What is the time complexity of Maximum Side Length of a Square with Sum Less than or Equal to Threshold?

The optimal solution for Maximum Side Length of a Square with Sum Less than or Equal to Threshold runs in O(n²) time and uses O(n²) space. The time complexity is O(n²) due to iterating through the prefix sums for each square size. The space complexity is O(n²) because we store the prefix sums in a separate matrix.

What is the brute force solution for Maximum Side Length of a Square with Sum Less than or Equal to Threshold?

The brute force approach for Maximum Side Length of a Square with Sum Less than or Equal to Threshold is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible square in the matrix to see if its sum is less than or equal to the threshold. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Maximum Side Length of a Square with Sum Less than or Equal to Threshold?

Maximum Side Length of a Square with Sum Less than or Equal to Threshold uses the following concepts: Array, Binary Search, Matrix, Prefix Sum. The recommended patterns to study are: Prefix Sum, Matrix.

What are the interview tips for Maximum Side Length of a Square with Sum Less than or Equal to Threshold?

Explain your thought process clearly; interviewers appreciate understanding your reasoning. Consider edge cases and test your solution with small examples. Use comments to clarify your code, especially in complex algorithms.

What are common mistakes when solving Maximum Side Length of a Square with Sum Less than or Equal to Threshold?

Not considering edge cases where no square can be formed. Failing to optimize the sum calculation, leading to timeouts.

#1292

Maximum Side Length of a Square with Sum Less than or Equal to Threshold

Medium

ArrayBinary SearchMatrixPrefix SumPrefix SumMatrix

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n²)

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Intuition

Time O(n²)Space O(n²)

Using prefix sums allows us to quickly calculate the sum of any square in constant time, making our solution much faster.

⚙️

Algorithm

3 steps

1Step 1: Create a prefix sum matrix where each cell (i, j) contains the sum of elements from (0,0) to (i,j).
2Step 2: Iterate through possible square sizes and check if the sum of each square can be computed using the prefix sum matrix.
3Step 3: If the sum of the square is less than or equal to the threshold, update the maximum side length.

solution.py17 lines

1# Full working Python code
2
3def maxSideLength(mat, threshold):
4    m, n = len(mat), len(mat[0])
5    prefix_sum = [[0] * (n + 1) for _ in range(m + 1)]
6    for i in range(1, m + 1):
7        for j in range(1, n + 1):
8            prefix_sum[i][j] = mat[i - 1][j - 1] + prefix_sum[i - 1][j] + prefix_sum[i][j - 1] - prefix_sum[i - 1][j - 1]
9    max_length = 0
10    for length in range(1, min(m, n) + 1):
11        for i in range(length, m + 1):
12            for j in range(length, n + 1):
13                square_sum = prefix_sum[i][j] - prefix_sum[i - length][j] - prefix_sum[i][j - length] + prefix_sum[i - length][j - length]
14                if square_sum <= threshold:
15                    max_length = max(max_length, length)
16    return max_length
17

ℹ

Complexity note: The time complexity is O(n²) due to iterating through the prefix sums for each square size. The space complexity is O(n²) because we store the prefix sums in a separate matrix.

1Using prefix sums allows for quick sum calculations of submatrices.
2Iterating through possible square sizes systematically helps ensure we find the maximum.

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