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 not import 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 matrix static 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 Code public 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 = 4 M = 4 # Function to check if a # pair with given # difference exist in the matrix def 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 Code n, m = 4 , 4 mat = [[ 5 , 2 , 3 , 4 ], [ 5 , 6 , 7 , 8 ], [ 9 , 10 , 11 , 12 ], [ 13 , 14 , 15 , 100 ]] # given difference k = 85 if 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 not using 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 matrix static 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 Code public 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 not let N = 4; let M = 4; // Function to check if a pair with given // difference exists in the matrix function 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 matrix let mat = [[5, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12], [13, 14, 15, 100]]; // Given difference let 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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