Open In App

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

Improve
Improve
Like Article
Like
Save
Share
Report

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 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;
}


Java




// 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


Python3




# 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


C#




// 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


PHP




<?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
?>


Javascript




<script>
 
// Javascript implementation of the approach
 
     
let N = 3;
let M = 7;
 
// Function that prints the original matrix
function 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)



Last Updated : 08 Jun, 2022
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads