Two sum: Difference between revisions
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{{out}} |
{{out}} |
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<pre>[1,3]</pre> |
<pre>[1,3]</pre> |
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=={{header|Pascal}}== |
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A little bit lengthy. Implemented an unsorted Version with quadratic runtime too and an extra test case that needs 29999*30000/2 ~ 450 mio summations. |
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<lang pascal> |
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program twosum; |
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{$IFDEF FPC}{$MODE DELPHI}{$ELSE}{$APPTYPE CONSOLE}{$ENDIF} |
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uses |
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sysutils; |
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type |
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tSolRec = record |
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SolRecI, |
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SolRecJ : NativeInt; |
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end; |
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tMyArray = array of NativeInt; |
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const |
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ConstArray :array[-17..-13] of NativeInt = (0, 2, 11, 19, 90); |
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function Check2SumUnSorted(const A :tMyArray; |
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sum:NativeInt; |
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var Sol:tSolRec):boolean; |
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//Check every possible sum A[0] + A[1..Max] than A[1] + A[2..Max]... |
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//quadratic runtime (max-1)*max/ 2 |
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var |
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i,j,tmpSum: NativeInt; |
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Begin |
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Sol.SolRecI:=0; |
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Sol.SolRecJ:=0; |
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i := low(A); |
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while i < High(A) do |
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Begin |
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tmpSum := Sum-a[i]; |
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j := i+1; |
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while(j <= High(A)) do |
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begin |
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if tmpSum = a[j] then |
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Begin |
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Sol.SolRecI:=i;Sol.SolRecJ:=j; |
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result := true; |
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EXIT; |
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end; |
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inc(j); |
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end; |
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inc(i); |
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end; |
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result := false; |
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end; |
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function Check2SumSorted(const A :tMyArray; |
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sum:NativeInt; |
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var Sol:tSolRec):boolean; |
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var |
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i,j,tmpSum: NativeInt; |
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Begin |
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Sol.SolRecI:=0; |
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Sol.SolRecJ:=0; |
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i := low(A); |
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j := High(A); |
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while(i < j) do |
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Begin |
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tmpSum := a[i] + a[j]; |
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if tmpSum = sum then |
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Begin |
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Sol.SolRecI:=i;Sol.SolRecJ:=j; |
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result := true; |
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EXIT; |
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end; |
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if tmpSum < sum then |
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begin |
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inc(i); |
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continue; |
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end; |
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//if tmpSum > sum then |
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dec(j); |
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end; |
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result := false; |
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end; |
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var |
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Sol :tSolRec; |
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CheckArr : tMyArray; |
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MySum,i : NativeInt; |
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Begin |
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randomize; |
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setlength(CheckArr,High(ConstArray)-Low(ConstArray)+1); |
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For i := 0 to SizeOf(ConstArray)-1 do |
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CheckArr[i] := ConstArray[i+low(ConstArray)]; |
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MySum := 21; |
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IF Check2SumSorted(CheckArr,MySum,Sol) then |
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writeln('[',Sol.SolRecI,',',Sol.SolRecJ,'] sum to ',MySum) |
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else |
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writeln('No solution found'); |
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//now test a bigger sorted array.. |
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setlength(CheckArr,30000); |
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For i := High(CheckArr) downto 0 do |
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CheckArr[i] := 0; |
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CheckArr[High(CheckArr)-1] := 1; |
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CheckArr[High(CheckArr)] := 2; |
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MySum := 3; |
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//runtime 1 second |
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IF Check2SumUnSorted(CheckArr,MySum,Sol) then |
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writeln('[',Sol.SolRecI,',',Sol.SolRecJ,'] sum to ',MySum) |
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else |
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writeln('No solution found'); |
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//runtime not measurable |
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IF Check2SumSorted(CheckArr,MySum,Sol) then |
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writeln('[',Sol.SolRecI,',',Sol.SolRecJ,'] sum to ',MySum) |
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else |
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writeln('No solution found'); |
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end. |
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</lang> |
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{{out}} |
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<pre>[1,3] sum to 21 |
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[29998,29999] sum to 3 |
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[29998,29999] sum to 3 |
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real 0m1.097s</pre> |
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=={{header|REXX}}== |
=={{header|REXX}}== |
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<lang rexx>list=0 2 11 19 90 |
<lang rexx>list=0 2 11 19 90 |
Revision as of 19:35, 4 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 that needs 29999*30000/2 ~ 450 mio summations. <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
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[0] + A[1..Max] than A[1] + A[2..Max]... //quadratic runtime (max-1)*max/ 2 var
i,j,tmpSum: NativeInt;
Begin
Sol.SolRecI:=0; Sol.SolRecJ:=0; i := low(A); while i < High(A) do Begin tmpSum := Sum-a[i]; j := i+1; while(j <= High(A)) do begin if tmpSum = a[j] then Begin Sol.SolRecI:=i;Sol.SolRecJ:=j; result := true; EXIT; end; inc(j); end; inc(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; result := false;
end;
var
Sol :tSolRec; CheckArr : tMyArray; MySum,i : NativeInt;
Begin
randomize; setlength(CheckArr,High(ConstArray)-Low(ConstArray)+1); For i := 0 to SizeOf(ConstArray)-1 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,30000); For i := High(CheckArr) downto 0 do CheckArr[i] := 0; CheckArr[High(CheckArr)-1] := 1; CheckArr[High(CheckArr)] := 2; MySum := 3; //runtime 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 [29998,29999] sum to 3 [29998,29999] sum to 3 real 0m1.097s
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()