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 **i ^{th}** row and

**B[j]**indicates the largest element in the

**j**column.

^{th}**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 0Input: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 **i ^{th}** row and the

**j**column and if the

^{th}**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

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