Check if a pair with given product exists in a Matrix

Given a 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++

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// 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 matrix
bool 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 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 product
    int k = 42;
  
    if (isPairWithProductK(mat, k)) {
        cout << "YES" << endl;
    }
    else
        cout << "NO" << endl;
  
    return 0;
}

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Java

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// Java code to check for pair with given product
// 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
    // 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 conributed by ihritik
}

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Python 3

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# Python 3 code to check for pair with 
# given product exists in the matrix or not
N = 4
M = 4
  
# Function to check if a pair with 
# given product exists in the matrix
def 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 code
if __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

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C#

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// C# code to check for pair with given product
// exists in the matrix or not
using 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 

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Output:

YES

Time Complexity: O(N*M)



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