Row with max 1s

Given a boolean 2D array, consisting of only 1’s and 0’s, where each row is sorted. Return the 0-based index of the first row that has the most number of 1s. If no such row exists, return -1.

Examples:

Input: arr[][] = [[0, 1, 1, 1],
               [0, 0, 1, 1],
               [1, 1, 1, 1],
               [0, 0, 0, 0]]
Output: 2
Explanation: Row 2 contains 4 1's (0-based indexing).

Input: arr[][] = [[0, 0], 
               [1, 1]]
Output: 1
Explanation: Row 1 contains 2 1's (0-based indexing).

Expected Time Complexity: O(n+m)
Expected Auxiliary Space: O(1)

Constraints:
1 ≤ number of rows, number of columns ≤ 103
0 ≤ arr[i][j] ≤ 1

Row with max 1s | Practice | GeeksforGeeks

Given a boolean 2D array, consisting of only 1's and 0's, where each row is sorted. Return the 0-based index of the…

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// User function template for C++
class Solution {
  public:
    int rowWithMax1s(vector<vector<int> > &arr) {
        // code here
        int m=-1,n=0;
        for(int i=0;i<arr.size();i++){
            int sum=0;
            for(int j=0;j<arr[0].size();j++){
                if(arr[i][j]==1){
                    sum+=arr[i][j];
                }
            }
            if(sum>n){
                n=max(n,sum);
                m=i;
            }
        }
        return m;
    }
};

#User function Template for python3
class Solution:

    def rowWithMax1s(self,arr):
        # code here
        max  =0 
        count=0
        rcount=0
        for i in arr:
            if max < sum (i):
                max = sum(i)
                rcount = count
            count+=1
        if max ==0:
            return -1;
        else:
            return rcount;

//{ Driver Code Starts
// Initial Template for Java

import java.io.*;
import java.util.*;

public class Main {

    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int tc = Integer.parseInt(br.readLine().trim());
        while (tc-- > 0) {
            String[] inputLine;
            inputLine = br.readLine().trim().split(" ");
            int n = Integer.parseInt(inputLine[0]);
            int m = Integer.parseInt(inputLine[1]);
            int[][] arr = new int[n][m];
            inputLine = br.readLine().trim().split(" ");

            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    arr[i][j] = Integer.parseInt(inputLine[i * m + j]);
                }
            }
            int ans = new Solution().rowWithMax1s(arr);
            System.out.println(ans);
        }
    }
}

// } Driver Code Ends


// User function Template for Java

class Solution {
    public int rowWithMax1s(int arr[][]) {
        // code here

        int m=-1 , n=0;
        for(int i=0 ; i <arr.length;i++){
            int sum=0;
            for(int j=0;j<arr[0].length;j++){
                if(arr[i][j]==1){
                    sum+=arr[i][j];
                }
            }
            if(sum > n){
                n=Math.max(n,sum);
                m=i;
            }
        }
        return m;
    }
}