How do you solve Increment Submatrices by One on LeetCode?

Increment Submatrices by One (LeetCode #2536) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Using a difference matrix allows for efficient range updates.

What is the brute force solution for Increment Submatrices by One?

The brute force approach for Increment Submatrices by One is Brute Force, with O(n² * q) time complexity and O(1) space complexity. The brute force approach directly applies each query to the matrix by iterating through the specified submatrix and incrementing each element. This is straightforward but inefficient for larger matrices. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Increment Submatrices by One?

Increment Submatrices by One uses the following concepts: Array, Matrix, Prefix Sum. The recommended patterns to study are: Difference Array, Prefix Sum.

What are the interview tips for Increment Submatrices by One?

Explain your thought process clearly when transitioning from brute force to optimal. Practice writing difference array problems to get comfortable with the concept. Always consider edge cases, such as minimum and maximum values for n and queries.

What are common mistakes when solving Increment Submatrices by One?

Not considering the boundaries when updating the difference matrix. Forgetting to handle the prefix sums correctly.

#2536

Increment Submatrices by One

Medium

ArrayMatrixPrefix SumDifference ArrayPrefix Sum

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

Instead of updating each element in the submatrix directly, we can use a difference array approach to mark the increments efficiently. This allows us to perform all updates in constant time and then compute the final matrix in linear time.

⚙️

Algorithm

3 steps

1Step 1: Initialize a difference matrix of size (n+1) x (n+1) filled with zeros.
2Step 2: For each query, increment the top-left corner and decrement the position just outside the bottom-right corner to mark the range for updates.
3Step 3: Convert the difference matrix back to the final matrix using a prefix sum approach.

solution.py15 lines

1def incrementSubmatrices(n, queries):
2    diff = [[0] * (n + 1) for _ in range(n + 1)]
3    for row1, col1, row2, col2 in queries:
4        diff[row1][col1] += 1
5        diff[row1][col2 + 1] -= 1
6        diff[row2 + 1][col1] -= 1
7        diff[row2 + 1][col2 + 1] += 1
8    mat = [[0] * n for _ in range(n)]
9    for i in range(n):
10        for j in range(n):
11            mat[i][j] = diff[i][j]
12            if i > 0: mat[i][j] += mat[i - 1][j]
13            if j > 0: mat[i][j] += mat[i][j - 1]
14            if i > 0 and j > 0: mat[i][j] -= mat[i - 1][j - 1]
15    return mat

ℹ

Complexity note: The time complexity is O(n²) for the final matrix computation, while the space complexity is O(n) due to the additional difference matrix used for updates.

1Using a difference matrix allows for efficient range updates.
2Prefix sums help in converting the difference matrix back to the final matrix.

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