# Pair with given sum in matrix

• Difficulty Level : Easy
• Last Updated : 23 May, 2021

Given a NxM matrix and a sum S. The task is to check if a pair with given Sum exists in the matrix or not.
Examples

```Input: mat[N][M] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16}};
sum = 31
Output: YES

Input: mat[N][M] = {{1, 2, 3, 4},
{5, 6, 7, 8}};
sum = 150
Output: NO```

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Approach:

• Take a hash to store all elements of the matrix in the hash.
• Start traversing through the matrix, and while traversing check if abs(sum-matrix_element) is present in the hash.
• If present, then return true, else insert the current matrix element into the hash.
• If all elements of the matrix are traversed and no pair is found, return false.

Below is the implementation of the above approach:

## C++

 `// CPP code to check for pair with given sum``#include ``using` `namespace` `std;` `#define N 4``#define M 4` `// Function to check if a pair with given sum``// exists in the matrix``bool` `isPairWithSum(``int` `mat[N][M], ``int` `sum)``{``    ``// hash to store elements``    ``unordered_set<``int``> s;` `    ``// looping through elements``    ``// if present in the matrix``    ``// return true, else push``    ``// the element in matrix``    ``for` `(``int` `i = 0; i < N; i++) {``        ``for` `(``int` `j = 0; j < M; j++) {``            ``if` `(s.find(sum - mat[i][j]) != s.end()) {``                ``return` `true``;``            ``}``            ``else` `{``                ``s.insert(mat[i][j]);``            ``}``        ``}``    ``}` `    ``return` `false``;``}` `// Driver code``int` `main()``{` `    ``// Input matrix``    ``int` `mat[N][M] = { { 1, 2, 3, 4 },``                      ``{ 5, 6, 7, 8 },``                      ``{ 9, 10, 11, 12 },``                      ``{ 13, 14, 15, 16 } };` `    ``// given sum``    ``int` `sum = 11;` `    ``if` `(isPairWithSum(mat, sum)) {``        ``cout << ``"YES"` `<< endl;``    ``}``    ``else``        ``cout << ``"NO"` `<< endl;` `    ``return` `0;``}`

## Java

 `// Java code to check for pair``// with given sum``import` `java.util.*;` `class` `GFG``{``    ` `// Function to check if a pair with``// given sum exists in the matrix``static` `final` `int` `N = ``4``;``static` `final` `int` `M = ``4``;``static` `boolean` `isPairWithSum(``int` `[][]mat,``                             ``int` `sum)``{``    ``// hash to store elements``    ``Set s = ``new` `HashSet();` `    ``// looping through elements``    ``// if present in the matrix``    ``// return true, else push``    ``// the element in matrix``    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``        ``for` `(``int` `j = ``0``; j < M; j++)``        ``{``            ``if` `(s.contains(sum - mat[i][j]))``            ``{``                ``return` `true``;``            ``}``            ``else``            ``{``                ``s.add(mat[i][j]);``            ``}``        ``}``    ``}` `    ``return` `false``;``}` `// Driver code``public` `static` `void` `main(String []args)``{` `    ``// Input matrix``    ``int` `[][]mat = { { ``1``, ``2``, ``3``, ``4` `},``                    ``{ ``5``, ``6``, ``7``, ``8` `},``                    ``{ ``9``, ``10``, ``11``, ``12` `},``                    ``{ ``13``, ``14``, ``15``, ``16` `} };` `    ``// given sum``    ``int` `sum = ``11``;` `    ``if` `(isPairWithSum(mat, sum))``    ``{``        ``System.out.println(``"YES"``);``    ``}``    ``else``        ``System.out.println(``"NO"``);``}``}` `// This code is contributed by ihritik`

## C#

 `// C# code to check for pair``// with given sum``using` `System;``using` `System.Collections.Generic;` `class` `GFG``{``    ` `// Function to check if a pair with``// given sum exists in the matrix``static` `readonly` `int` `N = 4;``static` `readonly` `int` `M = 4;``static` `bool` `isPairWithSum(``int` `[,]mat,``                            ``int` `sum)``{``    ``// hash to store elements``    ``HashSet<``int``> s = ``new` `HashSet<``int``>();` `    ``// looping through elements``    ``// if present in the matrix``    ``// return true, else push``    ``// the element in matrix``    ``for` `(``int` `i = 0; i < N; i++)``    ``{``        ``for` `(``int` `j = 0; j < M; j++)``        ``{``            ``if` `(s.Contains(sum - mat[i, j]))``            ``{``                ``return` `true``;``            ``}``            ``else``            ``{``                ``s.Add(mat[i, j]);``            ``}``        ``}``    ``}``    ``return` `false``;``}` `// Driver code``public` `static` `void` `Main(String []args)``{` `    ``// Input matrix``    ``int` `[,]mat = { { 1, 2, 3, 4 },``                    ``{ 5, 6, 7, 8 },``                    ``{ 9, 10, 11, 12 },``                    ``{ 13, 14, 15, 16 } };` `    ``// given sum``    ``int` `sum = 11;` `    ``if` `(isPairWithSum(mat, sum))``    ``{``        ``Console.WriteLine(``"YES"``);``    ``}``    ``else``        ``Console.WriteLine(``"NO"``);``}``}` `// This code contributed by Rajput-Ji`

## Python3

 `# python code to check for pair with given sum` `N``=` `4``M``=` `4` `# Function to check if a pair with given sum``# exists in the matrix``def` `isPairWithSum(mat, ``sum``):``    ``# hash to store elements``    ``s ``=` `set``()` `    ``# looping through elements``    ``# if present in the matrix``    ``# return true, else push``    ``# the element in matrix``    ``for` `i ``in` `range``(N):``        ``for` `j ``in` `range``(M):``            ``if` `(``sum` `-` `mat[i][j]) ``in` `s :``                ``return` `True``            ` `            ``else` `:``                ``s.add(mat[i][j])``            ` `    ``return` `False`  `# Driver code``if` `__name__ ``=``=` `'__main__'``:` `    ``# Input matrix``    ``mat ``=` `[[ ``1``, ``2``, ``3``, ``4` `],``                    ``[ ``5``, ``6``, ``7``, ``8``] ,``                    ``[ ``9``, ``10``, ``11``, ``12``] ,``                    ``[``13``, ``14``, ``15``, ``16``]] ` `    ``# given sum``    ``sum` `=` `11` `    ``if` `(isPairWithSum(mat, ``sum``)) :``        ``print``(``"YES"``)``    ` `    ``else``:``        ``print``(``"NO"``)` `    ` `# This code is contributed by AbhiThakur`

## PHP

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## Javascript

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Output:
`YES`

Time Complexity: O(N*M)

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