Row with max 1s Java

PROGRAM TO FIND THE ROW WITH MAXIMUM NUMBER OF 1s

import java.io.*;

  

class GFG {

    static int R = 4, C = 4;

    // Function to find the index of first index

    // of 1 in a boolean array arr[] 

    static int first(int arr[], int low, int high)

    {

        if (high >= low) {

            // Get the middle index

            int mid = low + (high - low) / 2;

  

            // Check if the element at middle index is first 1

            if ((mid == 0 || (arr[mid - 1] == 0)) && arr[mid] == 1)

                return mid;

  

            // If the element is 0, recur for right side

            else if (arr[mid] == 0)

                return first(arr, (mid + 1), high);

                  

            // If element is not first 1, recur for left side

            else 

                return first(arr, low, (mid - 1));

        }

        return -1;

    }

  

    // Function that returns index of row

    // with maximum number of 1s.

    static int rowWithMax1s(int mat[][])

    {

        // Initialize max values

        int max_row_index = 0, max = -1; 

  

        // Traverse for each row and count number of 

        // 1s by finding the index of first 1

        int i, index;

        for (i = 0; i < R; i++) {

            index = first(mat[i], 0, C - 1);

            if (index != -1 && C - index > max) {

                max = C - index;

                max_row_index = i;

            }

        }

  

        return max_row_index;

    }

    // Driver Code

    public static void main(String[] args)

    {

        int mat[][] = { { 0, 0, 0, 1 },

                        { 0, 1, 1, 1 },

                        { 1, 1, 1, 1 },

                        { 0, 0, 0, 0 } };

        System.out.println("Index of row with maximum 1s is "

                                            + rowWithMax1s(mat));

    }

}







OUTPUT

Index of row with maximum 1s is 2

CREDITS: https://www.geeksforgeeks.org/find-the-row-with-maximum-number-1s/