From 3617a14eb6632fd7cd8b11360c0485b8b9c52a02 Mon Sep 17 00:00:00 2001
From: gameloader <ggwqqo@163.com>
Date: Sun, 29 Sep 2024 10:22:21 +0800
Subject: [PATCH] leetcode update

---
 content/posts/leetcode.md | 84 +++++++++++++++++++++++++++++++++++++++
 1 file changed, 84 insertions(+)

diff --git a/content/posts/leetcode.md b/content/posts/leetcode.md
index 493fb8c..e393aec 100644
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@@ -14445,3 +14445,87 @@ public:
 };
 
 ```
+
+## day215 2024-09-29
+
+### 432. All O`one Data Structure
+
+Design a data structure to store the strings' count with the ability to return the strings with minimum and maximum counts.
+
+Implement the AllOne class:
+
+AllOne() Initializes the object of the data structure.
+inc(String key) Increments the count of the string key by 1. If key does not exist in the data structure, insert it with count 1.
+dec(String key) Decrements the count of the string key by 1. If the count of key is 0 after the decrement, remove it from the data structure. It is guaranteed that key exists in the data structure before the decrement.
+getMaxKey() Returns one of the keys with the maximal count. If no element exists, return an empty string "".
+getMinKey() Returns one of the keys with the minimum count. If no element exists, return an empty string "".
+Note that each function must run in O(1) average time complexity.
+
+![0929ecisQ42xBcws](https://testingcf.jsdelivr.net/gh/game-loader/picbase@master/uPic/0929ecisQ42xBcws.png)
+
+### 题解
+
+本题是一道难题，本题要快速获得出现频率最大的key和出现频率最小的key，就要将频率和对应的key记录下来。一方面我们要记录每个key和其对应的频率方便进行增减操作，即尽快完成inc和dec操作，另一方面我们也要记录不同频率和其对应的keys以便快速查找最大最小频率对应的key。记录每个key和其对应的频率可以通过map(可以使用unordered map，因为顺序不重要)实现，记录频率和其对应的key也可以通过map实现，区别在于因为一个频率可能对应多个字符串，则map需要将频率作为键，将一个字符串set作为值方便快速插入和删除该频率对应的字符串中的某个字符串。由于map默认是有序的，则找到最大值和最小值只需返回map的开头和结尾set中任一字符串即可。
+
+在增加某个key时，如果该key不存在，则插入这个key并将频率设置为1，同时查询频率map，若频率map中不存在1这个key，则插入1作为key，并在对应的set中插入这个字符串。存在则直接插入字符串。若增加的key存在，则在频率map中找到以对应频率为key的set，删掉这个字符串，并找频率+1对应的set，将这个字符串插入。减少某个key类似。获取最大最小key直接找到频率map的begin和rbegin，即有序map的开头和结尾返回set中任一字符串即可。
+
+注意如果将map和set的查询，删除，插入操作都视为均摊时间复杂度为O(1)的话，则本题我们实现的四个函数的时间复杂度可以视为O(1)，满足题目条件。
+
+### 代码
+
+```cpp
+class AllOne {
+private:
+    unordered_map<string, int> strmaps;
+    map<int, set<string>> fre;
+public:
+    AllOne() {
+    }
+
+    void inc(string key) {
+        // 如果 key 不存在，插入并设置频率为 1
+        if (strmaps.find(key) == strmaps.end()) {
+            strmaps[key] = 1;
+            fre[1].insert(key);
+        } else {
+            int count = strmaps[key];
+            strmaps[key]++;
+            fre[count].erase(key);
+            if (fre[count].empty()) {
+                fre.erase(count);
+            }
+            fre[count + 1].insert(key);
+        }
+    }
+
+    void dec(string key) {
+        int count = strmaps[key];
+        strmaps[key]--;
+        fre[count].erase(key);
+
+        if (fre[count].empty()) {
+            fre.erase(count);
+        }
+
+        if (strmaps[key] == 0) {
+            strmaps.erase(key);
+        } else {
+            fre[count - 1].insert(key);
+        }
+    }
+
+    string getMaxKey() {
+        if (fre.empty()) {
+            return "";
+        }
+        return *(fre.rbegin()->second.begin());
+    }
+
+    string getMinKey() {
+        if (fre.empty()) {
+            return "";
+        }
+        return *(fre.begin()->second.begin());
+    }
+};
+```