Leetcode 3212. Count Submatrices With Equal Frequency of X and Y

• Leetcode 3212. Count Submatrices With Equal Frequency of X and Y
  • 1. 解题思路
  • 2. 代码实现

• 题目链接：3212. Count Submatrices With Equal Frequency of X and Y

1. 解题思路

这一题我的思路就是一个二维累计数组的实现，具体来说，我们事先算出关于x和y的两个二维累积矩阵，然后遍历一下每一个位置，看看他们作为矩形右下角的位置时，对应的矩形内x和y的个数是否相同且不为0。

2. 代码实现

给出python代码实现如下：

class Solution:
    def numberOfSubmatrices(self, grid: List[List[str]]) -> int:
        n, m = len(grid), len(grid[0])
        
        def count(ch):
            cnt = [[0 for _ in range(m)] for _ in range(n)]
            cnt[0][0] = 0 if grid[0][0] != ch else 1
            for j in range(1, m):
                cnt[0][j] = cnt[0][j-1] if grid[0][j] != ch else cnt[0][j-1]+1
            for i in range(1, n):
                cnt[i][0] = cnt[i-1][0] if grid[i][0] != ch else cnt[i-1][0]+1
                for j in range(1, m):
                    cnt[i][j] = cnt[i-1][j] + cnt[i][j-1] - cnt[i-1][j-1]
                    if grid[i][j] == ch:
                        cnt[i][j] += 1
            return cnt
        
        cntx = count("X")
        cnty = count("Y")
        ans = 0
        for i in range(n):
            for j in range(m):
                if cntx[i][j] == cnty[i][j] and cntx[i][j] > 0:
                    ans += 1
        return ans

提交代码评测得到：耗时3560ms，占用内存108.8MB。