Two Sum (coding problem)

First, solve the problem on your own or just dry run it.

here is the leetcode problem

Two Sum - LeetCode

//this takes the O(n log n ) time complexity

// spsce complexity is O(1) which is constant space.

#include <iostream>

#include<bits/stdc++.h>

using namespace std;

int twonumberSum(int array[],int targetSum,int n)

{

sort(array, array + n);

int left = 0;

int right =n-1;

while (left < right) {

int currentSum = array[left] + array[right];

if (currentSum == targetSum) {

cout<<array[left]<<’ ‘;

cout<<array[right];

break;

}

else if (currentSum < targetSum) {

left++;

}

else if (currentSum > targetSum) {

right — ;

}

}

return {};

}

int main()

{

int array[]={-4,11,-1,8,1,6,3,5};

int n = sizeof(array) / sizeof(array[0]);

//cout<<”size is “<<n;

int targetSum=17;

twonumberSum(array,targetSum,n);

return 0;

}