# Find if a binary matrix exists with given row and column sums

• Difficulty Level : Medium
• Last Updated : 28 Jul, 2021

Given an array Row[] of size R where Row[i] is the sum of elements of the ith row and another array Column[] of size C where Column[i] is the sum of elements of the ith column. The task is to check if it is possible to construct a binary matrix of R * C dimension which satisfies given row sums and column sums. A binary matrix is a matrix which is filled with only 0’s and 1’s.
Sum means the number of 1’s in particular row or column.
Examples:

Input: Row[] = {2, 2, 2, 2, 2}, Column[] = {5, 5, 0, 0}
Output: YES
Matrix is
{1, 1, 0, 0}
{1, 1, 0, 0}
{1, 1, 0, 0}
{1, 1, 0, 0}
{1, 1, 0, 0}
Input: Row[] = {0, 0, 3} Column[] = {3, 0, 0}
Output: NO

Approach:

1. Key idea is that any cell in the matrix will contribute equally to both row and column sum, so sum of all the row sums must be equal to column sums.
2. Now, find the maximum of row sums, if this value is greater than the number of non zero column sums than matrix does not exist.
3. If the maximum of column sums is greater than the number of non zero row sums than matrix is not possible to construct.
4. If all the above 3 conditions is satisfied than matrix exists.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach``#include ``using` `namespace` `std;` `// Function to check if matrix exists``bool` `matrix_exist(``int` `row[], ``int` `column[], ``int` `r, ``int` `c)``{``    ``int` `row_sum = 0;``    ``int` `column_sum = 0;``    ``int` `row_max = -1;``    ``int` `column_max = -1;``    ``int` `row_non_zero = 0;``    ``int` `column_non_zero = 0;` `    ``// Store sum of rowsums, max of row sum``    ``// number of non zero row sums``    ``for` `(``int` `i = 0; i < r; i++) {``        ``row_sum += row[i];``        ``row_max = max(row_max, row[i]);``        ``if` `(row[i])``            ``row_non_zero++;``    ``}` `    ``// Store sum of column sums, max of column sum``    ``// number of non zero column sums``    ``for` `(``int` `i = 0; i < c; i++) {``        ``column_sum += column[i];``        ``column_max = max(column_max, column[i]);``        ``if` `(column[i])``            ``column_non_zero++;``    ``}` `    ``// Check condition 1, 2, 3``    ``if` `((row_sum != column_sum) ||``        ``(row_max > column_non_zero) ||``        ``(column_max > row_non_zero))``        ``return` `false``;` `    ``return` `true``;``}` `// Driver Code``int` `main()``{``    ``int` `row[] = { 2, 2, 2, 2, 2 };``    ``int` `column[] = { 5, 5, 0, 0 };``    ``int` `r = ``sizeof``(row) / ``sizeof``(row[0]);``    ``int` `c = ``sizeof``(column) / ``sizeof``(column[0]);` `    ``if` `(matrix_exist(row, column, r, c))``        ``cout << ``"YES\n"``;``    ``else``        ``cout << ``"NO\n"``;``}`

## Java

 `// Java implementation of above approach``import` `java.util.*;` `class` `GFG``{` `    ``// Function to check if matrix exists``    ``static` `boolean` `matrix_exist(``int` `row[], ``int` `column[],``                                        ``int` `r, ``int` `c)``    ``{``        ``int` `row_sum = ``0``;``        ``int` `column_sum = ``0``;``        ``int` `row_max = -``1``;``        ``int` `column_max = -``1``;``        ``int` `row_non_zero = ``0``;``        ``int` `column_non_zero = ``0``;` `        ``// Store sum of rowsums, max of row sum``        ``// number of non zero row sums``        ``for` `(``int` `i = ``0``; i < r; i++)``        ``{``            ``row_sum += row[i];``            ``row_max = Math.max(row_max, row[i]);``            ``if` `(row[i] > ``0``)``            ``{``                ``row_non_zero++;``            ``}``        ``}` `        ``// Store sum of column sums, max of column sum``        ``// number of non zero column sums``        ``for` `(``int` `i = ``0``; i < c; i++)``        ``{``            ``column_sum += column[i];``            ``column_max = Math.max(column_max, column[i]);``            ``if` `(column[i] > ``0``)``            ``{``                ``column_non_zero++;``            ``}``        ``}` `        ``// Check condition 1, 2, 3``        ``if` `((row_sum != column_sum)``                ``|| (row_max > column_non_zero)``                ``|| (column_max > row_non_zero))``        ``{``            ``return` `false``;``        ``}` `        ``return` `true``;``    ``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `row[] = { ``2``, ``2``, ``2``, ``2``, ``2` `};``    ``int` `column[] = { ``5``, ``5``, ``0``, ``0` `};``    ``int` `r = row.length;``    ``int` `c = column.length;` `    ``if` `(matrix_exist(row, column, r, c))``        ``System.out.println(``"Yes"``);``    ``else``        ``System.out.println(``"No"``);``}``}` `// This code has been contributed by 29AjayKumar`

## Python3

 `# Python implementation of the above approach` `# Function to check if matrix exists``def` `matrix_exist(row,  column,  r, c):` `    ``row_sum ``=` `0``    ``column_sum ``=` `0``    ``row_max ``=` `-``1``    ``column_max ``=` `-``1``    ``row_non_zero ``=` `0``    ``column_non_zero ``=` `0` `    ``# Store sum of rowsums, max of row sum``    ``# number of non zero row sums``    ``for` `i ``in` `range``(r):``        ``row_sum ``+``=` `row[i]``        ``row_max ``=` `max``(row_max, row[i])``        ``if` `(row[i]):``            ``row_non_zero ``=` `row_non_zero ``+` `1` `    ``# Store sum of column sums, max of column sum``    ``# number of non zero column sums``    ``for` `i ``in` `range``(c):``        ``column_sum ``=` `column_sum ``+` `column[i]``        ``column_max ``=` `max``(column_max, column[i])``        ``if` `(column[i]):``            ``column_non_zero ``=` `column_non_zero ``+` `1` `    ``# Check condition 1, 2, 3``    ``if` `((row_sum !``=` `column_sum)``        ``or` `(row_max > column_non_zero)``        ``or` `(column_max > row_non_zero)):``        ` `        ``return` `False` `    ``return` `True` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``row ``=` `[``2``, ``2``, ``2``, ``2``, ``2``]``    ``column ``=` `[``5``, ``5``, ``0``, ``0``]``    ``r ``=` `len``(row)``    ``c ``=` `len``(column)``    ``if` `matrix_exist(row, column, r, c):``        ``print``(``"YES"``)` `    ``else``:``        ``print``(``"NO"``)` `# this code is contributed by nirajgusain5`

## C#

 `// C# implementation of above approach``using` `System;` `public` `class` `GFG{` `    ``// Function to check if matrix exists``    ``static` `bool` `matrix_exist(``int``[] row, ``int``[] column,``                                        ``int` `r, ``int` `c)``    ``{``        ``int` `row_sum = 0;``        ``int` `column_sum = 0;``        ``int` `row_max = -1;``        ``int` `column_max = -1;``        ``int` `row_non_zero = 0;``        ``int` `column_non_zero = 0;` `        ``// Store sum of rowsums, max of row sum``        ``// number of non zero row sums``        ``for` `(``int` `i = 0; i < r; i++)``        ``{``            ``row_sum += row[i];``            ``row_max = Math.Max(row_max, row[i]);``            ``if` `(row[i] > 0)``            ``{``                ``row_non_zero++;``            ``}``        ``}` `        ``// Store sum of column sums, max of column sum``        ``// number of non zero column sums``        ``for` `(``int` `i = 0; i < c; i++)``        ``{``            ``column_sum += column[i];``            ``column_max = Math.Max(column_max, column[i]);``            ``if` `(column[i] > 0)``            ``{``                ``column_non_zero++;``            ``}``        ``}` `        ``// Check condition 1, 2, 3``        ``if` `((row_sum != column_sum)``                ``|| (row_max > column_non_zero)``                ``|| (column_max > row_non_zero))``        ``{``            ``return` `false``;``        ``}` `        ``return` `true``;``    ``}` `    ``// Driver Code``    ``static` `public` `void` `Main ()``    ``{``        ``int``[] row = { 2, 2, 2, 2, 2 };``        ``int``[] column = { 5, 5, 0, 0 };``        ``int` `r = row.Length;``        ``int` `c = column.Length;``    ` `        ``if` `(matrix_exist(row, column, r, c))``            ``Console.Write(``"YES"``);``        ``else``            ``Console.Write(``"NO"``);``    ``}``}` `// This code has been contributed by shubhamsingh10`

## Javascript

 ``
Output:
`YES`

Time Complexity : O(N)

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