Find the original matrix when largest element in a row and a column are given

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 3
#define M 7
  
// Function that prints the original matrix
void 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 code
int 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;
}

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Java

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// Java implementation of the approach
class GFG
{
      
static int N = 3;
static int M = 7;
  
// Function that prints the original matrix
static 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 code
public 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

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Python3

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# Python3 implementation of the approach
N = 3
M = 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

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
      
static int N = 3;
static int M = 7;
  
// Function that prints the original matrix
static 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 code
public 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

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PHP

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<?php 
// PHP implementation of the approach
  
$N = 3;
$M = 7;
  
  
// Function that prints the original matrix
function 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
?>

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Output:

2 0 0 0 2 0 0 
0 0 0 0 0 0 1 
2 3 0 0 0 0 0


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Striver(underscore)79 at Codechef and codeforces D

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