Given an NxM matrix and a product K. The task is to check if a pair with the given product 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}};
K = 42
Output: YES
Input: mat[N][M] = {{1, 2, 3, 4},
{5, 6, 7, 8}};
K = 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 the current element of the matrix is divisible by the given product and when the product K is divided by the current element, the dividend obtained is also present in the hash.
That is,
(k % mat[i][j] == 0) && (mp[k / mat[i][j]] > 0)
- If present, then return true, else insert current elements 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 product// 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// product exists in the matrixbool isPairWithProductK(int mat[N][M], int k){ // 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 ((k % mat[i][j] == 0) && (s.find(k / 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 product int k = 42; if (isPairWithProductK(mat, k)) { cout << "YES" << endl; } else cout << "NO" << endl; return 0;} |
Java
// Java code to check for pair with given product// 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 // product exists in the matrix static boolean isPairWithProductK(int mat[][], int k) { // 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 ((k % mat[i][j] == 0) && s.contains(k / 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 product int k = 42; if (isPairWithProductK(mat, k)) { System.out.println("YES"); } else System.out.println("NO"); } // This code is contributed by ihritik} |
Python 3
# Python 3 code to check for pair with # given product exists in the matrix or notN = 4M = 4# Function to check if a pair with # given product exists in the matrixdef isPairWithProductK(mat, k): # hash to store elements s = [] # 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 ((k % mat[i][j] == 0) and (k // mat[i][j]) in s): return True else : s.append(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 product k = 42 if (isPairWithProductK(mat, k)): print( "YES") else: print("NO")# This code is contributed by ita_c |
C#
// C# code to check for pair with given product// exists in the matrix or notusing System;using System.Collections.Generic;class GFG{ static readonly int N = 4; static readonly int M = 4; // Function to check if a pair with given // product exists in the matrix static bool isPairWithProductK(int [,]mat, int k) { // 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 ((k % mat[i, j] == 0) && s.Contains(k / mat[i, j])) { return true; } else { s.Add(mat[i, j]); } } } return false; } // Driver code public static void Main() { // Input matrix int [,]mat = { { 1, 2, 3, 4 }, { 5, 6, 7, 8 }, { 9, 10, 11, 12 }, { 13, 14, 15, 16 } }; // given product int k = 42; if (isPairWithProductK(mat, k)) { Console.WriteLine("YES"); } else Console.WriteLine("NO"); }}// This code has been contributed// by PrinciRaj1992 |
Javascript
<script>// Javascript code to check for pair with given product// exists in the matrix or not let N = 4; let M = 4; function isPairWithProductK(mat,k) { // 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 ((k % mat[i][j] == 0) && s.has(Math.floor(k / 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 product let k = 42; if (isPairWithProductK(mat, k)) { 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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