Water collected between towers

From Rosetta Code
Task
Water collected between towers
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:



AppleScript[edit]

Translation of: JavaScript
-- 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 lambda(x)
x > 0
end lambda
end script
 
-- minus :: Num a => a -> a -> a
script minus
on lambda(a, b)
a - b
end lambda
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 ------------------------------------------------------
 
-- 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
 
-- 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
 
-- 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 lambda(v, item i of xs, i, xs)
set end of lst to v
end repeat
return lst
end tell
end scanl
 
-- 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 lambda(v, item i of xs, i, xs)
set end of lst to v
end repeat
return reverse of lst
end tell
end scanr
 
-- zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]
on zipWith(f, xs, ys)
set nx to length of xs
set ny to length of ys
if nx < 1 or ny < 1 then
{}
else
set lng to cond(nx < ny, nx, ny)
set lst to {}
tell mReturn(f)
repeat with i from 1 to lng
set end of lst to lambda(item i of xs, item i of ys)
end repeat
return lst
end tell
end if
end zipWith
 
-- 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 lambda(item i of xs, i, xs)
end repeat
return lst
end tell
end map
 
-- 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 lambda(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter
 
-- sum :: Num a => [a] -> a
on sum(xs)
script add
on lambda(a, b)
a + b
end lambda
end script
 
foldl(add, 0, xs)
end sum
 
-- 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 lambda(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
 
-- 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 lambda : f
end script
end if
end mReturn
 
-- init :: [a] -> [a]
on init(xs)
if length of xs > 1 then
items 1 thru -2 of xs
else
{}
end if
end init
 
-- tail :: [a] -> [a]
on tail(xs)
if length of xs > 1 then
items 2 thru -1 of xs
else
{}
end if
end tail
 
-- 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
 
-- cond :: Bool -> a -> a -> a
on cond(bool, f, g)
if bool then
f
else
g
end if
end cond
Output:
{2, 14, 35, 0, 0, 0, 0}

AWK[edit]

 
# 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) }
 
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#[edit]

Version 1[edit]

Translation from Visual Basic .NET. See that version 1 entry for code comments and details.

using System;
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);
}
}
 
Output:
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.

Version 2[edit]

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.

// 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()));
}
}

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.

F#[edit]

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.

 
(*
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))
 
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

Go[edit]

 
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}))
}
Output:
2
14
35
0
0
0
0

Haskell[edit]

Following the approach of cdk's Haskell solution at Stack Overflow:

waterCollected :: [Int] -> Int
waterCollected xs =
sum $ -- Sum of water depths over each of:
filter (> 0) $ -- the columns that are covered by some water.
zipWith
(-) -- Where coverages are differences between
(zipWith
min -- water levels, (lower in each case of:
(scanl1 max xs) -- highest wall to left, and
(scanr1 max xs) -- highest wall to right)
)
xs -- and column tops.
 
main :: IO ()
main =
mapM_
(print . 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]
]
Output:
2
14
35
0
0
0
0

JavaScript[edit]

ES5[edit]

Translation of: Haskell
(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]
})();
Output:
[2, 14, 35, 0, 0, 0, 0]

ES6[edit]

Translation of: Haskell
(() => {
'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 ----------------------------------------
 
// 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]));
}
 
// 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)) : [];
 
// 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)) : [];
 
// 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
);
};
 
// 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()
);
};
 
// difference :: (Num a) => a -> a -> a
const difference = (a, b) => a - b;
 
// sum :: (Num a) => [a] -> a
const sum = xs => xs.reduce((a, x) => a + x, 0);
 
// 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;
 
 
// 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]
})();
Output:
[2, 14, 35, 0, 0, 0, 0]

Perl[edit]

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 ],
);
Output:
2 14 35 0 0 0 0

Perl 6[edit]

Translation of: Haskell
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 ],
;
Output:
(2 14 35 0 0 0 0)

Python[edit]

Based on the algorithm explained at Stack Overflow:

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]
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[edit]

#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)))

When run produces no output -- meaning that the tests have run successfully.

REXX[edit]

version 1[edit]

/* 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
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[edit]

/*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

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[edit]

This REXX version shows a scale and a representation of the towers and water collected.

/*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".*/
p.= /*P. stores plot versions of towers. */
do c=0 to # /*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 ) /*place the tower scale*/
else p.h=overlay(glyph , p.h, 10+c) /*build the tower. */
end /*h*/
end /*c*/
p.1=overlay(w.00 'units of rainwater collected', p.1, 15+#) /*append the text.*/
do z=t.0 by -1 to 0; say p.z /*display various tower floors & water.*/
end /*z*/
return

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[edit]

 
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

output

14
35
0
0
0
0

Scheme[edit]

 
(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)))
 
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[edit]

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
Output:
[2, 14, 35, 0, 0, 0, 0]

Tcl[edit]

Tcl makes for a surprisingly short and readable implementation, next to some of the more functional-oriented languages.

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
}
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[edit]

Version 1[edit]

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.

' 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
Output:
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.

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.

      ██
██
██≈≈██
██≈≈██
██████
████████
██████████
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.

Version 2[edit]

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.

    ''' <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

Regular version 2 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.

Sample of version 2 verbose output:

             ██
██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
██≈≈≈≈≈≈██≈≈≈≈≈≈≈≈≈≈≈≈≈≈██
██≈≈██≈≈██≈≈≈≈≈≈≈≈██≈≈████
██≈≈██≈≈██≈≈██≈≈≈≈██≈≈██████
██████≈≈██≈≈██≈≈≈≈██████████
████████████≈≈████████████████
████████████████████████████████ Block 3 holds 35 water units.
 
████████
████████
████████
████████
████████ Block 4 does not hold any water units.

zkl[edit]

Translation of: Haskell
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)
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();
Output:
L(2,14,35,0,0,0,0)