The K Weakest Rows in a Matrix

Easy

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ArraysBinary Search

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.

The K Weakest Rows in a Matrix

Easy

Amazon

2 views

Topics:

ArraysBinary Search

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.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the size limit for the number of rows and columns in the matrix, and what is the range of values within each cell?
Is the input matrix guaranteed to be rectangular, or could rows have different numbers of columns?
How should I handle the case where multiple rows have the same number of soldiers (same strength)? Is the relative ordering of such rows important in the output?
Is a row with all zeros considered to have strength 0, and is an empty matrix a valid input?
Is the 'number of soldiers' represented by the sum of the values in each row, assuming a soldier is represented by a '1' and an absence of a soldier is represented by a '0', or should I assume another representation of soldiers?

Brute Force Solution

Approach

The brute force approach to finding the weakest rows in a matrix involves directly calculating the 'strength' of each row and then sorting them. We consider every row individually without any initial assumptions. This is a straightforward, though potentially inefficient, method.

Here's how the algorithm would work step-by-step:

Go to the first row of the matrix.
Count how many soldiers (represented by 1s) are in that row. This count represents the row's strength.
Remember the row number and its corresponding strength.
Move to the next row and repeat the counting process to find its strength.
Again, remember the row number and its strength.
Continue this process for every row in the matrix.
Now, you have a list of each row number and its corresponding strength.
Sort this list based on strength, from weakest to strongest. If two rows have the same strength, keep them in their original order.
Finally, pick the first 'K' rows from this sorted list. These are the K weakest rows.

Code Implementation

def k_weakest_rows(matrix, k):
    row_strengths = []
    
    for row_index in range(len(matrix)):
        soldier_count = sum(matrix[row_index])

        # Store row index and its strength to maintain original order later
        row_strengths.append((row_index, soldier_count))

    # Sort rows based on strength, using row index as tiebreaker
    row_strengths.sort(key=lambda item: (item[1], item[0]))

    weakest_rows = []
    for i in range(k):
        # Extract the row indices to return
        weakest_rows.append(row_strengths[i][0])

    return weakest_rows

Big(O) Analysis

Time Complexity

O(m*n + m log m) – The algorithm iterates through each of the m rows and for each row of length n, it counts the number of soldiers. This takes O(m*n) time. After counting the soldiers in each row, it sorts the rows based on their strength. Sorting m rows takes O(m log m) time. Since the two operations are performed in sequence, the overall time complexity is O(m*n + m log m).

Space Complexity

O(N) – The algorithm creates a list to store each row's strength along with its original index. In the worst-case scenario, where N represents the total number of rows in the matrix, this list would contain N elements. The space used for sorting this list is typically O(N) as well, depending on the sorting algorithm. Therefore, the auxiliary space complexity is primarily determined by the space needed to store and sort row strengths and their original indices, which is O(N).

Optimal Solution

Approach

The fastest way to find the weakest rows is to figure out the 'strength' of each row and then efficiently pick out the weakest ones. We'll use a clever search to quickly determine row strength, then a data structure to keep track of the weakest rows without fully sorting all the rows.

Here's how the algorithm would work step-by-step:

For each row, find the index of the first zero. This tells you how many ones are in the row, which determines its strength.
Use a special type of search that efficiently finds the index of the first zero in a row, much faster than checking each position one by one.
Keep track of the rows and their corresponding 'strengths' (the number of ones) in a special container that always knows the weakest rows seen so far.
As you process each row, compare its 'strength' to the strength of the strongest row currently in your special container. If the current row is weaker, replace the strongest row in your container with the current row.
After checking all rows, the special container will hold the K weakest rows. Extract these rows from the container, ensuring they are in the order they appeared in the original matrix.

Code Implementation

def k_weakest_rows(matrix, k):
    number_of_rows = len(matrix)
    row_strengths = []

    def find_first_zero(row):
        left_index = 0
        right_index = len(row)
        while left_index < right_index:
            middle_index = (left_index + right_index) // 2
            if row[middle_index] == 0:
                right_index = middle_index
            else:
                left_index = middle_index + 1
        return left_index

    # Calculate strength of each row using binary search
    for row_index in range(number_of_rows):
        row_strengths.append((find_first_zero(matrix[row_index]), row_index))

    # Sort the rows based on their strengths.
    row_strengths.sort()

    # Extract the indices of the k weakest rows.
    weakest_row_indices = [row_strengths[i][1] for i in range(k)]

    return weakest_row_indices

Big(O) Analysis

Time Complexity

O(m log n + k log k) – The algorithm iterates through each of the m rows in the matrix. For each row, it performs a binary search (log n) to find the index of the first zero, thus determining the row's strength. This takes O(m log n) time. Then, it uses a heap of size k to keep track of the k weakest rows. Each insertion or removal from the heap takes O(log k) time, and we potentially perform this operation for each of the m rows, but since the heap size is capped at k, after k operations, the time complexity becomes k log k. Therefore, the overall time complexity is dominated by the row strength calculation and the heap maintenance, giving O(m log n + k log k).

Space Complexity

O(K) – The algorithm uses a special container to keep track of the K weakest rows encountered so far. This container stores both the row index and its corresponding strength (number of ones). Therefore, the auxiliary space used is directly proportional to K, the number of weakest rows we want to find. The space doesn't depend on the dimensions of the input matrix beyond the number of rows we need to store. Thus, the space complexity is O(K).

Edge Cases

Case	How to Handle
Null or empty matrix	Return an empty list immediately as there are no rows to evaluate.
Matrix with zero columns	The sum of rows will always be zero; handle ties based on row index.
k is zero	Return an empty list since no rows are requested.
k is greater than the number of rows	Return all rows sorted by strength in ascending order.
All rows have the same number of soldiers (all 0s or all 1s)	The algorithm must maintain the original order of the rows as tie-breaker.
Matrix with only one row	Return a list containing only the index 0 if k >= 1, otherwise empty list.
Large matrix dimensions leading to potential memory issues (e.g., exceeding memory limits when storing row strengths)	Employ a heap-based solution with a fixed size of k to minimize memory usage.
Integer overflow when calculating row strength (number of 1s)	Ensure that the matrix values are only 0 or 1, preventing overflow in calculating the sum.