Pair Sum in Array Equal to Number

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Find pair of elements from an array whose sum is equal to a given number X


Time Complexity: O(NlogN)
Space Complexity: O(1)

Let say there are numbers

-1 5 7 -8 12 16 -30 -4
And we need to find a pair whose sum is say 17

Then the efficient way is shown below

1. Arrange the numbers in ascending order (Sort it )

2. Run a loop this way :
  • Let say one person stand at start of array and the second person stand at end

  • They add up the numbers on which they are standing                        

                    => -30 +16 =14

  • If their number is equal to the given number X then they enjoy

  • Else if the sum is lesser than the given number they ask the first person standing at start of array to move one step towards the end because if he travels right the number increases and hence the sum.

  • Similarly if the sum is more the person at the end is asked to move towards start by one step

  • We repeat until the persons collide as after that the same pairs will be obtained.

  • If no such pair is found then both become sad.

Important thing is the Sort() function.We should use Merge or Heap sort as they take O(NLogN) time.Orelse we can use sort() STL function. with parameters "sort(arr,arr+size)"


Input:   -3 ,-4 , 10 , 0 ,3 ,-2 ,15 , 3
Sum:  7
Answer:  -3 and 10

#include <bits/stdc++.h>
using namespace std;

int main()
	int arr[] = {-3,-4,10,0,3,-2,15,3};
	int size_of_array = sizeof(arr)/sizeof(arr[0]);
	//RequiredSum is number to which the pair should sum to
	int RequiredSum = 7;
	sort(arr,arr + size_of_array); //sort the array

	int startIndex = 0 ,  endIndex = size_of_array - 1 , sum =0; //variables pointing on their respective indices and sum to store sum of the pair

	while(startIndex <endIndex) //We require a pair so 2 elements and hence both elements should be of different indices
		sum = arr[startIndex] + arr[endIndex];
		if( sum == RequiredSum)
			cout << "The numbers are " << arr[startIndex] <<" and " << arr[endIndex] <<endl;
			return 0;

		else if(sum < RequiredSum) //if sum is less then we need to increase the smaller one
			startIndex ++;
		else //if the sum if more we need to decrease the larger number
			endIndex --;
	cout << "No such pair exists.";
	return 0;
Try It


Time Complexity - O(NlogN)
Space Complexity - O(N)

Let say there are numbers

-1 5 7 -8 12 16 -30 -4

and we need to find a pair whose sum is say 23

This concept is based on the Mapping approach.

1. Create a map such that each element is added.
2. If there exists a pair that sums to X then both the elements are present in map.
3. So we loop through the array and do this

  • Find if the (X – present) element is present in map
  • If present then print the number.

**You can STL map data structure for this purpose.Or else you can create an array for indexing where index are the value of the array element itself.(But using array has a disadvantage that its size is required.)


Input:   -3 ,-4 , 10 , 0 ,3 ,-2 ,15 , 3
Sum:  7
Answer:  -3 and 10

#include <bits/stdc++.h>
using namespace std;

map <int,int> m;
map<int,int>::iterator it;

int main()
	int arr[] = {-3,-4,10,0,3,-2,15,3};
	int size_of_array = sizeof(arr)/sizeof(arr[0]);
	//cout << "Enter the number to which the pair should sum to"<<endl;
       //Let Sum = 7
	int x = 7;

	//cin >> x; // the number to which the sum of pair of elements must be equal

	for (int i = 0; i < size_of_array;++i)
		//Scan and add elements into the map
		it = m.find(arr[i]);
		if(it == m.end())
			m.insert(make_pair(arr[i],1)); //Add the element in the map and set the count to 1 that represents it is present

	for (int i = 0; i < size_of_array;i++)
			it = m.find((x - arr[i]));  //If we have two numbers say m and n that sums to x then 
		//if we have m and if we find n in the map then we got the numbers.

			if(it != m.end()) //If it exists then we got the pair
				pair<int,int> p = *it; //Obtain the pair so as to reference the 2nd number
				cout << "The numbers are " << arr[i] <<" and " << p.first <<endl;
				return 0;
	cout << "No such pair exists.";
	return 0;
Try It

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