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# Counting sets of 1s and 0s in a binary matrix

• Difficulty Level : Easy
• Last Updated : 14 Dec, 2022

Given n × m binary matrix, count the number of sets where a set can be formed one or more same values in a row or column.

Examples:

```Input: 1 0 1
0 1 0
Output: 8
Explanation: There are six one-element sets
(three 1s and three 0s). There are two two-
element sets, the first one consists of the
first and the third cells of the first row.
The second one consists of the first and the
third cells of the second row.

Input: 1 0
1 1
Output: 6```

The number of non-empty subsets of x elements is 2x – 1. We traverse every row and calculate numbers of 1’s and 0’s cells. For every u zeros and v ones, total sets is 2u – 1 + 2v – 1. We then traverse all columns and compute same values and compute overall sum. We finally subtract m x n from the overall sum as single elements are considered twice.

Implementation:

## CPP

 `// CPP program to compute number of sets` `// in a binary matrix.` `#include ` `using` `namespace` `std;`   `const` `int` `m = 3; ``// no of columns` `const` `int` `n = 2; ``// no of rows`   `// function to calculate the number of` `// non empty sets of cell` `long` `long` `countSets(``int` `a[n][m])` `{` `    ``// stores the final answer` `    ``long` `res = 0;`   `    ``// traverses row-wise` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``int` `u = 0, v = 0;` `        ``for` `(``int` `j = 0; j < m; j++) {` `            ``if` `(a[i][j] == 1)` `                ``u++;` `            ``else` `                ``v++;` `        ``}` `        ``res += ``pow``(2, u) - 1 + ``pow``(2, v) - 1;` `    ``}`   `    ``// traverses column wise` `    ``for` `(``int` `i = 0; i < m; i++) {` `        ``int` `u = 0, v = 0;` `        ``for` `(``int` `j = 0; j < n; j++) {` `            ``if` `(a[j][i] == 1)` `                ``u++;` `            ``else` `                ``v++;` `        ``}` `        ``res += ``pow``(2, u) - 1 + ``pow``(2, v) - 1;` `    ``}`   `    ``// at the end subtract n*m as no of` `    ``// single sets have been added twice.` `    ``return` `res - (n * m);` `}`   `// driver program to test the above function.` `int` `main()` `{`   `    ``int` `a[] = { { 1, 0, 1 }, { 0, 1, 0 } };`   `    ``cout << countSets(a);`   `    ``return` `0;` `}`

## Java

 `// Java program to compute number of sets` `// in a binary matrix.` `import` `java.util.*;` `class` `GFG {` `static` `final` `int` `m = ``3``; ``// no of columns` `static` `final` `int` `n = ``2``; ``// no of rows`   `// function to calculate the number of` `// non empty sets of cell` `static` `long` `countSets(``int` `a[][]) {`   `    ``// stores the final answer` `    ``long` `res = ``0``;`   `    ``// traverses row-wise` `    ``for` `(``int` `i = ``0``; i < n; i++) {` `    ``int` `u = ``0``, v = ``0``;` `    ``for` `(``int` `j = ``0``; j < m; j++) {` `        ``if` `(a[i][j] == ``1``)` `        ``u++;` `        ``else` `        ``v++;` `    ``}` `    ``res += Math.pow(``2``, u) - ``1` `+ Math.pow(``2``, v) - ``1``;` `    ``}`   `    ``// traverses column wise` `    ``for` `(``int` `i = ``0``; i < m; i++) {` `    ``int` `u = ``0``, v = ``0``;` `    ``for` `(``int` `j = ``0``; j < n; j++) {` `        ``if` `(a[j][i] == ``1``)` `        ``u++;` `        ``else` `        ``v++;` `    ``}` `    ``res += Math.pow(``2``, u) - ``1` `+ Math.pow(``2``, v) - ``1``;` `    ``}`   `    ``// at the end subtract n*m as no of` `    ``// single sets have been added twice.` `    ``return` `res - (n * m);` `}`   `// Driver code` `public` `static` `void` `main(String[] args) {` `    ``int` `a[][] = {{``1``, ``0``, ``1``}, {``0``, ``1``, ``0``}};`   `    ``System.out.print(countSets(a));` `}` `}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python3 program to compute number of sets ` `# in a binary matrix. ` `m ``=` `3` `# no of columns ` `n ``=` `2` `# no of rows `   `# function to calculate the number of ` `# non empty sets of cell ` `def` `countSets(a):` `    `  `    ``# stores the final answer ` `    ``res ``=` `0` `    `  `    ``# traverses row-wise ` `    ``for` `i ``in` `range``(n):` `        ``u ``=` `0` `        ``v ``=` `0` `        ``for` `j ``in` `range``(m):` `            ``if` `a[i][j]:` `                ``u ``+``=` `1` `            ``else``: ` `                ``v ``+``=` `1` `        ``res ``+``=` `pow``(``2``, u) ``-` `1` `+` `pow``(``2``, v) ``-` `1` `    `  `    ``# traverses column wise ` `    ``for` `i ``in` `range``(m):` `        `  `        ``u ``=` `0` `        ``v ``=` `0` `        ``for` `j ``in` `range``(n):` `            ``if` `a[j][i]:` `                ``u ``+``=` `1` `            ``else``: ` `                ``v ``+``=` `1` `        ``res ``+``=` `pow``(``2``, u) ``-` `1` `+` `pow``(``2``, v) ``-` `1` `    `  `    ``# at the end subtract n*m as no of ` `    ``# single sets have been added twice. ` `    ``return` `res ``-` `(n``*``m) `   `# Driver program to test the above function. ` `a ``=` `[[``1``, ``0``, ``1``],[``0``, ``1``, ``0``]] `   `print``(countSets(a))`   `# This code is contributed by shubhamsingh10`

## C#

 `// C# program to compute number of` `// sets in a binary matrix.` `using` `System;`   `class` `GFG {` `    `  `    ``static` `int` `m = 3; ``// no of columns` `    ``static` `int` `n = 2; ``// no of rows` `    `  `    ``// function to calculate the number of` `    ``// non empty sets of cell` `    ``static` `long` `countSets(``int` `[,]a)` `    ``{` `    `  `        ``// stores the final answer` `        ``long` `res = 0;` `    `  `        ``// Traverses row-wise` `        ``for` `(``int` `i = 0; i < n; i++)` `        ``{` `            ``int` `u = 0, v = 0;` `            `  `            ``for` `(``int` `j = 0; j < m; j++)` `            ``{` `                ``if` `(a[i,j] == 1)` `                    ``u++;` `                ``else` `                    ``v++;` `            ``}` `            ``res += (``long``)(Math.Pow(2, u) - 1` `                       ``+ Math.Pow(2, v)) - 1;` `        ``}` `    `  `        ``// Traverses column wise` `        ``for` `(``int` `i = 0; i < m; i++)` `        ``{` `            ``int` `u = 0, v = 0;` `            `  `            ``for` `(``int` `j = 0; j < n; j++)` `            ``{` `                ``if` `(a[j,i] == 1)` `                    ``u++;` `                ``else` `                    ``v++;` `            ``}` `            ``res += (``long``)(Math.Pow(2, u) - 1` `                       ``+ Math.Pow(2, v)) - 1;` `        ``}` `    `  `        ``// at the end subtract n*m as no of` `        ``// single sets have been added twice.` `        ``return` `res - (n * m);` `    ``}` `    `  `    ``// Driver code` `    ``public` `static` `void` `Main()` `    ``{` `        ``int` `[,]a = {{1, 0, 1}, {0, 1, 0}};` `    `  `        ``Console.WriteLine(countSets(a));` `    ``}` `}`   `// This code is contributed by vt_m.`

## PHP

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## Javascript

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Output

`8`

Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.
Auxiliary Space: O(1), as we are not using any extra space.

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