What is the time complexity of Count Submatrices with Top-Left Element and Sum Less Than k?

The optimal solution for Count Submatrices with Top-Left Element and Sum Less Than k runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to sorting the sums for each submatrix, while the space complexity is O(n) for storing prefix sums.

What is the brute force solution for Count Submatrices with Top-Left Element and Sum Less Than k?

The brute force approach for Count Submatrices with Top-Left Element and Sum Less Than k is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible submatrix that starts from the top-left element of the grid. We calculate the sum of each submatrix and count how many of them have a sum less than or equal to k. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Count Submatrices with Top-Left Element and Sum Less Than k?

Count Submatrices with Top-Left Element and Sum Less Than k uses the following concepts: Array, Matrix, Prefix Sum. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Submatrices with Top-Left Element and Sum Less Than k?

Explain your thought process clearly while coding. Discuss the trade-offs of your approaches. Practice writing both brute-force and optimal solutions.

What are common mistakes when solving Count Submatrices with Top-Left Element and Sum Less Than k?

Not using prefix sums, leading to inefficient sum calculations. Forgetting to handle edge cases where k is very small.

#3070

Count Submatrices with Top-Left Element and Sum Less Than k

Medium

ArrayMatrixPrefix SumHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

The optimal approach uses prefix sums and a data structure to efficiently count submatrices with sums less than k. By leveraging prefix sums, we can quickly calculate the sum of any submatrix.

⚙️

Algorithm

5 steps

1Step 1: Compute a prefix sum matrix for the grid.
2Step 2: Iterate through all possible bottom-right corners of submatrices starting from (0,0).
3Step 3: For each bottom-right corner, calculate the sum using the prefix sum matrix.
4Step 4: Use a sorted data structure to count how many prefix sums are less than k.
5Step 5: Return the total count of valid submatrices.

solution.py21 lines

1# Full working Python code
2import bisect
3
4def countSubmatrices(grid, k):
5    m, n = len(grid), len(grid[0])
6    prefix_sum = [[0] * (n + 1) for _ in range(m + 1)]
7    for i in range(m):
8        for j in range(n):
9            prefix_sum[i + 1][j + 1] = grid[i][j] + prefix_sum[i][j + 1] + prefix_sum[i + 1][j] - prefix_sum[i][j]
10    count = 0
11    for i in range(m):
12        for j in range(n):
13            sums = []
14            for x in range(i + 1):
15                current_sum = prefix_sum[i + 1][j + 1] - prefix_sum[x][j + 1]
16                sums.append(current_sum)
17            sums.sort()
18            for s in sums:
19                if s < k:
20                    count += 1
21    return count

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the sums for each submatrix, while the space complexity is O(n) for storing prefix sums.

1Using prefix sums allows for efficient sum calculations of submatrices.
2Sorting helps quickly count how many sums are less than k.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.