Two sum: Difference between revisions
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<pre>[1,3]</pre>
=={{header|Pascal}}==
A little bit lengthy. Implemented an unsorted Version with quadratic runtime too and an extra test case with 60000 elements that needs
<lang pascal>
program twosum;{$IFDEF FPC}{$MODE DELPHI}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF}
uses
sysutils;
Line 68 ⟶ 66:
tMyArray = array of NativeInt;
const
// just a gag using unusual index limits
ConstArray :array[-17..-13] of NativeInt = (0, 2, 11, 19, 90);
Line 73 ⟶ 72:
sum:NativeInt;
var Sol:tSolRec):boolean;
//Check every possible sum A[
//than A[max-1] + A[max-2..0] etc pp.
//quadratic runtime: maximal (max-1)*max/ 2 checks
//High(A) always checked for dynamic array, even const
//therefore run High(A) to low(A) = is always 0 for dynamic array
var
i,j,tmpSum: NativeInt;
Line 80 ⟶ 82:
Sol.SolRecI:=0;
Sol.SolRecJ:=0;
i :=
while i
Begin
tmpSum :=
j := i
while
begin
if tmpSum = a[j] then
Begin
Sol.SolRecI:=
result := true;
EXIT;
end;
end;
end;
result := false;
Line 127 ⟶ 129:
dec(j);
end;
writeln(i:10,j:10);
result := false;
end;
Line 138 ⟶ 141:
randomize;
setlength(CheckArr,High(ConstArray)-Low(ConstArray)+1);
For i :=
CheckArr[i] := ConstArray[i+low(ConstArray)];
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//now test a bigger sorted array..
setlength(CheckArr,
For i := High(CheckArr) downto 0 do
CheckArr[i] :=
MySum := CheckArr[
writeln(#13#10,'Now checking array of ',length(CheckArr),
' elements',#13#10);
//runtime about 1 second▼
▲ //runtime 1 second
IF Check2SumUnSorted(CheckArr,MySum,Sol) then
writeln('[',Sol.SolRecI,',',Sol.SolRecJ,'] sum to ',MySum)
Line 166 ⟶ 168:
else
writeln('No solution found');
end.</lang>
</lang>▼
{{out}}
<pre>[1,3] sum to 21
[29998,29999] sum to 3▼
[29998,29999] sum to 3▼
Now checking array of 60000 elements
real 0m1.097s</pre>▼
=={{header|REXX}}==
<lang rexx>list=0 2 11 19 90
|
Revision as of 07:56, 5 October 2016
- Task
Given a sorted array of single positive integers, is it possible to find a pair of integers from that array that sum up to a given sum? If so, return indices of the two integers or an empty array if not.
- Example
Given numbers = [0, 2, 11, 19, 90], sum = 21,
Because numbers[1] + numbers[3] = 2 + 19 = 21,
return [1, 3].
- Source
Stack Overflow: Find pair of numbers in array that add to given sum
C#
<lang csharp>public static int[] TwoSum(int[] numbers, int sum) {
var map = new Dictionary<int, int>(); for (var i = 0; i < numbers.Length; i++) { var key = sum - numbers[i]; if (map.ContainsKey(key)) return new[] { map[key], i }; map.Add(numbers[i], i); } return Array.Empty<int>();
}</lang>
- Output:
[0,3]
ooRexx
<lang oorexx>a=.array~of(0, 2, 11, 19, 90) x=21 do i=1 To a~items
If a[i]>x Then Leave Do j=i+1 To a~items s=a[i] s+=a[j] Select When s=x Then Leave i When s>x Then Leave j Otherwise Nop End End End
If s=x Then Do
i-=1 /* array a's index starts with 1, so adjust */ j-=1 Say '['i','j']' End
Else
Say '[] - no items found'</lang>
- Output:
[1,3]
Pascal
A little bit lengthy. Implemented an unsorted Version with quadratic runtime too and an extra test case with 60000 elements that needs 59999*59998/2 ~ 1.8 billion checks ( ~2 cpu-cycles/check ). <lang pascal> program twosum;{$IFDEF FPC}{$MODE DELPHI}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF} uses
sysutils;
type
tSolRec = record SolRecI, SolRecJ : NativeInt; end; tMyArray = array of NativeInt;
const // just a gag using unusual index limits
ConstArray :array[-17..-13] of NativeInt = (0, 2, 11, 19, 90);
function Check2SumUnSorted(const A :tMyArray;
sum:NativeInt; var Sol:tSolRec):boolean;
//Check every possible sum A[max] + A[max-1..0] //than A[max-1] + A[max-2..0] etc pp. //quadratic runtime: maximal (max-1)*max/ 2 checks //High(A) always checked for dynamic array, even const //therefore run High(A) to low(A) = is always 0 for dynamic array var
i,j,tmpSum: NativeInt;
Begin
Sol.SolRecI:=0; Sol.SolRecJ:=0; i := High(A); while i > low(A) do Begin tmpSum := sum-a[i]; j := i-1; while j >= low(A) do begin if tmpSum = a[j] then Begin Sol.SolRecI:=j;Sol.SolRecJ:=i; result := true; EXIT; end; dec(j); end; dec(i); end; result := false;
end;
function Check2SumSorted(const A :tMyArray;
sum:NativeInt; var Sol:tSolRec):boolean;
var
i,j,tmpSum: NativeInt;
Begin
Sol.SolRecI:=0; Sol.SolRecJ:=0; i := low(A); j := High(A); while(i < j) do Begin tmpSum := a[i] + a[j]; if tmpSum = sum then Begin Sol.SolRecI:=i;Sol.SolRecJ:=j; result := true; EXIT; end; if tmpSum < sum then begin inc(i); continue; end; //if tmpSum > sum then dec(j); end; writeln(i:10,j:10); result := false;
end;
var
Sol :tSolRec; CheckArr : tMyArray; MySum,i : NativeInt;
Begin
randomize; setlength(CheckArr,High(ConstArray)-Low(ConstArray)+1); For i := High(CheckArr) downto low(CheckArr) do CheckArr[i] := ConstArray[i+low(ConstArray)];
MySum := 21; IF Check2SumSorted(CheckArr,MySum,Sol) then writeln('[',Sol.SolRecI,',',Sol.SolRecJ,'] sum to ',MySum) else writeln('No solution found'); //now test a bigger sorted array.. setlength(CheckArr,60000); For i := High(CheckArr) downto 0 do CheckArr[i] := i; MySum := CheckArr[Low(CheckArr)]+CheckArr[Low(CheckArr)+1]; writeln(#13#10,'Now checking array of ',length(CheckArr), ' elements',#13#10); //runtime about 1 second IF Check2SumUnSorted(CheckArr,MySum,Sol) then writeln('[',Sol.SolRecI,',',Sol.SolRecJ,'] sum to ',MySum) else writeln('No solution found');
//runtime not measurable IF Check2SumSorted(CheckArr,MySum,Sol) then writeln('[',Sol.SolRecI,',',Sol.SolRecJ,'] sum to ',MySum) else writeln('No solution found');
end.</lang>
- Output:
[1,3] sum to 21 Now checking array of 60000 elements [0,1] sum to 1 [0,1] sum to 1 real 0m1.035s
REXX
<lang rexx>list=0 2 11 19 90 Do i=0 By 1 Until list=
Parse Var list a.i list End
n=i-1 x=21 do i=1 To n
If a.i>x Then Leave Do j=i+1 To n s=a.i s=s+a.j Select When s=x Then Leave i When s>x Then Leave j Otherwise Nop End End End
If s=x Then
Say '['i','j']'
Else
Say '[] - no items found'</lang>
- Output:
[1,3]
zkl
The sorted O(n) no external storage solution: <lang zkl>fcn twoSum(sum,ns){
i,j:=0,ns.len()-1; while(i<j){ if((s:=ns[i] + ns[j]) == sum) return(i,j); else if(s<sum) i+=1; else if(s>sum) j-=1; }
}</lang> <lang zkl>twoSum2(21,T(0,2,11,19,90)).println(); twoSum2(25,T(0,2,11,19,90)).println();</lang>
- Output:
L(1,3) False
The unsorted O(n!) all solutions solution: <lang zkl>fcn twoSum2(sum,ns){
Utils.Helpers.combosKW(2,ns).filter('wrap([(a,b)]){ a+b==sum }) // lazy combos .apply('wrap([(a,b)]){ return(ns.index(a),ns.index(b)) })
}</lang> <lang zkl>twoSum2(21,T(0,2,11,19,90,21)).println(); twoSum2(25,T(0,2,11,19,90,21)).println();</lang>
- Output:
L(L(0,5),L(1,3)) L()