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You are given an m x n binary matrix mat of 1s (representing soldiers) and 0s (representing civilians). The soldiers are positioned in the front and the civilians in the back of each row. That is, all the 1s will appear to the left of all the 0s in each row.

A row i is weaker than a row j if one of the following is true:

  1. The number of soldiers in row i is less than the number of soldiers in row j.
  2. 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.

Clarifying Questions

  1. Input Constraints:
    • What is the range of m and n (dimensions of the matrix)?
    • Is the matrix always valid, containing only 1s and 0s with all 1s preceding 0s in each row?
  2. Output Constraints:
    • Do we always need to return exactly k rows, even if k is larger than the number of rows?
  3. Special Cases:
    • What happens if multiple rows have the same number of soldiers?
  4. Input Example:
    • What kind of test cases should we consider for correct implementation (e.g. edge cases)?

Strategy

The overall strategy to solve the problem will be as follows:

  1. Count the Soldiers: For each row in the matrix, count the number of 1’s. Since the rows are sorted with 1’s followed by 0’s, we can use binary search to find the count of 1’s efficiently.

  2. Pair and Sort: Pair each row index with the count of soldiers and then sort these pairs. Sorting should be first by the count of soldiers and then by the row index if counts are equal.

  3. Extract Indices: Extract and return the first k indices from the sorted list of pairs.

Code

Here is the implementation in Python:

from typing import List
import heapq

def kWeakestRows(mat: List[List[int]], k: int) -> List[int]:
    # Function to count the number of soldiers in a row using binary search
    def count_soldiers(row):
        low, high = 0, len(row)
        while low < high:
            mid = (low + high) // 2
            if row[mid] == 1:
                low = mid + 1
            else:
                high = mid
        return low
    
    # Create a list of (number_of_soldiers, row_index) pairs
    soldiers_counts = [(count_soldiers(row), idx) for idx, row in enumerate(mat)]
    
    # Using a heap to efficiently get the k smallest elements
    heapq.heapify(soldiers_counts)
    
    # Extract the indices of the k weakest rows
    return [heapq.heappop(soldiers_counts)[1] for _ in range(k)]

# Example usage
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
print(kWeakestRows(mat, k))  # Output: [2, 0, 3]

Time Complexity

  1. Counting Soldiers: For each row, the worst-case time complexity to count soldiers using binary search is O(log n).
  2. Pairing and Sorting: Creating the list of pairs takes O(m), where m is the number of rows in the matrix. Sorting the rows based on the count of soldiers takes O(m log m).
  3. Heap Operations: Extracting the k smallest elements from the heap takes O(k log m).

Overall Time Complexity: O(m log n + m log m + k log m)

In practice, m log m is usually the dominant factor unless k is very large.

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