update: 添加问题“3212.统计X和Y频数相等的子矩阵数量”的代码和题解 (#1452)

3212: AC.cpp (#1451) - AC,15.57%,85.38% + (en)

Signed-off-by: LetMeFly666 <Tisfy@qq.com>

Author: LetMeFly666

.commitmsg

@@ -1 +1 @@
-3070: AC.cpp (#1449) - AC,91.79%,100.00% + (en)
+3212: AC.cpp (#1451) - AC,15.57%,85.38% + (en)

Codes/3212-count-submatrices-with-equal-frequency-of-x-and-y.cpp

@@ -0,0 +1,42 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2026-03-19 22:10:19
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2026-03-19 22:19:31
+ */
+#ifdef _DEBUG
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+private:
+    inline int v(int x) {
+        return x == 'X' ? 1 : x == 'Y' ? -1 : 0;
+    }
+public:
+    int numberOfSubmatrices(vector<vector<char>>& grid) {
+        int ans = 0;
+        int n = grid.size(), m = grid[0].size();
+        vector<vector<bool>> hasX(n, vector<bool>(m));  // has x or y
+        vector<vector<int>> diff(n, vector<int>(m));
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                if (i && j) {
+                    hasX[i][j] = hasX[i - 1][j] | hasX[i][j - 1] | (grid[i][j] != '.');
+                    diff[i][j] = diff[i - 1][j] + diff[i][j - 1] - diff[i - 1][j - 1] + v(grid[i][j]);
+                } else if (i) {
+                    hasX[i][j] = hasX[i - 1][j] | (grid[i][j] != '.');
+                    diff[i][j] = diff[i - 1][j] + v(grid[i][j]);
+                } else if (j) {
+                    hasX[i][j] = hasX[i][j - 1] | (grid[i][j] != '.');
+                    diff[i][j] = diff[i][j - 1] + v(grid[i][j]);
+                } else {
+                    hasX[i][j] = grid[i][j] != '.';
+                    diff[i][j] = v(grid[i][j]);
+                }
+                ans += (!diff[i][j] && hasX[i][j]);
+            }
+        }
+        return ans;
+    }
+};

README.md

@@ -1060,6 +1060,7 @@
 |3206.交替组I|简单|<a href="https://leetcode.cn/problems/alternating-groups-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/26/LeetCode%203206.%E4%BA%A4%E6%9B%BF%E7%BB%84I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144071026" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/alternating-groups-i/solutions/3001910/letmefly-3206jiao-ti-zu-ibian-li-by-tisf-euqx/" target="_blank">LeetCode题解</a>|
 |3208.交替组II|中等|<a href="https://leetcode.cn/problems/alternating-groups-ii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/11/28/LeetCode%203208.%E4%BA%A4%E6%9B%BF%E7%BB%84II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144123453" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/alternating-groups-ii/solutions/3004235/letmefly-3208jiao-ti-zu-iihua-dong-chuan-5bcc/" target="_blank">LeetCode题解</a>|
 |3211.生成不含相邻零的二进制字符串|中等|<a href="https://leetcode.cn/problems/generate-binary-strings-without-adjacent-zeros/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/10/29/LeetCode%203211.%E7%94%9F%E6%88%90%E4%B8%8D%E5%90%AB%E7%9B%B8%E9%82%BB%E9%9B%B6%E7%9A%84%E4%BA%8C%E8%BF%9B%E5%88%B6%E5%AD%97%E7%AC%A6%E4%B8%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143352841" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/generate-binary-strings-without-adjacent-zeros/solutions/2970587/letmefly-3211sheng-cheng-bu-han-xiang-li-sdh3/" target="_blank">LeetCode题解</a>|
+|3212.统计X和Y频数相等的子矩阵数量|中等|<a href="https://leetcode.cn/problems/count-submatrices-with-equal-frequency-of-x-and-y/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2026/03/19/LeetCode%203212.%E7%BB%9F%E8%AE%A1X%E5%92%8CY%E9%A2%91%E6%95%B0%E7%9B%B8%E7%AD%89%E7%9A%84%E5%AD%90%E7%9F%A9%E9%98%B5%E6%95%B0%E9%87%8F/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/159253457" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-submatrices-with-equal-frequency-of-x-and-y/solutions/3931242/letmefly-3212tong-ji-x-he-y-pin-shu-xian-x2ai/" target="_blank">LeetCode题解</a>|
 |3216.交换后字典序最小的字符串|简单|<a href="https://leetcode.cn/problems/lexicographically-smallest-string-after-a-swap/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/10/30/LeetCode%203216.%E4%BA%A4%E6%8D%A2%E5%90%8E%E5%AD%97%E5%85%B8%E5%BA%8F%E6%9C%80%E5%B0%8F%E7%9A%84%E5%AD%97%E7%AC%A6%E4%B8%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/143362223" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/lexicographically-smallest-string-after-a-swap/solutions/2971048/letmefly-3216jiao-huan-hou-zi-dian-xu-zu-nxp9/" target="_blank">LeetCode题解</a>|
 |3217.从链表中移除在数组中存在的节点|中等|<a href="https://leetcode.cn/problems/delete-nodes-from-linked-list-present-in-array/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/11/01/LeetCode%203217.%E4%BB%8E%E9%93%BE%E8%A1%A8%E4%B8%AD%E7%A7%BB%E9%99%A4%E5%9C%A8%E6%95%B0%E7%BB%84%E4%B8%AD%E5%AD%98%E5%9C%A8%E7%9A%84%E8%8A%82%E7%82%B9/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/154259704" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/delete-nodes-from-linked-list-present-in-array/solutions/3821125/letmefly-3217cong-lian-biao-zhong-yi-chu-2mcy/" target="_blank">LeetCode题解</a>|
 |3218.切蛋糕的最小总开销I|中等|<a href="https://leetcode.cn/problems/minimum-cost-for-cutting-cake-i/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/12/25/LeetCode%203218.%E5%88%87%E8%9B%8B%E7%B3%95%E7%9A%84%E6%9C%80%E5%B0%8F%E6%80%BB%E5%BC%80%E9%94%80I/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/144728332" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-cost-for-cutting-cake-i/solutions/3030374/letmefly-3218qie-dan-gao-de-zui-xiao-zon-a1t6/" target="_blank">LeetCode题解</a>|

Solutions/LeetCode 3212.统计X和Y频数相等的子矩阵数量.md

@@ -0,0 +1,124 @@
+---
+title: 3212.统计 X 和 Y 频数相等的子矩阵数量：前缀和
+date: 2026-03-19 22:20:10
+tags: [题解, LeetCode, 中等, 数组, 矩阵, 前缀和]
+categories: [题解, LeetCode]
+index_img: https://assets.leetcode.com/uploads/2024/06/07/examplems.png
+---
+
+# 【LetMeFly】3212.统计 X 和 Y 频数相等的子矩阵数量：前缀和
+
+力扣题目链接：[https://leetcode.cn/problems/count-submatrices-with-equal-frequency-of-x-and-y/](https://leetcode.cn/problems/count-submatrices-with-equal-frequency-of-x-and-y/)
+
+<p>给你一个二维字符矩阵 <code>grid</code>，其中 <code>grid[i][j]</code> 可能是 <code>'X'</code>、<code>'Y'</code> 或 <code>'.'</code>，返回满足以下条件的<span data-keyword="submatrix">子矩阵</span>数量：</p>
+
+<ul>
+	<li>包含 <code>grid[0][0]</code></li>
+	<li><code>'X'</code> 和 <code>'Y'</code> 的频数相等。</li>
+	<li>至少包含一个 <code>'X'</code>。</li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong class="example">示例 1：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">grid = [["X","Y","."],["Y",".","."]]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">3</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2024/06/07/examplems.png" style="padding: 10px; background: rgb(255, 255, 255); border-radius: 0.5rem; width: 175px; height: 350px;" /></strong></p>
+</div>
+
+<p><strong class="example">示例 2：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">grid = [["X","X"],["X","Y"]]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">0</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>不存在满足 <code>'X'</code> 和 <code>'Y'</code> 频数相等的子矩阵。</p>
+</div>
+
+<p><strong class="example">示例 3：</strong></p>
+
+<div class="example-block">
+<p><strong>输入：</strong> <span class="example-io">grid = [[".","."],[".","."]]</span></p>
+
+<p><strong>输出：</strong> <span class="example-io">0</span></p>
+
+<p><strong>解释：</strong></p>
+
+<p>不存在满足至少包含一个 <code>'X'</code> 的子矩阵。</p>
+</div>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= grid.length, grid[i].length &lt;= 1000</code></li>
+	<li><code>grid[i][j]</code> 可能是 <code>'X'</code>、<code>'Y'</code> 或 <code>'.'</code>.</li>
+</ul>
+
+
+> 前置注意事项：本题的子矩阵是必须包含左上角的矩阵。
+
+## 解题方法：前缀和
+
+类似昨天的每日一题[3070.元素和小于等于 k 的子矩阵的数目](https://blog.letmefly.xyz/2026/03/18/LeetCode%203070.%E5%85%83%E7%B4%A0%E5%92%8C%E5%B0%8F%E4%BA%8E%E7%AD%89%E4%BA%8Ek%E7%9A%84%E5%AD%90%E7%9F%A9%E9%98%B5%E7%9A%84%E6%95%B0%E7%9B%AE/)，使用一个$diff$数组存储从$(0,0)$到$(i,j)$的`X - Y`的数量，再额外使用一个数组存储从$(0,0)$到$(i,j)$是否存在`X`(或`Y`)。
+
+遍历过程中维护并更新这两个数组即可。
+
++ 时间复杂度$O(mn)$
++ 空间复杂度$O(mn)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @LastEditTime: 2026-03-19 22:19:31
+ */
+class Solution {
+private:
+    inline int v(int x) {
+        return x == 'X' ? 1 : x == 'Y' ? -1 : 0;
+    }
+public:
+    int numberOfSubmatrices(vector<vector<char>>& grid) {
+        int ans = 0;
+        int n = grid.size(), m = grid[0].size();
+        vector<vector<bool>> hasX(n, vector<bool>(m));  // has x or y
+        vector<vector<int>> diff(n, vector<int>(m));
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j < m; j++) {
+                if (i && j) {
+                    hasX[i][j] = hasX[i - 1][j] | hasX[i][j - 1] | (grid[i][j] != '.');
+                    diff[i][j] = diff[i - 1][j] + diff[i][j - 1] - diff[i - 1][j - 1] + v(grid[i][j]);
+                } else if (i) {
+                    hasX[i][j] = hasX[i - 1][j] | (grid[i][j] != '.');
+                    diff[i][j] = diff[i - 1][j] + v(grid[i][j]);
+                } else if (j) {
+                    hasX[i][j] = hasX[i][j - 1] | (grid[i][j] != '.');
+                    diff[i][j] = diff[i][j - 1] + v(grid[i][j]);
+                } else {
+                    hasX[i][j] = grid[i][j] != '.';
+                    diff[i][j] = v(grid[i][j]);
+                }
+                ans += (!diff[i][j] && hasX[i][j]);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/159253457)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2026/03/19/LeetCode%203212.%E7%BB%9F%E8%AE%A1X%E5%92%8CY%E9%A2%91%E6%95%B0%E7%9B%B8%E7%AD%89%E7%9A%84%E5%AD%90%E7%9F%A9%E9%98%B5%E6%95%B0%E9%87%8F/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)