Given an N×M matrix and a difference K. The task is to check if a pair with the given absolute difference 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, 100}};
K = 85
Output: YES
Input: mat[N][M] = {{1, 2, 3, 4},
{5, 6, 7, 8}};
K = 150
Output: NO
Approach:
- Initialize a hash-map to keep track of already visited elements of the matrix.
- Iterate over the matrix and check if the difference between the current element and k is already present in the hash-map. If yes then the current element and the difference between the elements and k are the desired pair.
- Otherwise, add the current element to the map.
Below is the implementation of the above approach:
C++
// CPP code to check for pair with given // difference exists in the matrix or not #include <bits/stdc++.h> using namespace std; #define N 4 #define M 4 // Function to check if a pair with given // difference exists in the matrix bool isPairWithDiff(int mat[N][M], int k) { unordered_set <int> ump ; // Loop to iterate over the matrix for( int i = 0 ; i < N ; i++ ) { for( int j =0 ; j < M ; j++ ) { if( mat[i][j] > k ) { int m = mat[i][j] - k ; if( ump.find(m) != ump.end() ) { return true ; } } else { int m = k - mat[i][j] ; if( ump.find(m) != ump.end() ) { return true ; } } ump.insert(mat[i][j]); } } return false;} // Driver Code int main() { // Input matrix int mat[N][M] ={ { 5, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 100 } }; // Given difference int k = 85; if (isPairWithDiff(mat, k)) cout << "YES" << endl; else cout << "NO" << endl; return 0; } |
Java
// Java code to check for pair with given// difference exists in the matrix or notimport java.util.*;class GFG {static final int N = 4;static final int M = 4;// Function to check if a pair with given// difference exists in the matrixstatic boolean isPairWithDiff(int mat[][], int k){ // Store elements in a hash HashSet<Integer> s = new HashSet<Integer>(); // Loop to iterate over the // elements of the matrix for (int i = 0; i < N; i++) { for (int j = 0; j < M; j++) { if( mat[i][j] > k ) { int m = mat[i][j] - k ; if(s.contains(m)) { return true ; } } else { int m = k - mat[i][j] ; if(s.contains(m)) { return true ; } } s.add(mat[i][j]); } } return false;}// Driver Codepublic static void main(String[] args){ // Input matrix int mat[][] = { { 5, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 100 } }; // Given difference int k = 85; System.out.println( isPairWithDiff(mat, k) == true ? "YES" : "NO"); }}// This code contributed by Rajput-Ji |
Python3
# Python 3 program to check for pairs # with given difference exits in the # matrix or not N = 4M = 4# Function to check if a # pair with given# difference exist in the matrixdef isPairWithDiff(mat, k): # Store elements in a hash s = set() # Store elements in dict for i in range(N): for j in range(M): if mat[i][j] > k: m = mat[i][j] - k if m in s: return True else: m = k - mat[i][j] if m in s: return True s.add(mat[i][j]) return False # Driver Coden, m = 4, 4mat = [[5, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 100]] # given differencek = 85if isPairWithDiff(mat, k): print("Yes")else: print("No")# This code is contributed by# Mohit kumar 29 (IIIT gwalior) |
C#
// C# code to check for pair with given// difference exists in the matrix or notusing System;using System.Collections.Generic; class GFG {static int N = 4;static int M = 4;// Function to check if a pair with given// difference exists in the matrixstatic Boolean isPairWithDiff( int [,]mat, int k){ // Store elements in a hash HashSet<int> s = new HashSet<int>(); for (int i = 0; i < N; i++) { for (int j = 0; j < M; j++) { if( mat[i, j] > k ) { int m = mat[i, j] - k ; if(s.Contains(m)) { return true ; } } else { int m = k - mat[i, j]; if(s.Contains(m)) { return true ; } } s.Add(mat[i, j]); } } return false;}// Driver Codepublic static void Main(String[] args){ // Input matrix int [,]mat = { { 5, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 100 } }; // Given difference int k = 85; Console.WriteLine( isPairWithDiff(mat, k) == true ? "YES" : "NO"); }}/* This code contributed by PrinciRaj1992 */ |
Javascript
<script>// Javascript code to check for pair with given// difference exists in the matrix or notlet N = 4;let M = 4;// Function to check if a pair with given// difference exists in the matrixfunction isPairWithDiff(mat,k){ // Store elements in a hash let s = []; // Loop to iterate over the // elements of the matrix for (let i = 0; i < N; i++) { for (let j = 0; j < M; j++) { if( mat[i][j] > k ) { let m = mat[i][j] - k ; if(s.includes(m)) { return true ; } } else { let m = k - mat[i][j] ; if(s.includes(m)) { return true ; } } s.push(mat[i][j]); } } return false;}// Driver Code // Input matrixlet mat = [[5, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 100]];// Given differencelet k = 85;document.write( isPairWithDiff(mat, k) == true ? "YES" : "NO");// This code is contributed by patel2127</script> |
Output
YES
Complexity Analysis:
- Time Complexity: O(N*M)
- Auxiliary Space: O(N*M)
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