Given an NxN square matrix M[][]. The task is to check whether the matrix M is a zero division matrix or not. A matrix is said to be a zero division matrix only if the 2 or more results of floor division of the product of column-wise element to the product of row-wise element is zero.
Examples:
Input: M[ ][ ] = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ]Output: 1
Explanation: Refer to image for explanation.
Input: M[ ][ ] = [ [ 1, 2, 3 ],
[ 6, 8, 2 ],
[ 7, 8, 9 ] ]
Output: 0
Approach: This problem is implementation-based. Follow the steps below to solve the given problem.
- Take a matrix M[ ][ ] and calculate the product of row-wise and column-wise elements.
- Store product of each row and column in R[ ] and C[ ] array respectively.
- Now take floor division of index-wise element of C[ ] array by R[ ] array.
- If we get two or more zero then it is zero division grid and return 1 else return 0.
- In case the denominator has zero then it is case of zero division error return -1.
Below is the implementation of the above approach.
C++
// C++ program for above approach#include <bits/stdc++.h>using namespace std;// Function to check zero division matrixint ZeroDiv(vector<vector<int> >& M, int& N){ // R and C array to store product vector<int> R, C; // Iterate through the matrix // to take product for (int i = 0; i < N; i++) { int r = 1; int c = 1; for (int j = 0; j < N; j++) { r *= M[i][j]; c *= M[j][i]; } // Appending product of each row // and column to R and C array. R.push_back(r); C.push_back(c); } // z to count number of zero int z = 0; // Try block to check zero division error try { for (int i = 0; i < N; i++) if (C[i] / R[i] == 0) z += 1; } // If zero division error occur return -1 catch (int x) { cout << x << "\n"; return -1; } // If z greater than equal to 2 return 1 if (z >= 2) { return 1; } // Else return 0 else { return 0; }}// Driver Codeint main(){ // # Matrix M vector<vector<int> > M = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } }; // N length of matrix M int N = M.size(); // Call zerodiv function int x = ZeroDiv(M, N); // Print the answer cout << x << "\n";}// This code is contributed by Taranpreet |
Java
// JAVA program for above approachimport java.util.*;class GFG{ // Function to check zero division matrix public static int ZeroDiv(ArrayList<ArrayList<Integer> > M, int N) { // R and C array to store product ArrayList<Integer> R = new ArrayList<Integer>(); ArrayList<Integer> C = new ArrayList<Integer>(); // Iterate through the matrix // to take product for (int i = 0; i < N; i++) { int r = 1; int c = 1; for (int j = 0; j < N; j++) { r = r * M.get(i).get(j); c = c * M.get(i).get(j); } // Appending product of each row // and column to R and C array. R.add(r); C.add(c); } // z to count number of zero int z = 0; // Try block to check zero division error try { for (int i = 0; i < N; i++) if (C.get(i) % R.get(i) == 0) z = z + 1; } // If zero division error occur return -1 catch (Exception e) { System.out.println("Something went wrong" + e); return -1; } // If z greater than equal to 2 return 1 if (z >= 2) { return 1; } // Else return 0 else { return 0; } } // Driver Code public static void main(String[] args) { // # Matrix M ArrayList<ArrayList<Integer> > M = new ArrayList<ArrayList<Integer> >(); ArrayList<Integer> temp1 = new ArrayList<>(Arrays.asList(1, 2, 3)); ArrayList<Integer> temp2 = new ArrayList<>(Arrays.asList(4, 5, 6)); ArrayList<Integer> temp3 = new ArrayList<>(Arrays.asList(7, 8, 9)); M.add(temp1); M.add(temp2); M.add(temp3); // N length of matrix M int N = M.size(); // Call zerodiv function int x = ZeroDiv(M, N); // Print the answer System.out.println(x); }}// This code is contributed by Taranpreet |
Python3
# Python program for above approach# Function to check zero division matrixdef ZeroDiv(M, N): # R and C array to store product R = [] C = [] # Iterate through the matrix # to take product for i in range(N): r = 1 c = 1 for j in range(N): r *= M[i][j] c *= M[j][i] # Appending product of each row # and column to R and C array. R.append(r) C.append(c) # z to count number of zero z = 0 # Try block to check zero division error try: for i in range(N): if C[i]//R[i] == 0: z += 1 # If zero division error occur return -1 except: return -1 # If z greater than equal to 2 return 1 if z >= 2: return 1 # Else return 0 else: return 0# Driver Codeif __name__ == "__main__": # Matrix M M = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] # N length of matrix M N = len(M) # Call zerodiv function x = ZeroDiv(M, N) # Print the answer print(x) |
C#
// C# program for above approachusing System;using System.Collections.Generic;class GFG{ // Function to check zero division matrix public static int ZeroDiv(List<List<int> > M, int N) { // R and C array to store product List<int> R = new List<int>(); List<int> C = new List<int>(); // Iterate through the matrix // to take product for (int i = 0; i < N; i++) { int r = 1; int c = 1; for (int j = 0; j < N; j++) { r = r * M[i][j]; c = c * M[i][j]; } // Appending product of each row // and column to R and C array. R.Add(r); C.Add(c); } // z to count number of zero int z = 0; // Try block to check zero division error try { for (int i = 0; i < N; i++) if (C[i] % R[i] == 0) z = z + 1; } // If zero division error occur return -1 catch (Exception e) { Console.WriteLine("Something went wrong" + e); return -1; } // If z greater than equal to 2 return 1 if (z >= 2) { return 1; } // Else return 0 else { return 0; } } // Driver Code public static void Main(string[] args) { // # Matrix M List<List<int> > M = new List<List<int> >(); List<int> temp1 = new List<int>(){1, 2, 3}; List<int> temp2 = new List<int>(){4, 5, 6}; List<int> temp3 = new List<int>(){7, 8, 9}; M.Add(temp1); M.Add(temp2); M.Add(temp3); // N length of matrix M int N = M.Count; // Call zerodiv function int x = ZeroDiv(M, N); // Print the answer Console.WriteLine(x); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript code for the above approach // Function to check zero division matrix function ZeroDiv(M, N) { // R and C array to store product let R = [] let C = [] // Iterate through the matrix // to take product for (let i = 0; i < N; i++) { r = 1 c = 1 for (let j = 0; j < N; j++) { r *= M[i][j] c *= M[j][i] } // Appending product of each row // and column to R and C array. R.push(r) C.push(c) } // z to count number of zero z = 0 // Try block to check zero division error try { for (let i = 0; i < N; i++) if (Math.floor(C[i] / R[i]) == 0) z += 1 } // If zero division error occur return -1 catch { return -1 } // If z greater than equal to 2 return 1 if (z >= 2) return 1 // Else return 0 else return 0 } // Driver Code // Matrix M let M = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] // N length of matrix M let N = M.length // Call zerodiv function let x = ZeroDiv(M, N) // Print the answer document.write(x) // This code is contributed by Potta Lokesh </script> |
Output
1
Time Complexity: O(N*N)
Auxiliary Space: O(N)
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