Day 79 of 100 days dsa coding challenge

Taking on a new challenge: solving GeeksforGeeks POTD daily and sharing my solutions! 💻🔥
The goal: sharpen problem-solving skills, level up coding, and learn something new every day. Follow my journey! 🚀

100DaysOfCode #CodingChallenge #ProblemSolving #GeeksforGeeks #DeveloperJourney

Problem:

https://www.geeksforgeeks.org/problems/row-with-max-1s0023/1

Row with max 1s

Difficulty: Medium Accuracy: 33.09%

You are given a 2D binary array arr[][] consisting of only 1s and 0s. Each row of the array is sorted in non-decreasing order. Your task is to find and return the index of the first row that contains the maximum number of 1s. If no such row exists, return -1.
Note:
• The array follows 0-based indexing.
• The number of rows and columns in the array are denoted by n and m respectively.

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 the most number of 1s (4 1s). Hence, the output is 2.
Input: arr[][] = [[0,0], [1,1]]
Output: 1
Explanation: Row 1 contains the most number of 1s (2 1s). Hence, the output is 1.
Input: arr[][] = [[0,0], [0,0]]
Output: -1
Explanation: No row contains any 1s, so the output is -1.
Constraints:
1 ≤ arr.size(), arr[i].size() ≤ 103
0 ≤ arr[i][j] ≤ 1

Solution:
class Solution:
def rowWithMax1s(self, arr):
n, m = len(arr), len(arr[0])
i, j = 0, m - 1
ans = -1

    while i < n and j >= 0:
        if arr[i][j] == 1:
            ans = i
            j -= 1
        else:
            i += 1

    return ans