The K Weakest Rows in a Matrix

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.

For example:

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]

Explain your approach and provide code with its time and space complexity. Also discuss different edge cases that can occur with this problem and how to handle them.

Naive Solution

We can iterate through each row, count the number of soldiers (1s), and store the row index along with the count. Then, we sort these pairs based on the soldier count (and row index as a tie-breaker). Finally, we extract the first k row indices from the sorted list.

def k_weakest_rows_naive(mat, k):
    rows = []
    for i, row in enumerate(mat):
        count = sum(row)
        rows.append((count, i))
    
    rows.sort()
    
    result = [row[1] for row in rows[:k]]
    return result

Optimal Solution

To optimize the soldier counting process, we can use binary search. Since the soldiers are always in front, we can binary search for the index of the first civilian (0) in each row. This gives us the count of soldiers in that row. Then the rest of the algorithm is the same as in the naive solution, but with improved time complexity for soldier counting.

def k_weakest_rows_optimal(mat, k):
    def count_soldiers(row):
        left, right = 0, len(row)
        while left < right:
            mid = (left + right) // 2
            if row[mid] == 0:
                right = mid
            else:
                left = mid + 1
        return left

    rows = []
    for i, row in enumerate(mat):
        count = count_soldiers(row)
        rows.append((count, i))
    
    rows.sort()
    
    result = [row[1] for row in rows[:k]]
    return result

Big(O) Runtime Analysis

Naive Solution

• Counting Soldiers: O(m * n), where 'm' is the number of rows and 'n' is the number of columns. This is because we iterate through each cell in the matrix.
• Sorting: O(m * log m), where 'm' is the number of rows. This is because we sort the rows based on the number of soldiers.
• Overall: O(m * n + m * log m)

Optimal Solution

• Counting Soldiers (Binary Search): O(m * log n), where 'm' is the number of rows and 'n' is the number of columns. This is because we perform a binary search on each row.
• Sorting: O(m * log m), where 'm' is the number of rows. This is because we sort the rows based on the number of soldiers.
• Overall: O(m * log n + m * log m)

In cases where n > m, the optimal solution provides a significant improvement in runtime efficiency.

Big(O) Space Usage Analysis

Naive Solution

• Rows Array: O(m), where 'm' is the number of rows. This is because we store the soldier count and row index for each row.
• Overall: O(m)

Optimal Solution

• Rows Array: O(m), where 'm' is the number of rows. This is because we store the soldier count and row index for each row.
• Overall: O(m)

Both solutions have the same space complexity because they both store information for each row.

Edge Cases

1. Empty Matrix: If the input matrix is empty (m = 0), return an empty list.
2. Empty Rows: If a row is empty (n = 0), consider the soldier count to be 0.
3. All Soldiers or All Civilians: The binary search in the optimal solution should correctly handle rows with all soldiers (all 1s) or all civilians (all 0s).
4. k > m: If k is greater than the number of rows, return all rows sorted by weakness.

Here's the updated code to handle these edge cases:

def k_weakest_rows_optimal_edge_cases(mat, k):
    if not mat:
        return []

    def count_soldiers(row):
        if not row:
            return 0
        left, right = 0, len(row)
        while left < right:
            mid = (left + right) // 2
            if row[mid] == 0:
                right = mid
            else:
                left = mid + 1
        return left

    rows = []
    for i, row in enumerate(mat):
        count = count_soldiers(row)
        rows.append((count, i))
    
    rows.sort()
    
    result = [row[1] for row in rows[:min(k, len(mat))]]
    return result