Given two arrays A[] and B[] of N and M integers respectively. Also given is a N X M binary matrix where 1 indicates that there was a positive integer in the original matrix and 0 indicates that the position is filled with 0 in the original matrix. The task is to form back the original matrix such that A[i] indicates the largest element in the ith row and B[j] indicates the largest element in the jth column.
Examples:
Input: A[] = {2, 1, 3}, B[] = {2, 3, 0, 0, 2, 0, 1},
matrix[] = {{1, 0, 0, 0, 1, 0, 0},
{0, 0, 0, 0, 0, 0, 1},
{1, 1, 0, 0, 0, 0, 0}}
Output:
2 0 0 0 2 0 0
0 0 0 0 0 0 1
2 3 0 0 0 0 0
Input: A[] = {2, 4}, B[] = {4, 2},
matrix[] = {{1, 1},
{1, 1}}
Output:
2 2
4 2
Approach: Iterate for every index (i, j) in the matrix, and if mat[i][j] == 1, then fill the position with min(A[i], B[j]). This is because the current element is part of the ith row and the jth column and if the max(A[i], B[j]) was chosen then one of conditions couldn’t be fulfilled i.e. the chosen element could exceed either the maximum element required in the current row or the current column.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define N 3#define M 7// Function that prints the original matrixvoid printOriginalMatrix(int a[], int b[], int mat[N][M]){ // Iterate in the row for (int i = 0; i < N; i++) { // Iterate in the column for (int j = 0; j < M; j++) { // If previously existed an element if (mat[i][j] == 1) cout << min(a[i], b[j]) << " "; else cout << 0 << " "; } cout << endl; }}// Driver codeint main(){ int a[] = { 2, 1, 3 }; int b[] = { 2, 3, 0, 0, 2, 0, 1 }; int mat[N][M] = { { 1, 0, 0, 0, 1, 0, 0 }, { 0, 0, 0, 0, 0, 0, 1 }, { 1, 1, 0, 0, 0, 0, 0 } }; printOriginalMatrix(a, b, mat); return 0;} |
Java
// Java implementation of the approachclass GFG{ static int N = 3;static int M = 7;// Function that prints the original matrixstatic void printOriginalMatrix(int a[], int b[], int[][] mat){ // Iterate in the row for (int i = 0; i < N; i++) { // Iterate in the column for (int j = 0; j < M; j++) { // If previously existed an element if (mat[i][j] == 1) System.out.print(Math.min(a[i], b[j]) + " "); else System.out.print("0" + " "); } System.out.println(); }}// Driver codepublic static void main(String[] args){ int a[] = { 2, 1, 3 }; int b[] = { 2, 3, 0, 0, 2, 0, 1 }; int[][] mat = {{ 1, 0, 0, 0, 1, 0, 0 }, { 0, 0, 0, 0, 0, 0, 1 }, { 1, 1, 0, 0, 0, 0, 0 }}; printOriginalMatrix(a, b, mat);}}// This code is contributed by Code_Mech |
Python3
# Python3 implementation of the approachN = 3M = 7# Function that prints the original matrix def printOriginalMatrix(a, b, mat) : # Iterate in the row for i in range(N) : # Iterate in the column for j in range(M) : # If previously existed an element if (mat[i][j] == 1) : print(min(a[i], b[j]), end = " "); else : print(0, end = " "); print()# Driver code if __name__ == "__main__" : a = [ 2, 1, 3 ] b = [ 2, 3, 0, 0, 2, 0, 1 ] mat = [[ 1, 0, 0, 0, 1, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 1 ], [ 1, 1, 0, 0, 0, 0, 0 ]]; printOriginalMatrix(a, b, mat); # This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;class GFG{ static int N = 3;static int M = 7;// Function that prints the original matrixstatic void printOriginalMatrix(int[] a, int[] b, int[,] mat){ // Iterate in the row for (int i = 0; i < N; i++) { // Iterate in the column for (int j = 0; j < M; j++) { // If previously existed an element if (mat[i,j] == 1) Console.Write(Math.Min(a[i], b[j]) + " "); else Console.Write("0" + " "); } Console.WriteLine(); }}// Driver codepublic static void Main(){ int[] a = { 2, 1, 3 }; int[] b = { 2, 3, 0, 0, 2, 0, 1 }; int[,] mat = {{ 1, 0, 0, 0, 1, 0, 0 }, { 0, 0, 0, 0, 0, 0, 1 }, { 1, 1, 0, 0, 0, 0, 0 }}; printOriginalMatrix(a, b, mat);}}// This code is contributed by Code_Mech |
PHP
<?php // PHP implementation of the approach$N = 3;$M = 7;// Function that prints the original matrixfunction printOriginalMatrix($a, $b, $mat){ // Iterate in the row for ($i = 0; $i < $GLOBALS['N']; $i++) { // Iterate in the column for ($j = 0; $j < $GLOBALS['M']; $j++) { // If previously existed an element if ($mat[$i][$j] == 1) echo min($a[$i], $b[$j])." "; else echo "0"." "; } echo "\r\n"; } }// Driver code$a = array( 2, 1, 3 );$b = array(2, 3, 0, 0, 2, 0, 1 );$mat = array( array( 1, 0, 0, 0, 1, 0, 0 ), array( 0, 0, 0, 0, 0, 0, 1 ), array( 1, 1, 0, 0, 0, 0, 0 ));printOriginalMatrix($a, $b, $mat);// This code is contributed by Shashank_Sharma?> |
Javascript
<script>// Javascript implementation of the approach let N = 3;let M = 7;// Function that prints the original matrixfunction printOriginalMatrix(a,b,mat){ // Iterate in the row for (let i = 0; i < N; i++) { // Iterate in the column for (let j = 0; j < M; j++) { // If previously existed an element if (mat[i][j] == 1) document.write(Math.min(a[i], b[j]) + " "); else document.write("0" + " "); } document.write("<br>"); }}// Driver code let a = [ 2, 1, 3 ]; let b = [ 2, 3, 0, 0, 2, 0, 1 ]; let mat = [[ 1, 0, 0, 0, 1, 0, 0 ], [ 0, 0, 0, 0, 0, 0, 1 ], [ 1, 1, 0, 0, 0, 0, 0 ]]; printOriginalMatrix(a, b, mat); // This code is contributed Bobby</script> |
Output:
2 0 0 0 2 0 0 0 0 0 0 0 0 1 2 3 0 0 0 0 0
Time Complexity: O(N * M)
Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
