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Find the original matrix when largest element in a row and a column are given
• Last Updated : 27 Apr, 2021

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 ``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

 ``

## Javascript

 ``
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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