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‎Problemset/0432-All O`one Data Structure/README.md

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# 432. All O`one Data Structure
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Design a data structure to store the strings' count with the ability to return the strings with minimum and maximum counts.
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Implement the `AllOne` class:
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* `AllOne()` Initializes the object of the data structure.
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* `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`.
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* `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.
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* `getMaxKey()` Returns one of the keys with the maximal count. If no element exists, return an empty string `""`.
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* `getMinKey()` Returns one of the keys with the minimum count. If no element exists, return an empty string `""`.
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**Note** that each function must run in `O(1)` average time complexity.
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#### Example 1:
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<pre>
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<strong>Input:</strong>
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["AllOne", "inc", "inc", "getMaxKey", "getMinKey", "inc", "getMaxKey", "getMinKey"]
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[[], ["hello"], ["hello"], [], [], ["leet"], [], []]
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<strong>Output:</strong>
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[null, null, null, "hello", "hello", null, "hello", "leet"]
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<strong>Explanation:</strong>
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AllOne allOne = new AllOne();
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allOne.inc("hello");
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allOne.inc("hello");
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allOne.getMaxKey(); // return "hello"
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allOne.getMinKey(); // return "hello"
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allOne.inc("leet");
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allOne.getMaxKey(); // return "hello"
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allOne.getMinKey(); // return "leet"
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</pre>
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#### Constraints:
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* `1 <= key.length <= 10`
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* `key` consists of lowercase English letters.
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* It is guaranteed that for each call to `dec`, `key` is existing in the data structure.
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* At most <code>5 * 10<sup>4</sup></code> calls will be made to `inc`, `dec`, `getMaxKey`, and `getMinKey`.
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## Solutions (Python)
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### 1. Solution
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```Python
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class AllOne:
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def __init__(self):
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self.keys = []
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self.count = {}
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self.index = {}
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self.range = {}
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def inc(self, key: str) -> None:
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if key not in self.count:
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self.keys.append(key)
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self.count[key] = 1
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self.index[key] = len(self.keys) - 1
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if 1 not in self.range:
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self.range[1] = [self.index[key], self.index[key]]
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self.range[1][1] = self.index[key]
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else:
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count0 = self.count[key]
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count1 = count0 + 1
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i = self.range[count0][0]
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j = self.index[key]
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self.index[self.keys[i]], self.index[self.keys[j]] = j, i
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self.keys[i], self.keys[j] = self.keys[j], self.keys[i]
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self.count[key] += 1
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self.range[count0][0] += 1
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if self.range[count0][0] > self.range[count0][1]:
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self.range.pop(count0)
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if count1 not in self.range:
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self.range[count1] = [i, i]
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self.range[count1][1] = i
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def dec(self, key: str) -> None:
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if self.count[key] == 1:
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self.index[self.keys[-1]] = self.index[key]
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self.keys[self.index[key]] = self.keys[-1]
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self.keys.pop()
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self.count.pop(key)
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self.index.pop(key)
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self.range[1][1] -= 1
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if self.range[1][0] > self.range[1][1]:
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self.range.pop(1)
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else:
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count0 = self.count[key]
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count1 = count0 - 1
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i = self.range[count0][1]
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j = self.index[key]
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self.index[self.keys[i]], self.index[self.keys[j]] = j, i
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self.keys[i], self.keys[j] = self.keys[j], self.keys[i]
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self.count[key] -= 1
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self.range[count0][1] -= 1
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if self.range[count0][0] > self.range[count0][1]:
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self.range.pop(count0)
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if count1 not in self.range:
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self.range[count1] = [i, i]
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self.range[count1][0] = i
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def getMaxKey(self) -> str:
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return self.keys[0] if self.keys != [] else ""
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def getMinKey(self) -> str:
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return self.keys[-1] if self.keys != [] else ""
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# Your AllOne object will be instantiated and called as such:
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# obj = AllOne()
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# obj.inc(key)
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# obj.dec(key)
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# param_3 = obj.getMaxKey()
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# param_4 = obj.getMinKey()
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```

‎Problemset/0432-All O`one Data Structure/README_CN.md

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# 432. 全 O(1) 的数据结构
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请你设计一个用于存储字符串计数的数据结构，并能够返回计数最小和最大的字符串。
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实现 `AllOne` 类：
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* `AllOne()` 初始化数据结构的对象。
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* `inc(String key)` 字符串 `key` 的计数增加 `1` 。如果数据结构中尚不存在 `key` ，那么插入计数为 `1` 的 `key` 。
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* `dec(String key)` 字符串 `key` 的计数减少 `1` 。如果 `key` 的计数在减少后为 `0` ，那么需要将这个 `key` 从数据结构中删除。测试用例保证：在减少计数前，`key` 存在于数据结构中。
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* `getMaxKey()` 返回任意一个计数最大的字符串。如果没有元素存在，返回一个空字符串 `""` 。
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* `getMinKey()` 返回任意一个计数最小的字符串。如果没有元素存在，返回一个空字符串 `""` 。
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注意：每个函数都应当满足 `O(1)` 平均时间复杂度。
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#### 示例 1:
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<pre>
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<strong>输入:</strong>
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["AllOne", "inc", "inc", "getMaxKey", "getMinKey", "inc", "getMaxKey", "getMinKey"]
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[[], ["hello"], ["hello"], [], [], ["leet"], [], []]
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<strong>输出:</strong>
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[null, null, null, "hello", "hello", null, "hello", "leet"]
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<strong>解释:</strong>
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AllOne allOne = new AllOne();
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allOne.inc("hello");
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allOne.inc("hello");
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allOne.getMaxKey(); // 返回 "hello"
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allOne.getMinKey(); // 返回 "hello"
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allOne.inc("leet");
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allOne.getMaxKey(); // 返回 "hello"
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allOne.getMinKey(); // 返回 "leet"
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</pre>
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#### 提示:
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* `1 <= key.length <= 10`
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* `key` 由小写英文字母组成
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* 测试用例保证：在每次调用 `dec` 时，数据结构中总存在 `key`
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* 最多调用 `inc`、`dec`、`getMaxKey` 和 `getMinKey` 方法 <code>5 * 10<sup>4</sup></code> 次
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## 题解 (Python)
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### 1. 题解
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```Python
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class AllOne:
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def __init__(self):
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self.keys = []
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self.count = {}
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self.index = {}
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self.range = {}
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def inc(self, key: str) -> None:
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if key not in self.count:
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self.keys.append(key)
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self.count[key] = 1
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self.index[key] = len(self.keys) - 1
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if 1 not in self.range:
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self.range[1] = [self.index[key], self.index[key]]
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self.range[1][1] = self.index[key]
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else:
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count0 = self.count[key]
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count1 = count0 + 1
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i = self.range[count0][0]
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j = self.index[key]
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self.index[self.keys[i]], self.index[self.keys[j]] = j, i
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self.keys[i], self.keys[j] = self.keys[j], self.keys[i]
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self.count[key] += 1
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self.range[count0][0] += 1
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if self.range[count0][0] > self.range[count0][1]:
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self.range.pop(count0)
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if count1 not in self.range:
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self.range[count1] = [i, i]
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self.range[count1][1] = i
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def dec(self, key: str) -> None:
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if self.count[key] == 1:
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self.index[self.keys[-1]] = self.index[key]
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self.keys[self.index[key]] = self.keys[-1]
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self.keys.pop()
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self.count.pop(key)
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self.index.pop(key)
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self.range[1][1] -= 1
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if self.range[1][0] > self.range[1][1]:
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self.range.pop(1)
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else:
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count0 = self.count[key]
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count1 = count0 - 1
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i = self.range[count0][1]
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j = self.index[key]
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self.index[self.keys[i]], self.index[self.keys[j]] = j, i
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self.keys[i], self.keys[j] = self.keys[j], self.keys[i]
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self.count[key] -= 1
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self.range[count0][1] -= 1
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if self.range[count0][0] > self.range[count0][1]:
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self.range.pop(count0)
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if count1 not in self.range:
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self.range[count1] = [i, i]
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self.range[count1][0] = i
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def getMaxKey(self) -> str:
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return self.keys[0] if self.keys != [] else ""
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def getMinKey(self) -> str:
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return self.keys[-1] if self.keys != [] else ""
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# Your AllOne object will be instantiated and called as such:
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# obj = AllOne()
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# obj.inc(key)
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# obj.dec(key)
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# param_3 = obj.getMaxKey()
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# param_4 = obj.getMinKey()
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```

‎Problemset/0432-All O`one Data Structure/Solution.py

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class AllOne:
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def __init__(self):
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self.keys = []
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self.count = {}
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self.index = {}
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self.range = {}
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def inc(self, key: str) -> None:
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if key not in self.count:
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self.keys.append(key)
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self.count[key] = 1
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self.index[key] = len(self.keys) - 1
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if 1 not in self.range:
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self.range[1] = [self.index[key], self.index[key]]
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self.range[1][1] = self.index[key]
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else:
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count0 = self.count[key]
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count1 = count0 + 1
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i = self.range[count0][0]
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j = self.index[key]
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self.index[self.keys[i]], self.index[self.keys[j]] = j, i
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self.keys[i], self.keys[j] = self.keys[j], self.keys[i]
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self.count[key] += 1
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self.range[count0][0] += 1
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if self.range[count0][0] > self.range[count0][1]:
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self.range.pop(count0)
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if count1 not in self.range:
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self.range[count1] = [i, i]
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self.range[count1][1] = i
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def dec(self, key: str) -> None:
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if self.count[key] == 1:
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self.index[self.keys[-1]] = self.index[key]
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self.keys[self.index[key]] = self.keys[-1]
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self.keys.pop()
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self.count.pop(key)
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self.index.pop(key)
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self.range[1][1] -= 1
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if self.range[1][0] > self.range[1][1]:
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self.range.pop(1)
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else:
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count0 = self.count[key]
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count1 = count0 - 1
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i = self.range[count0][1]
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j = self.index[key]
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self.index[self.keys[i]], self.index[self.keys[j]] = j, i
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self.keys[i], self.keys[j] = self.keys[j], self.keys[i]
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self.count[key] -= 1
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self.range[count0][1] -= 1
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if self.range[count0][0] > self.range[count0][1]:
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self.range.pop(count0)
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if count1 not in self.range:
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self.range[count1] = [i, i]
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self.range[count1][0] = i
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def getMaxKey(self) -> str:
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return self.keys[0] if self.keys != [] else ""
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def getMinKey(self) -> str:
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return self.keys[-1] if self.keys != [] else ""
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# Your AllOne object will be instantiated and called as such:
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# obj = AllOne()
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# obj.inc(key)
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# obj.dec(key)
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# param_3 = obj.getMaxKey()
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# param_4 = obj.getMinKey()

‎README.md

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[427][427l] |[Construct Quad Tree][427] |![py]
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[429][429l] |[N-ary Tree Level Order Traversal][429] |![py]&nbsp;&nbsp;![rb]
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[430][430l] |[Flatten a Multilevel Doubly Linked List][430] |![rb]
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[432][432l] |[All O`one Data Structure][432] |![py]
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[433][433l] |[Minimum Genetic Mutation][433] |![py]&nbsp;&nbsp;![rb]
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[434][434l] |[Number of Segments in a String][434] |![py]
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[435][435l] |[Non-overlapping Intervals][435] |![rs]
@@ -1697,6 +1698,7 @@
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[427]:Problemset/0427-Construct%20Quad%20Tree/README.md#427-construct-quad-tree
16981699
[429]:Problemset/0429-N-ary%20Tree%20Level%20Order%20Traversal/README.md#429-n-ary-tree-level-order-traversal
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[430]:Problemset/0430-Flatten%20a%20Multilevel%20Doubly%20Linked%20List/README.md#430-flatten-a-multilevel-doubly-linked-list
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[432]:Problemset/0432-All%20O`one%20Data%20Structure/README.md#432-all-oone-data-structure
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[433]:Problemset/0433-Minimum%20Genetic%20Mutation/README.md#433-minimum-genetic-mutation
17011703
[434]:Problemset/0434-Number%20of%20Segments%20in%20a%20String/README.md#434-number-of-segments-in-a-string
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[435]:Problemset/0435-Non-overlapping%20Intervals/README.md#435-non-overlapping-intervals
@@ -3119,6 +3121,7 @@
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[427l]:https://leetcode.com/problems/construct-quad-tree/
31203122
[429l]:https://leetcode.com/problems/n-ary-tree-level-order-traversal/
31213123
[430l]:https://leetcode.com/problems/flatten-a-multilevel-doubly-linked-list/
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[432l]:https://leetcode.com/problems/all-oone-data-structure/
31223125
[433l]:https://leetcode.com/problems/minimum-genetic-mutation/
31233126
[434l]:https://leetcode.com/problems/number-of-segments-in-a-string/
31243127
[435l]:https://leetcode.com/problems/non-overlapping-intervals/

‎README_CN.md

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273273
[427][427l] |[建立四叉树][427] |![py]
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[429][429l] |[N叉树的层序遍历][429] |![py]&nbsp;&nbsp;![rb]
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[430][430l] |[扁平化多级双向链表][430] |![rb]
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[432][432l] |[全 O(1) 的数据结构] |![py]
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[433][433l] |[最小基因变化][433] |![py]&nbsp;&nbsp;![rb]
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[434][434l] |[字符串中的单词数][434] |![py]
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[435][435l] |[无重叠区间][435] |![rs]
@@ -1697,6 +1698,7 @@
16971698
[427]:Problemset/0427-Construct%20Quad%20Tree/README_CN.md#427-建立四叉树
16981699
[429]:Problemset/0429-N-ary%20Tree%20Level%20Order%20Traversal/README_CN.md#429-n叉树的层序遍历
16991700
[430]:Problemset/0430-Flatten%20a%20Multilevel%20Doubly%20Linked%20List/README_CN.md#430-扁平化多级双向链表
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[432]:Problemset/0432-All%20O`one%20Data%20Structure/README_CN.md#432-全-O1-的数据结构
17001702
[433]:Problemset/0433-Minimum%20Genetic%20Mutation/README_CN.md#433-最小基因变化
17011703
[434]:Problemset/0434-Number%20of%20Segments%20in%20a%20String/README_CN.md#434-字符串中的单词数
17021704
[435]:Problemset/0435-Non-overlapping%20Intervals/README_CN.md#435-无重叠区间
@@ -3119,6 +3121,7 @@
31193121
[427l]:https://leetcode.cn/problems/construct-quad-tree/
31203122
[429l]:https://leetcode.cn/problems/n-ary-tree-level-order-traversal/
31213123
[430l]:https://leetcode.cn/problems/flatten-a-multilevel-doubly-linked-list/
3124+
[432l]:https://leetcode.cn/problems/all-oone-data-structure/
31223125
[433l]:https://leetcode.cn/problems/minimum-genetic-mutation/
31233126
[434l]:https://leetcode.cn/problems/number-of-segments-in-a-string/
31243127
[435l]:https://leetcode.cn/problems/non-overlapping-intervals/

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