# Count 1s in binary matrix having remaining indices of its row and column filled with 0s

• Last Updated : 31 May, 2022

Given a binary matrix, mat[][] of size M * N, the task is to count the number of 1s from the given binary matrix whose corresponding row and column consists of 0s only in the remaining indices.

Examples:

Input: mat[][] = {{1, 0, 0}, {0, 0, 1}, {0, 0, 0}}
Output:
Explanation:
The only two cells satisfying the conditions are (0, 0) and (1, 2).
Therefore, the count is 2.

Input: mat[][] = {{1, 0}, {1, 1}}
Output:

Naive Approach: The simplest approach is to iterate over the matrix and check the given condition for all 1s present in the given matrix by traversing its corresponding row and column. Increase count of all 1s satisfying the condition. Finally, print the count as the required answer.

Time Complexity: O(M*N2
Auxiliary Space: O(M + N)

Efficient Approach: The above approach can be optimized based on the idea that the sum of such rows and columns will be only 1. Follow the steps below to solve the problem:

1. Initialize two arrays, rows[] and cols[], to store the sum of each row and each column of the matrix respectively.
2. Initialize a variable, say cnt, to store the count of 1s satisfying given condition.
3. Traverse the matrix for every mat[i][j] = 1, check if rows[i] and cols[j] is 1.If found to be true, then increment cnt.
4. After completing the above steps, print the final value of count.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to count required 1s``// from the given matrix``int` `numSpecial(vector >& mat)``{` `    ``// Stores the dimensions of the mat[][]``    ``int` `m = mat.size(), n = mat[0].size();` `    ``int` `rows[m];``    ``int` `cols[n];` `    ``int` `i, j;` `    ``// Calculate sum of rows``    ``for` `(i = 0; i < m; i++) {``        ``rows[i] = 0;``        ``for` `(j = 0; j < n; j++)``            ``rows[i] += mat[i][j];``    ``}` `    ``// Calculate sum of columns``    ``for` `(i = 0; i < n; i++) {` `        ``cols[i] = 0;``        ``for` `(j = 0; j < m; j++)``            ``cols[i] += mat[j][i];``    ``}` `    ``// Stores required count of 1s``    ``int` `cnt = 0;``    ``for` `(i = 0; i < m; i++) {``        ``for` `(j = 0; j < n; j++) {` `            ``// If current cell is 1``            ``// and sum of row and column is 1``            ``if` `(mat[i][j] == 1 && rows[i] == 1``                ``&& cols[j] == 1)` `                ``// Increment count of 1s``                ``cnt++;``        ``}``    ``}` `    ``// Return the final count``    ``return` `cnt;``}` `// Driver Code``int` `main()``{``    ``// Given matrix``    ``vector > mat``        ``= { { 1, 0, 0 }, { 0, 0, 1 }, { 0, 0, 0 } };` `    ``// Function Call``    ``cout << numSpecial(mat) << endl;` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``class` `GFG{``    ` `// Function to count required 1s``// from the given matrix``static` `int` `numSpecial(``int` `[][]mat)``{``    ` `    ``// Stores the dimensions of the mat[][]``    ``int` `m = mat.length;``    ``int` `n = mat[``0``].length;`` ` `    ``int` `[]rows = ``new` `int``[m];``    ``int` `[]cols = ``new` `int``[n];`` ` `    ``int` `i, j;`` ` `    ``// Calculate sum of rows``    ``for``(i = ``0``; i < m; i++)``    ``{``        ``rows[i] = ``0``;``        ` `        ``for``(j = ``0``; j < n; j++)``            ``rows[i] += mat[i][j];``    ``}`` ` `    ``// Calculate sum of columns``    ``for``(i = ``0``; i < n; i++)``    ``{``        ``cols[i] = ``0``;``        ` `        ``for``(j = ``0``; j < m; j++)``            ``cols[i] += mat[j][i];``    ``}`` ` `    ``// Stores required count of 1s``    ``int` `cnt = ``0``;``    ` `    ``for``(i = ``0``; i < m; i++)``    ``{``        ``for``(j = ``0``; j < n; j++)``        ``{``            ` `            ``// If current cell is 1``            ``// and sum of row and column is 1``            ``if` `(mat[i][j] == ``1` `&&``                  ``rows[i] == ``1` `&&``                  ``cols[j] == ``1``)`` ` `                ``// Increment count of 1s``                ``cnt++;``        ``}``    ``}``    ` `    ``// Return the final count``    ``return` `cnt;``}`` ` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ` `    ``// Given matrix``    ``int` `[][]mat = { { ``1``, ``0``, ``0` `},``                    ``{ ``0``, ``0``, ``1` `},``                    ``{ ``0``, ``0``, ``0` `} };`` ` `    ``// Function call``    ``System.out.print(numSpecial(mat) + ``"\n"``);``}``}`` ` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program for the above approach` `# Function to count required 1s``# from the given matrix``def` `numSpecial(mat):``    ` `    ``# Stores the dimensions``    ``# of the mat``    ``m ``=` `len``(mat)``    ``n ``=` `len``(mat[``0``])` `    ``rows ``=` `[``0``] ``*` `m``    ``cols ``=` `[``0``] ``*` `n` `    ``i, j ``=` `0``, ``0` `    ``# Calculate sum of rows``    ``for` `i ``in` `range``(m):``        ``rows[i] ``=` `0` `        ``for` `j ``in` `range``(n):``            ``rows[i] ``+``=` `mat[i][j]` `    ``# Calculate sum of columns``    ``for` `i ``in` `range``(n):``        ``cols[i] ``=` `0` `        ``for` `j ``in` `range``(m):``            ``cols[i] ``+``=` `mat[j][i]` `    ``# Stores required count of 1s``    ``cnt ``=` `0` `    ``for` `i ``in` `range``(m):``        ``for` `j ``in` `range``(n):` `            ``# If current cell is 1``            ``# and sum of row and column is 1``            ``if` `(mat[i][j] ``=``=` `1` `and``                  ``rows[i] ``=``=` `1` `and``                  ``cols[j] ``=``=` `1``):` `                ``# Increment count of 1s``                ``cnt ``+``=` `1` `    ``# Return the final count``    ``return` `cnt` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given matrix``    ``mat ``=` `[ [ ``1``, ``0``, ``0` `],``            ``[ ``0``, ``0``, ``1` `],``            ``[ ``0``, ``0``, ``0` `] ]` `    ``# Function call``    ``print``(numSpecial(mat))` `# This code is contributed by Amit Katiyar`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{``    ` `// Function to count required 1s``// from the given matrix``static` `int` `numSpecial(``int` `[,]mat)``{``    ` `    ``// Stores the dimensions of the [,]mat``    ``int` `m = mat.GetLength(0);``    ``int` `n = mat.GetLength(1);`` ` `    ``int` `[]rows = ``new` `int``[m];``    ``int` `[]cols = ``new` `int``[n];`` ` `    ``int` `i, j;`` ` `    ``// Calculate sum of rows``    ``for``(i = 0; i < m; i++)``    ``{``        ``rows[i] = 0;``        ` `        ``for``(j = 0; j < n; j++)``            ``rows[i] += mat[i, j];``    ``}`` ` `    ``// Calculate sum of columns``    ``for``(i = 0; i < n; i++)``    ``{``        ``cols[i] = 0;``        ` `        ``for``(j = 0; j < m; j++)``            ``cols[i] += mat[j, i];``    ``}`` ` `    ``// Stores required count of 1s``    ``int` `cnt = 0;``    ` `    ``for``(i = 0; i < m; i++)``    ``{``        ``for``(j = 0; j < n; j++)``        ``{``            ` `            ``// If current cell is 1 and``            ``// sum of row and column is 1``            ``if` `(mat[i, j] == 1 &&``                  ``rows[i] == 1 &&``                  ``cols[j] == 1)`` ` `                ``// Increment count of 1s``                ``cnt++;``        ``}``    ``}``    ` `    ``// Return the readonly count``    ``return` `cnt;``}`` ` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ` `    ``// Given matrix``    ``int` `[,]mat = { { 1, 0, 0 },``                   ``{ 0, 0, 1 },``                   ``{ 0, 0, 0 } };`` ` `    ``// Function call``    ``Console.Write(numSpecial(mat) + ``"\n"``);``}``}`` ` `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output:

`2`

Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.

Auxiliary Space: O(N+M), as we are using extra space for two arrays row and col.

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