Introduction

In this blog post, we will explore how to solve the LeetCode problem "2643" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given a m x n binary matrix mat, find the 0-indexed position of the row that contains the maximum count of ones, and the number of ones in that row. In case there are multiple rows that have the maximum count of ones, the row with the smallest row number should be selected. Return an array containing the index of the row, and the number of ones in it. Example 1: Input: mat = [[0,1],[1,0]] Output: [0,1] Explanation: Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1). So, the answer is [0,1]. Example 2: Input: mat = [[0,0,0],[0,1,1]] Output: [1,2] Explanation: The row indexed 1 has the maximum count of ones (2). So we return its index, 1, and the count. So, the answer is [1,2]. Example 3: Input: mat = [[0,0],[1,1],[0,0]] Output: [1,2] Explanation: The row indexed 1 has the maximum count of ones (2). So the answer is [1,2]. Constraints: m == mat.length n == mat[i].length 1 <= m, n <= 100 mat[i][j] is either 0 or 1.

Explanation

• Iterate through each row of the matrix.
  • Count the number of ones in each row.
  • Keep track of the row index and the maximum count of ones encountered so far, updating whenever a row with a larger count or the same count but smaller index is found.

• Runtime Complexity: O(m*n), Storage Complexity: O(1)

Code

    def rowAndMaximumOnes(mat):
    """
    Finds the 0-indexed position of the row that contains the maximum count of ones,
    and the number of ones in that row.

    Args:
        mat: A m x n binary matrix.

    Returns:
        An array containing the index of the row, and the number of ones in it.
    """

    max_ones = -1
    row_index = -1
    for i in range(len(mat)):
        current_ones = sum(mat[i])
        if current_ones > max_ones:
            max_ones = current_ones
            row_index = i
        elif current_ones == max_ones and i < row_index:
            row_index = i
    return [row_index, max_ones]