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
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 <iostream>using namespace std;const int col = 4;const int row = 3;// Function to return the count of// columns having odd number of 1sint 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 codeint 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 approachclass GFG{static int col = 4;static int row = 3;// Function to return the count of// columns having odd number of 1sstatic 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 codepublic 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 approachcol = 4row = 3# Function to return the count of# columns having odd number of 1sdef 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 codearr = [[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 |
Javascript
<script>// JavaScript implementation of the approachconst col = 4;const row = 3;// Function to return the count of// columns having odd number of 1sfunction countOddColumn(arr) { // To store the sum of every column let sum = new Array(col).fill(0); // For every column for (let i = 0; i < col; i++) { // Sum of all the element // of the current column for (let j = 0; j < row; j++) { sum[i] += arr[j][i]; } } // To store the required count let count = 0; for (let i = 0; i < col; i++) { // If the sum of the current // column is odd if (sum[i] % 2 == 1) { count++; } } return count;}// Driver codelet arr = [[0, 0, 1, 0],[1, 0, 0, 1],[1, 1, 1, 0]];document.write(countOddColumn((arr)))</script> |
Output:
2
Time Complexity: O(row * col).
Auxiliary Space: O(col).
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
