(blk) - Len(Replace(blk, "≈≈", ""))) \ 2, lf)
Next
End Sub</lang>
- Output:
<lang>Block 1 retains 2 water units.
Block 2 retains 14 water units. Block 3 retains 35 water units. Block 4 retains 0 water units. Block 5 retains 0 water units. Block 6 retains 0 water units. Block 7 retains 0 water units.</lang> Verbose output shows towers with water ("Almost equal to" characters) left in the "wells" between towers. Just supply any command-line parameter to see it. Use no command line parameters to see the plain output above. <lang> ██
██ ██≈≈██ ██≈≈██ ██████ ████████
██████████ Block 1 retains 2 water units.
██
██
██≈≈≈≈≈≈≈≈██
██≈≈██≈≈≈≈██
██≈≈██≈≈██≈≈████ ██≈≈██≈≈████████ ██████≈≈████████ ████████████████≈≈██ ████████████████████ Block 2 retains 14 water units.
██
██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
██≈≈≈≈≈≈██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
██≈≈██≈≈██≈≈≈≈≈≈≈≈██≈≈████
██≈≈██≈≈██≈≈██≈≈≈≈██≈≈██████
██████≈≈██≈≈██≈≈≈≈██████████
████████████≈≈████████████████ ████████████████████████████████ Block 3 retains 35 water units.
████████ ████████ ████████ ████████ ████████ Block 4 retains 0 water units.
██ ████ ██████
████████ ████████ ████████ ████████ ████████ Block 5 retains 0 water units.
██ ██████ ████████ ████████ ████████ ████████ ████████ ████████ Block 6 retains 0 water units.
██ ██ ██ ██████
██████████ ██████████ ██████████ ██████████ ██████████ ██████████ Block 7 retains 0 water units.</lang>
Version 2
Method: More conventional "scanning" method. A Char array is used, but no Replace() statements. Output is similar to version 1, although there is now a left margin of three spaces, the results statement is immediately to the right of the string representation of the tower blocks (instead of underneath), the verb is "hold(s)" instead of "retains", and there is a special string when the results indicate zero.
<lang vbnet> <summary>
wide - Widens the aspect ratio of a linefeed separated string.
</summary>
<param name="src">A string representing a block of towers.</param>
<param name="margin">Optional padding for area to the left.</param>
<returns>A double-wide version of the string.</returns>
Function wide(src As String, Optional margin As String = "") As String
Dim res As String = margin : For Each ch As Char In src
res += If(ch < " ", ch & margin, ch + ch) : Next : Return res
End Function
<summary>
cntChar - Counts characters, also custom formats the output.
</summary>
<param name="src">The string to count characters in.</param>
<param name="ch">The character to be counted.</param>
<param name="verb">Verb to include in format. Expecting "hold",
but can work with "retain" or "have".</param>
<returns>The count of chars found in a string, and formats a verb.</returns>
Function cntChar(src As String, ch As Char, verb As String) As String
Dim cnt As Integer = 0
For Each c As Char In src : cnt += If(c = ch, 1, 0) : Next
Return If(cnt = 0, "does not " & verb & " any",
verb.Substring(0, If(verb = "have", 2, 4)) & "s " & cnt.ToString())
End Function
<summary>
report - Produces a report of the number of rain units found in
a block of towers, optionally showing the towers.
Autoincrements the blkID for each report.
</summary>
<param name="tea">An int array with tower elevations.</param>
<param name="blkID">An int of the block of towers ID.</param>
<param name="verb">The verb to use in the description.
Defaults to "has / have".</param>
<param name="showIt">When true, the report includes a string representation
of the block of towers.</param>
<returns>A string containing the amount of rain units, optionally preceeded by
a string representation of the towers holding any water.</returns>
Function report(tea As Integer(), ' Tower elevation array.
ByRef blkID As Integer, ' Block ID for the description.
Optional verb As String = "have", ' Verb to use in the description.
Optional showIt As Boolean = False) As String ' Show representaion.
Dim block As String = "", ' The block of towers.
lf As String = vbLf, ' The separator between floors.
rTwrPos As Integer ' The position of the rightmost tower of this floor.
Do
For rTwrPos = tea.Length - 1 To 0 Step -1 ' Determine the rightmost tower
If tea(rTwrPos) > 0 Then Exit For ' postition on this floor.
Next
If rTwrPos < 0 Then Exit Do ' When no towers remain, exit the do loop.
' init the floor to a space filled Char array, as wide as the block of towers.
Dim floor As Char() = New String(" ", tea.Length).ToCharArray()
Dim bpf As Integer = 0 ' The count of blocks found per floor.
For column As Integer = 0 To rTwrPos ' Scan from left to right.
If tea(column) > 0 Then ' If a tower exists here,
floor(column) = "█" ' mark the floor with a block,
tea(column) -= 1 ' drop the tower elevation by one,
bpf += 1 ' and advance the block count.
ElseIf bpf > 0 Then ' Otherwise, see if a tower is present to the left.
floor(column) = "≈" ' OK to fill with water.
End If
Next
If bpf > If(showIt, 0, 1) Then ' Continue the building only when needed.
' If not showing blocks, discontinue building when a single tower remains.
' build tower blocks string with each floor added to top.
block = New String(floor) & If(block = "", "", lf) & block
Else
Exit Do ' Ran out of towers, so exit the do loop.
End If
Loop While True ' Depending on previous break statements to terminate the do loop.
blkID += 1 ' increment block ID counter.
' format report and return it.
Return If(showIt, String.Format(vbLf & "{0}", wide(block, " ")), "") &
String.Format(" Block {0} {1} water units.", blkID, cntChar(block, "≈", verb))
End Function
<summary>
Main routine.
With one command line parameter, it shows tower blocks,
with no command line parameters, it shows a plain report
</summary>
Sub Main()
Dim shoTow As Boolean = Environment.GetCommandLineArgs().Count > 1 ' Show towers.
Dim blkCntr As Integer = 0 ' Block ID for reports.
Dim verb As String = "hold" ' "retain" or "have" can be used instead of "hold".
Dim tea As Integer()() = {New Integer() {1, 5, 3, 7, 2}, ' Tower elevation data.
New Integer() {5, 3, 7, 2, 6, 4, 5, 9, 1, 2},
New Integer() {2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1},
New Integer() {5, 5, 5, 5}, New Integer() {5, 6, 7, 8},
New Integer() {8, 7, 7, 6}, New Integer() {6, 7, 10, 7, 6}}
For Each block As Integer() In tea
' Produce report for each block of towers.
Console.WriteLine(report(block, blkCntr, verb, shoTow))
Next
End Sub</lang>
Regular version 2 output: <lang> Block 1 holds 2 water units.
Block 2 holds 14 water units. Block 3 holds 35 water units. Block 4 does not hold any water units. Block 5 does not hold any water units. Block 6 does not hold any water units. Block 7 does not hold any water units.</lang>
Sample of version 2 verbose output: <lang> ██
██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
██≈≈≈≈≈≈██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
██≈≈██≈≈██≈≈≈≈≈≈≈≈██≈≈████
██≈≈██≈≈██≈≈██≈≈≈≈██≈≈██████
██████≈≈██≈≈██≈≈≈≈██████████
████████████≈≈████████████████
████████████████████████████████ Block 3 holds 35 water units.
████████ ████████ ████████ ████████ ████████ Block 4 does not hold any water units.</lang>
zkl
<lang zkl>fcn waterCollected(walls){
// compile max wall heights from left to right and right to left
// then each pair is left/right wall of that cell.
// Then the min of each wall pair == water height for that cell
scanl(walls,(0).max) // scan to right, f is max(0,a,b)
.zipWith((0).MAX.min, // f is MAX.min(a,b) == min(a,b)
scanl(walls.reverse(),(0).max).reverse()) // right to left
// now subtract the wall height from the water level and add 'em up
.zipWith('-,walls).filter('>(0)).sum(0);
} fcn scanl(xs,f,i=0){ // aka reduce but save list of results
xs.reduce('wrap(s,x,a){ s=f(s,x); a.append(s); s },i,ss:=List());
ss
} // scanl((1,5,3,7,2),max,0) --> (1,5,5,7,7)</lang> <lang zkl>T( T(1, 5, 3, 7, 2), T(5, 3, 7, 2, 6, 4, 5, 9, 1, 2),
T(2, 6, 3, 5, 2, 8, 1, 4, 2, 2, 5, 3, 5, 7, 4, 1), T(5, 5, 5, 5), T(5, 6, 7, 8),T(8, 7, 7, 6), T(6, 7, 10, 7, 6) )
.pump(List, waterCollected).println();</lang>
- Output:
L(2,14,35,0,0,0,0)