add(coding/2349): design-a-number-container-system

source/coding/2349-design-a-number-container-system.md

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+---
+title: 2349. Design a Number Container System
+notebook: coding
+tags:
+- medium
+date: 2025-02-08 17:43:23
+updated: 2025-02-08 17:43:23
+---
+## Problem
+
+Design a number container system that can do the following:
+
+- **Insert** or **Replace** a number at the given index in the system.
+- **Return** the smallest index for the given number in the system.
+
+Implement the `NumberContainers` class:
+
+- `NumberContainers()` Initializes the number container system.
+- `void change(int index, int number)` Fills the container at `index` with the `number`. If there is already a number at that `index`, replace it.
+- `int find(int number)` Returns the smallest index for the given `number`, or `-1` if there is no index that is filled by `number` in the system.
+
+<https://leetcode.com/problems/design-a-number-container-system/>
+
+**Example 1:**
+
+> Input
+> `["NumberContainers", "find", "change", "change", "change", "change", "find", "change", "find"]`
+> `[[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]]`
+> Output
+> `[null, -1, null, null, null, null, 1, null, 2]`
+> Explanation
+>
+> ``` cpp
+> NumberContainers nc = new NumberContainers();
+> nc.find(10); // There is no index that is filled with number 10. Therefore, we return -1.
+> nc.change(2, 10); // Your container at index 2 will be filled with number 10.
+> nc.change(1, 10); // Your container at index 1 will be filled with number 10.
+> nc.change(3, 10); // Your container at index 3 will be filled with number 10.
+> nc.change(5, 10); // Your container at index 5 will be filled with number 10.
+> nc.find(10); // Number 10 is at the indices 1, 2, 3, and 5. Since the smallest index that is filled with 10 is 1, we return 1.
+> nc.change(1, 20); // Your container at index 1 will be filled with number 20. Note that index 1 was filled with 10 and then replaced with 20.
+> nc.find(10); // Number 10 is at the indices 2, 3, and 5. The smallest index that is filled with 10 is 2. Therefore, we return 2.
+> ```
+
+**Constraints:**
+
+- `1 <= index, number <= 10⁹`
+- At most `10⁵` calls will be made **in total** to `change` and `find`.
+
+## Test Cases
+
+``` python
+class NumberContainers:
+
+    def __init__(self):
+
+
+    def change(self, index: int, number: int) -> None:
+
+
+    def find(self, number: int) -> int:
+
+
+
+# Your NumberContainers object will be instantiated and called as such:
+# obj = NumberContainers()
+# obj.change(index,number)
+# param_2 = obj.find(number)
+```
+
+{% asset_code coding/2349-design-a-number-container-system/solution_test.py %}
+
+## Thoughts
+
+跟 [3160. Find the Number of Distinct Colors Among the Balls](3160-find-the-number-of-distinct-colors-among-the-balls) 差不多。index 相当于球的编号，number 相当于颜色。主要的区别是本题需要记录每个 number 对应的 index 数组，以便随时可以查询此 number 所在的所有 index 的最小值。
+
+同样用字典记录每个 index 最新的 number。再用字典记录每个 number 的所有 index，用有序数组记录，这里直接用 Python 的 list。需要增加或删除的时候，先用二分法查找再执行插入或删除操作。
+
+构造函数的时间复杂度 `O(1)`；`change` 方法的时间复杂度 `O(n)`；`find` 方法的时间复杂度 `O(n)`。空间复杂度 `O(n)`。其中 n 是 `change` 的调用次数。
+
+## Code
+
+{% asset_code coding/2349-design-a-number-container-system/solution.py %}

source/coding/2349-design-a-number-container-system/solution.py

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+from bisect import bisect_left
+from collections import defaultdict
+
+
+class NumberContainers:
+
+    def __init__(self):
+        self._numbers: dict[int, int] = {} # index -> number
+        self._indices: dict[int, list[int]] = defaultdict(list) # number -> sorted list of indices
+
+    def change(self, index: int, number: int) -> None:
+        if index in self._numbers and self._numbers[index] == number:
+            return
+
+        if index in self._numbers:
+            old_number = self._numbers[index]
+            indices = self._indices[old_number]
+            del indices[bisect_left(indices, index)]
+
+        self._numbers[index] = number
+        indices = self._indices[number]
+        indices.insert(bisect_left(indices, index), index)
+
+    def find(self, number: int) -> int:
+        if number not in self._indices:
+            return -1
+
+        indices = self._indices[number]
+        if not indices:
+            return -1
+
+        return indices[0]
+
+
+# Your NumberContainers object will be instantiated and called as such:
+# obj = NumberContainers()
+# obj.change(index,number)
+# param_2 = obj.find(number)

source/coding/2349-design-a-number-container-system/solution_test.py

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+import pytest
+
+from solution import NumberContainers
+
+null = None
+
+
+@pytest.mark.parametrize('actions, params, expects', [
+    (
+        ["NumberContainers", "find", "change", "change", "change", "change", "find", "change", "find"],
+        [[], [10], [2, 10], [1, 10], [3, 10], [5, 10], [10], [1, 20], [10]],
+        [null, -1, null, null, null, null, 1, null, 2],
+    ),
+])
+@pytest.mark.parametrize('clazz', [NumberContainers])
+def test_solution(clazz, actions, params, expects):
+    sol = None
+    for action, args, expected in zip(actions, params, expects):
+        if action == 'NumberContainers':
+            sol = clazz(*args)
+        else:
+            assert getattr(sol, action)(*args) == expected