Count Submatrices With Equal Frequency of X and Y - Problem

Given a 2D character matrix grid, where grid[i][j] is either 'X', 'Y', or '.', return the number of submatrices that contain:

grid[0][0] (top-left corner)
An equal frequency of 'X' and 'Y'
At least one 'X'

A submatrix is defined by its top-left corner at (0, 0) and bottom-right corner at (i, j) where 0 ≤ i < m and 0 ≤ j < n.

Input & Output

Example 1 — Basic Case

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

› Output: 3

💡 Note: Valid submatrices: (0,0) to (0,1) has X=1,Y=1; (0,0) to (1,2) has X=2,Y=2; (0,0) to (1,1) has X=1,Y=2 (invalid)

Example 2 — Single Cell

$ Input: grid = [["X"]]

› Output: 0

💡 Note: Only submatrix is (0,0) to (0,0) with X=1,Y=0. Not equal counts, so 0 valid submatrices

Example 3 — No X Characters

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

› Output: 0

💡 Note: No submatrix contains at least one X, so result is 0

Constraints

1 ≤ grid.length, grid[i].length ≤ 100
grid[i][j] is either 'X', 'Y', or '.'

Visualization

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Asked in

G Google 25 M Microsoft 18 A Amazon 15

Key Insight: This problem requires checking all submatrices starting from (0,0) for equal X and Y counts with at least one X. The optimal approach uses 2D prefix sums to preprocess the grid and answer range queries in O(1) time. While the brute force solution works for small inputs, the prefix sum optimization provides the best balance of efficiency and clarity with O(mn) time and space complexity.

Common Approaches

✓ Check Every Zone Manually (Brute Force)

⏱️ Time: O(m²n²) Space: O(1)

For each possible bottom-right corner (i,j), iterate through all cells in the submatrix from (0,0) to (i,j) and count occurrences of 'X' and 'Y'. Check if counts are equal and at least one 'X' exists.

Player Balance Tracker (Hash Map)

⏱️ Time: O(mn) Space: O(mn)

Convert the 2D problem into tracking balance values where balance = count(X) - count(Y). Use a hash map to store frequency of each balance value encountered while scanning the matrix row by row.

Smart Scoreboards (Prefix Sum Optimization)

⏱️ Time: O(mn) Space: O(mn)

Build prefix sum arrays for both X and Y characters to answer range sum queries in constant time. For each possible submatrix, use prefix sums to get X and Y counts instantly and check if they're equal with at least one X.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

✓ Linear Growth

Space Complexity

⚡ Linearithmic Space

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Smart Actions

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