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
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 <bits/stdc++.h>using namespace std;#define N 4#define M 4// Function to check if a pair with given sum// exists in the matrixbool 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 codeint 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 sumimport java.util.*;class GFG{ // Function to check if a pair with // given sum exists in the matrixstatic final int N = 4;static final int M = 4;static boolean isPairWithSum(int [][]mat, int sum){ // hash to store elements Set<Integer> s = new HashSet<Integer>(); // 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 codepublic 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 sumusing System;using System.Collections.Generic;class GFG{ // Function to check if a pair with // given sum exists in the matrixstatic 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 codepublic 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 sumN= 4M= 4# Function to check if a pair with given sum# exists in the matrixdef 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 codeif __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
<?php // PHP code to check for pair with given sum $N = 4;$M = 4; // Function to check if a pair with given sum// exists in the matrixfunction isPairWithSum(&$mat, $sum){ global $N,$M; // array to store elements $s = array(); // looping through elements // if present in the matrix // return true, else push // the element in matrix for ($i = 0; $i < $N; $i++) { for ($j = 0; $j < $M; $j++) { if (in_array($sum - $mat[$i][$j],$s) != end($s)) { return true; } else { array_push($s,$mat[$i][$j]); } } } return false;} // Driver code // Input matrix $mat = array(array( 1, 2, 3, 4 ), array( 5, 6, 7, 8 ), array( 9, 10, 11, 12 ), array(13, 14, 15, 16 )); // given sum $sum = 11; if (isPairWithSum($mat, $sum)) { echo "YES" ."\n"; } else echo "NO" ."\n"; return 0;?> |
Javascript
<script>// Javascript code to check for pair// with given sum // Function to check if a pair with // given sum exists in the matrix let N = 4; let M = 4; function isPairWithSum(mat,sum) { // hash to store elements let s = new Set(); // looping through elements // if present in the matrix // return true, else push // the element in matrix for (let i = 0; i < N; i++) { for (let j = 0; j < M; j++) { if (s.has(sum - mat[i][j])) { return true; } else { s.add(mat[i][j]); } } } return false; } // Driver code // Input matrix let mat = [[ 1, 2, 3, 4 ], [ 5, 6, 7, 8] , [ 9, 10, 11, 12] , [13, 14, 15, 16]] ; // given sum let sum = 11; if (isPairWithSum(mat, sum)) { document.write("YES"); } else document.write("NO"); // This code is contributed by rag2127</script> |
Output
YES
Time Complexity: O(N*M)
Auxiliary Space: O(N*M)
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