#3212

Medium

LeetCode Problem Workspace

Count Submatrices With Equal Frequency of X and Y

Count the number of submatrices with equal frequency of 'X' and 'Y' in a 2D character grid.

Problem Statement

You are given a 2D character matrix grid, where each element is either 'X', 'Y', or '.'. The task is to return the number of submatrices where the frequency of 'X' and 'Y' are equal. A submatrix is any contiguous block of the matrix that can vary in size from 1x1 to the entire grid.

To solve this problem, you need to find an efficient approach that avoids brute force checking all possible submatrices. A smart transformation of the grid combined with prefix sums can be helpful in solving this problem optimally.

Examples

Example 1

Input: grid = [["X","Y","."],["Y",".","."]]

Output: 3

Example 2

Input: grid = [["X","X"],["X","Y"]]

Output: 0

No submatrix has an equal frequency of 'X' and 'Y' .

Example 3

Input: grid = [[".","."],[".","."]]

Output: 0

No submatrix has at least one 'X' .

Constraints

1 <= grid.length, grid[i].length <= 1000
grid[i][j] is either 'X', 'Y', or '.'.

Solution Approach

Transform the Grid

Replace each 'X' in the grid with 1, each 'Y' with -1, and '.' with 0. This transformation will allow us to focus on finding submatrices whose sums are zero, representing equal counts of 'X' and 'Y'.

Use Prefix Sum Arrays

Utilize a prefix sum array to quickly calculate the sum of any submatrix. By storing the sum of elements up to a given position, we can efficiently determine the sum of any submatrix without re-summing each time.

Count Zero-Sum Submatrices

After transforming the grid and using prefix sums, iterate over all possible pairs of rows. For each pair, treat the submatrix columns as a 1D array and count the number of zero-sum subarrays. This will give the number of valid submatrices for that row pair.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final approach. Using prefix sums, the complexity for each pair of rows is reduced to O(n), and iterating over all pairs of rows leads to a time complexity of O(n^2) for an n x m grid. The space complexity is O(n*m) due to the storage of the prefix sum array.

What Interviewers Usually Probe

Check for a transformation approach involving numerical mapping of 'X', 'Y', and '.' to facilitate zero-sum subarray counting.
Look for efficient usage of prefix sums to avoid brute-force submatrix enumeration.
Ensure that the candidate identifies and handles edge cases such as grids with only '.' or one type of character.

Common Pitfalls or Variants

Common pitfalls

Using brute force to check every possible submatrix will result in a time complexity that is too high for larger grids.
Forgetting to transform the grid correctly, which can lead to incorrect results when checking submatrices.
Not properly handling edge cases such as grids with no 'X' or 'Y' characters, which can cause miscounts.

Follow-up variants

What if the grid size is fixed, say 10x10? Can you optimize for this case?
What happens if the grid contains additional characters, such as 'A' or 'Z'? How would you adjust the approach?
How would you adapt the solution if instead of equal counts of 'X' and 'Y', you needed a specific ratio of 'X' to 'Y'?

FAQ

How do I handle a grid with no 'X' or 'Y' characters?

In cases where there are no 'X' or 'Y', the result will always be 0 since there are no valid submatrices with equal frequencies of 'X' and 'Y'.

What is the purpose of transforming 'X' to 1, 'Y' to -1, and '.' to 0?

This transformation simplifies the problem by converting it to one of finding subarrays that sum to 0, which makes use of efficient techniques like prefix sums.

Can I use brute force to solve this problem?

While brute force is possible, it is inefficient, especially for large grids, and would lead to a time complexity of O(n^4) which is too slow.

How do I efficiently compute the sum of submatrices?

You can use a prefix sum array to compute the sum of any submatrix in constant time, reducing the time complexity significantly.

What if the matrix size increases to 1000x1000?

For large grids, it's essential to use optimized techniques like prefix sums and avoid brute-force enumeration of submatrices to ensure the solution remains efficient.

terminal

Solution

Solution 1: 2D Prefix Sum

According to the problem description, we only need to calculate the prefix sums $s[i][j][0]$ and $s[i][j][1]$ for each position $(i, j)$, which represent the number of characters `X` and `Y` in the submatrix from $(0, 0)$ to $(i, j)$, respectively. If $s[i][j][0] > 0$ and $s[i][j][0] = s[i][j][1]$, it means the condition is met, and we increment the answer by one.

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class Solution:
    def numberOfSubmatrices(self, grid: List[List[str]]) -> int:
        m, n = len(grid), len(grid[0])
        s = [[[0] * 2 for _ in range(n + 1)] for _ in range(m + 1)]
        ans = 0
        for i, row in enumerate(grid, 1):
            for j, x in enumerate(row, 1):
                s[i][j][0] = s[i - 1][j][0] + s[i][j - 1][0] - s[i - 1][j - 1][0]
                s[i][j][1] = s[i - 1][j][1] + s[i][j - 1][1] - s[i - 1][j - 1][1]
                if x != ".":
                    s[i][j][ord(x) & 1] += 1
                if s[i][j][0] > 0 and s[i][j][0] == s[i][j][1]:
                    ans += 1
        return ans

Continue Practicing

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