# Count of columns with odd number of 1s

Given an N * M 2D binary matrix, the task is to find the count of columns having odd number of 1s.

Examples:

Input: mat[][] = {
{0, 0, 1, 0},
{1, 0, 0, 1},
{1, 1, 1, 0}}
Output: 2
Column 2 and 4 are the only columns
having odd number of 1’s.

Input: mat[][] = {
{1, 1, 0, 0, 1, 1},
{0, 1, 0, 1, 0, 0},
{1, 1, 1, 0, 1, 0}}
Output: 4

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Find the sum of all the columns of the matrix separately, the columns having an odd sum are the columns which have an odd number of 1s.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `const` `int` `col = 4; ` `const` `int` `row = 3; ` ` `  `// Function to return the count of ` `// columns having odd number of 1s ` `int` `countOddColumn(``int` `arr[row][col]) ` `{ ` ` `  `    ``// To store the sum of every column ` `    ``int` `sum[col] = { 0 }; ` ` `  `    ``// For every column ` `    ``for` `(``int` `i = 0; i < col; i++) { ` ` `  `        ``// Sum of all the element ` `        ``// of the current column ` `        ``for` `(``int` `j = 0; j < row; j++) { ` `            ``sum[i] += arr[j][i]; ` `        ``} ` `    ``} ` ` `  `    ``// To store the required count ` `    ``int` `count = 0; ` ` `  `    ``for` `(``int` `i = 0; i < col; i++) { ` ` `  `        ``// If the sum of the current ` `        ``// column is odd ` `        ``if` `(sum[i] % 2 == 1) { ` `            ``count++; ` `        ``} ` `    ``} ` ` `  `    ``return` `count; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[row][col] = { { 0, 0, 1, 0 }, ` `                          ``{ 1, 0, 0, 1 }, ` `                          ``{ 1, 1, 1, 0 } }; ` ` `  `    ``cout << countOddColumn((arr)); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG ` `{ ` `static` `int` `col = ``4``; ` `static` `int` `row = ``3``; ` ` `  `// Function to return the count of ` `// columns having odd number of 1s ` `static` `int` `countOddColumn(``int` `arr[][]) ` `{ ` ` `  `    ``// To store the sum of every column ` `    ``int` `[]sum = ``new` `int``[col]; ` ` `  `    ``// For every column ` `    ``for` `(``int` `i = ``0``; i < col; i++) ` `    ``{ ` ` `  `        ``// Sum of all the element ` `        ``// of the current column ` `        ``for` `(``int` `j = ``0``; j < row; j++)  ` `        ``{ ` `            ``sum[i] += arr[j][i]; ` `        ``} ` `    ``} ` ` `  `    ``// To store the required count ` `    ``int` `count = ``0``; ` ` `  `    ``for` `(``int` `i = ``0``; i < col; i++) ` `    ``{ ` ` `  `        ``// If the sum of the current ` `        ``// column is odd ` `        ``if` `(sum[i] % ``2` `== ``1``)  ` `        ``{ ` `            ``count++; ` `        ``} ` `    ``} ` `    ``return` `count; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String []args)  ` `{ ` `    ``int` `arr[][] = {{ ``0``, ``0``, ``1``, ``0` `}, ` `                   ``{ ``1``, ``0``, ``0``, ``1` `}, ` `                   ``{ ``1``, ``1``, ``1``, ``0` `}}; ` ` `  `    ``System.out.println(countOddColumn((arr))); ` `} ` `} ` ` `  `// This code is contributed by Rajput-Ji `

## Python3

 `# Python3 implementation of the approach ` `col ``=` `4` `row ``=` `3` ` `  `# Function to return the count of ` `# columns having odd number of 1s ` `def` `countOddColumn(arr): ` ` `  `    ``# To store the sum of every column ` `    ``sum` `=` `[``0` `for` `i ``in` `range``(col)] ` ` `  `    ``# For every column ` `    ``for` `i ``in` `range``(col): ` ` `  `        ``# Sum of all the element ` `        ``# of the current column ` `        ``for` `j ``in` `range``(row): ` `            ``sum``[i] ``+``=` `arr[j][i] ` ` `  `    ``# To store the required count ` `    ``count ``=` `0` ` `  `    ``for` `i ``in` `range``(col): ` ` `  `        ``# If the sum of the current ` `        ``# column is odd ` `        ``if` `(``sum``[i] ``%` `2` `=``=` `1``): ` `            ``count ``+``=` `1` ` `  `    ``return` `count ` ` `  `# Driver code ` `arr ``=` `[[``0``, ``0``, ``1``, ``0``], ` `       ``[``1``, ``0``, ``0``, ``1``], ` `       ``[``1``, ``1``, ``1``, ``0``]] ` ` `  `print``(countOddColumn((arr))) ` ` `  `# This code is contributed by Mohit Kumar `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG  ` `{  ` `    ``static` `int` `col = 4;  ` `    ``static` `int` `row = 3;  ` `     `  `    ``// Function to return the count of  ` `    ``// columns having odd number of 1s  ` `    ``static` `int` `countOddColumn(``int` `[,]arr)  ` `    ``{  ` `     `  `        ``// To store the sum of every column  ` `        ``int` `[]sum = ``new` `int``[col];  ` `     `  `        ``// For every column  ` `        ``for` `(``int` `i = 0; i < col; i++)  ` `        ``{  ` `     `  `            ``// Sum of all the element  ` `            ``// of the current column  ` `            ``for` `(``int` `j = 0; j < row; j++)  ` `            ``{  ` `                ``sum[i] += arr[j, i];  ` `            ``}  ` `        ``}  ` `     `  `        ``// To store the required count  ` `        ``int` `count = 0;  ` `     `  `        ``for` `(``int` `i = 0; i < col; i++)  ` `        ``{  ` `     `  `            ``// If the sum of the current  ` `            ``// column is odd  ` `            ``if` `(sum[i] % 2 == 1)  ` `            ``{  ` `                ``count++;  ` `            ``}  ` `        ``}  ` `        ``return` `count;  ` `    ``}  ` `     `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` `        ``int` `[,]arr = {{ 0, 0, 1, 0 },  ` `                      ``{ 1, 0, 0, 1 },  ` `                      ``{ 1, 1, 1, 0 }};  ` `     `  `        ``Console.WriteLine(countOddColumn((arr)));  ` `    ``}  ` `} ` ` `  `// This code is contributed by kanugargng `

Output:

```2
```

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