POTD Solutions | 09 Nov’ 23 | Predict the Column

View all POTD Solutions

Welcome to the daily solutions of our PROBLEM OF THE DAY (POTD). We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Matrix but will also help you build up problem-solving skills.

POTD Solutions 9 November 2023

C++

class Solution {
    columnWithMaxZeros(arr, N) {
        // Initialize variables to keep track of the maximum count and the corresponding column index.
        let max_count = 0;
        let ans = -1;

        // Iterate over each column.
        for (let i = 0; i < N; i++) {
            // Initialize count for the current column.
            let count = 0;

            // Iterate over each row in the current column.
            for (let j = 0; j < N; j++) {
                // Counting zeros in the current column.
                if (arr[j][i] === 0) {
                    count++;
                }
            }

            // Update max_count and ans if the current column has more zeros.
            if (count > max_count) {
                max_count = count;
                ans = i;
            }
        }

        // Return the column index with the maximum number of zeros.
        return ans;
    }
}

Java

class Solution {
    int columnWithMaxZeros(int arr[][], int N)
    {
        int idx = -1, val = 0;
        for (int j = 0; j < N; j++) {
            int sum = 0;
            for (int i = 0; i < N; i++) {
                sum += 1 - arr[i][j];
            }
            if (val < sum) {
                val = sum;
                idx = j;
            }
        }
        return idx;
    }
}

Python3

class Solution:
    def columnWithMaxZeros(self, arr, N):
        max_count = 0
        ans = -1

        for i in range(N):
            count = 0
            for j in range(N):
                # Counting zeroes
                if arr[j][i] == 0:
                    count += 1

            # Updating max_count and ans as required
            if count > max_count:
                max_count = count
                ans = i

        return ans

C#

using System;
using System.Collections.Generic;

class Solution
{
    /*
     * Function to find the column with the maximum number of zeros
     * N: Number of rows and columns in the array
     */
    public int ColumnWithMaxZeros(List<List<int>> arr, int N)
    {
        // Initialize variables to store the maximum count of zeros and the corresponding column index
        int maxCount = 0, columnIndexWithMaxZeros = -1;

        // Iterate over each column
        for (int i = 0; i < N; i++)
        {
            // Initialize count for zeros in the current column
            int count = 0;

            // Iterate over each row in the current column
            for (int j = 0; j < N; j++)
            {
                // Counting zeros in the current column
                if (arr[j][i] == 0)
                {
                    count++;
                }
            }

            // Update maxCount and columnIndexWithMaxZeros if the current column has more zeros
            if (count > maxCount)
            {
                maxCount = count;
                columnIndexWithMaxZeros = i;
            }
        }

        // Return the column index with the maximum number of zeros
        return columnIndexWithMaxZeros;
    }
}

Javascript

class Solution {
    columnWithMaxZeros(arr, N) {
        // Initialize variables to keep track of the maximum count and the corresponding column index.
        let maxCount = 0;
        let ans = -1;

        // Iterate over each column.
        for (let i = 0; i < N; i++) {
            // Initialize count for the current column.
            let count = 0;

            // Iterate over each row in the current column.
            for (let j = 0; j < N; j++) {
                // Counting zeros in the current column.
                if (arr[j][i] === 0) {
                    count++;
                }
            }

            // Update maxCount and ans if the current column has more zeros.
            if (count > maxCount) {
                maxCount = count;
                ans = i;
            }
        }

        // Return the column index with the maximum number of zeros.
        return ans;
    }
}

Time Complexity: O(N X N), where N is the number of rows or columns in the matrix
Auxiliary Space: O(1)