Introduction

In this blog post, we will explore how to solve the LeetCode problem "1337" 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

You are given an m x n binary matrix mat of 1's (representing soldiers) and 0's (representing civilians). The soldiers are positioned in front of the civilians. That is, all the 1's will appear to the left of all the 0's in each row. A row i is weaker than a row j if one of the following is true: The number of soldiers in row i is less than the number of soldiers in row j. Both rows have the same number of soldiers and i < j. Return the indices of the k weakest rows in the matrix ordered from weakest to strongest. Example 1: Input: mat = [[1,1,0,0,0], [1,1,1,1,0], [1,0,0,0,0], [1,1,0,0,0], [1,1,1,1,1]], k = 3 Output: [2,0,3] Explanation: The number of soldiers in each row is: - Row 0: 2 - Row 1: 4 - Row 2: 1 - Row 3: 2 - Row 4: 5 The rows ordered from weakest to strongest are [2,0,3,1,4]. Example 2: Input: mat = [[1,0,0,0], [1,1,1,1], [1,0,0,0], [1,0,0,0]], k = 2 Output: [0,2] Explanation: The number of soldiers in each row is: - Row 0: 1 - Row 1: 4 - Row 2: 1 - Row 3: 1 The rows ordered from weakest to strongest are [0,2,3,1]. Constraints: m == mat.length n == mat[i].length 2 <= n, m <= 100 1 <= k <= m matrix[i][j] is either 0 or 1.

Explanation

• Calculate Soldier Count: For each row, efficiently determine the number of soldiers (1s) using binary search, leveraging the sorted nature of each row.
  • Store and Sort: Store the soldier count along with the row index as a tuple. Sort these tuples based on soldier count (primary key) and row index (secondary key).
  • Extract Top K: Extract the row indices from the sorted tuples for the k weakest rows.

• Runtime Complexity: O(m log n + m log m), where m is the number of rows and n is the number of columns. Storage Complexity: O(m)

Code

    def kWeakestRows(mat, k):
    """
    Finds the indices of the k weakest rows in a binary matrix.

    Args:
        mat: A list of lists of integers representing the binary matrix.
        k: The number of weakest rows to return.

    Returns:
        A list of integers representing the indices of the k weakest rows,
        ordered from weakest to strongest.
    """

    def count_soldiers(row):
        """
        Counts the number of soldiers (1s) in a row using binary search.
        """
        left, right = 0, len(row) - 1
        while left <= right:
            mid = (left + right) // 2
            if row[mid] == 1:
                left = mid + 1
            else:
                right = mid - 1
        return left

    row_strengths = []
    for i, row in enumerate(mat):
        row_strengths.append((count_soldiers(row), i))

    row_strengths.sort()  # Sort based on soldier count and then row index

    weakest_rows = [row_strengths[i][1] for i in range(k)]
    return weakest_rows