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‎Problemset/2343-Query Kth Smallest Trimmed Number/README.md

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# 2343. Query Kth Smallest Trimmed Number
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You are given a **0-indexed** array of strings `nums`, where each string is of **equal length** and consists of only digits.
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You are also given a **0-indexed** 2D integer array `queries` where <code>queries[i] = [k<sub>i</sub>, trim<sub>i</sub>]</code>. For each `queries[i]`, you need to:
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* **Trim** each number in `nums` to its **rightmost** <code>trim<sub>i</sub></code> digits.
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* Determine the **index** of the <code>k<sub>i</sub><sup>th</sup></code> smallest trimmed number in `nums`. If two trimmed numbers are equal, the number with the **lower** index is considered to be smaller.
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* Reset each number in `nums` to its original length.
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Return *an array* `answer` *of the same length as* `queries`, *where* `answer[i]` *is the answer to the* <code>i<sup>th</sup></code> *query*.
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**Note**:
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* To trim to the rightmost `x` digits means to keep removing the leftmost digit, until only `x` digits remain.
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* Strings in `nums` may contain leading zeros.
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#### Example 1:
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<pre>
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<strong>Input:</strong> nums = ["102","473","251","814"], queries = [[1,1],[2,3],[4,2],[1,2]]
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<strong>Output:</strong> [2,2,1,0]
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<strong>Explanation:</strong>
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1. After trimming to the last digit, nums = ["2","3","1","4"]. The smallest number is 1 at index 2.
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2. Trimmed to the last 3 digits, nums is unchanged. The 2nd smallest number is 251 at index 2.
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3. Trimmed to the last 2 digits, nums = ["02","73","51","14"]. The 4th smallest number is 73.
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4. Trimmed to the last 2 digits, the smallest number is 2 at index 0.
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Note that the trimmed number "02" is evaluated as 2.
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</pre>
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#### Example 2:
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<pre>
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<strong>Input:</strong> nums = ["24","37","96","04"], queries = [[2,1],[2,2]]
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<strong>Output:</strong> [3,0]
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<strong>Explanation:</strong>
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1. Trimmed to the last digit, nums = ["4","7","6","4"]. The 2nd smallest number is 4 at index 3.
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There are two occurrences of 4, but the one at index 0 is considered smaller than the one at index 3.
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2. Trimmed to the last 2 digits, nums is unchanged. The 2nd smallest number is 24.
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</pre>
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#### Constraints:
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* `1 <= nums.length <= 100`
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* `1 <= nums[i].length <= 100`
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* `nums[i]` consists of only digits.
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* All `nums[i].length` are **equal**.
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* `1 <= queries.length <= 100`
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* `queries[i].length == 2`
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* <code>1 <= k<sub>i</sub> <= nums.length</code>
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* <code>1 <= trim<sub>i</sub> <= nums[i].length</code>
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**Follow up:** Could you use the **Radix Sort Algorithm** to solve this problem? What will be the complexity of that solution?
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## Solutions (Python)
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### 1. Solution
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```Python
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class Solution:
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def smallestTrimmedNumbers(self, nums: List[str], queries: List[List[int]]) -> List[int]:
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nums = [(nums[i], i) for i in range(len(nums))]
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trims = {}
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answer = [0] * len(queries)
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for i in range(len(queries)):
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if queries[i][1] not in trims:
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trims[queries[i][1]] = []
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trims[queries[i][1]].append((i, queries[i][0] - 1))
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for trim in range(1, len(nums[0][0]) + 1):
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count = [0] * 10
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tmp = [("", 0)] * len(nums)
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for num, _ in nums:
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count[ord(num[-trim]) - 48] += 1
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for i in range(1, 10):
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count[i] += count[i - 1]
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for num, i in nums[::-1]:
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count[ord(num[-trim]) - 48] -= 1
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tmp[count[ord(num[-trim]) - 48]] = (num, i)
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for i, k in trims.get(trim, []):
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answer[i] = tmp[k][1]
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nums = tmp
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return answer
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```

‎Problemset/2343-Query Kth Smallest Trimmed Number/README_CN.md

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# 2343. 裁剪数字后查询第 K 小的数字
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给你一个下标从 **0** 开始的字符串数组 `nums` ，其中每个字符串 **长度相等** 且只包含数字。
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再给你一个下标从 **0** 开始的二维整数数组 `queries` ，其中 <code>queries[i] = [k<sub>i</sub>, trim<sub>i</sub>]</code> 。对于每个 `queries[i]` ，你需要：
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* 将 `nums` 中每个数字 **裁剪** 到剩下 **最右边** <code>trim<sub>i</sub></code> 个数位。
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* 在裁剪过后的数字中，找到 `nums` 中第 <code>k<sub>i</sub></code> 小数字对应的 **下标** 。如果两个裁剪后数字一样大，那么下标 **更小** 的数字视为更小的数字。
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* 将 `nums` 中每个数字恢复到原本字符串。
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请你返回一个长度与 `queries` 相等的数组 `answer`，其中 `answer[i]`是第 `i` 次查询的结果。
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**提示：**
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* 裁剪到剩下最右边 `x` 个数位的意思是不断删除最左边的数位，直到剩下 `x` 个数位。
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* `nums` 中的字符串可能会有前导 0 。
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#### 示例 1:
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<pre>
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<strong>输入:</strong> nums = ["102","473","251","814"], queries = [[1,1],[2,3],[4,2],[1,2]]
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<strong>输出:</strong> [2,2,1,0]
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<strong>解释:</strong>
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1. 裁剪到只剩 1 个数位后，nums = ["2","3","1","4"] 。最小的数字是 1 ，下标为 2 。
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2. 裁剪到剩 3 个数位后，nums 没有变化。第 2 小的数字是 251 ，下标为 2 。
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3. 裁剪到剩 2 个数位后，nums = ["02","73","51","14"] 。第 4 小的数字是 73 ，下标为 1 。
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4. 裁剪到剩 2 个数位后，最小数字是 2 ，下标为 0 。
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注意，裁剪后数字 "02" 值为 2 。
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</pre>
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#### 示例 2:
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<pre>
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<strong>输入:</strong> nums = ["24","37","96","04"], queries = [[2,1],[2,2]]
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<strong>输出:</strong> [3,0]
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<strong>解释:</strong>
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1. 裁剪到剩 1 个数位，nums = ["4","7","6","4"] 。第 2 小的数字是 4 ，下标为 3 。
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有两个 4 ，下标为 0 的 4 视为小于下标为 3 的 4 。
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2. 裁剪到剩 2 个数位，nums 不变。第二小的数字是 24 ，下标为 0 。
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</pre>
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#### 提示:
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* `1 <= nums.length <= 100`
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* `1 <= nums[i].length <= 100`
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* `nums[i]` 只包含数字。
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* 所有 `nums[i].length` 的长度 **相同** 。
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* `1 <= queries.length <= 100`
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* `queries[i].length == 2`
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* <code>1 <= k<sub>i</sub> <= nums.length</code>
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* <code>1 <= trim<sub>i</sub> <= nums[i].length</code>
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**进阶：**你能使用 **基数排序算法** 解决此问题吗？这种解法的复杂度又是多少？
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## 题解 (Python)
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### 1. 题解
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```Python
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class Solution:
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def smallestTrimmedNumbers(self, nums: List[str], queries: List[List[int]]) -> List[int]:
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nums = [(nums[i], i) for i in range(len(nums))]
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trims = {}
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answer = [0] * len(queries)
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for i in range(len(queries)):
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if queries[i][1] not in trims:
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trims[queries[i][1]] = []
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trims[queries[i][1]].append((i, queries[i][0] - 1))
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for trim in range(1, len(nums[0][0]) + 1):
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count = [0] * 10
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tmp = [("", 0)] * len(nums)
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for num, _ in nums:
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count[ord(num[-trim]) - 48] += 1
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for i in range(1, 10):
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count[i] += count[i - 1]
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for num, i in nums[::-1]:
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count[ord(num[-trim]) - 48] -= 1
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tmp[count[ord(num[-trim]) - 48]] = (num, i)
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for i, k in trims.get(trim, []):
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answer[i] = tmp[k][1]
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nums = tmp
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return answer
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```

‎Problemset/2343-Query Kth Smallest Trimmed Number/Solution.py

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class Solution:
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def smallestTrimmedNumbers(self, nums: List[str], queries: List[List[int]]) -> List[int]:
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nums = [(nums[i], i) for i in range(len(nums))]
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trims = {}
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answer = [0] * len(queries)
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for i in range(len(queries)):
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if queries[i][1] not in trims:
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trims[queries[i][1]] = []
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trims[queries[i][1]].append((i, queries[i][0] - 1))
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for trim in range(1, len(nums[0][0]) + 1):
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count = [0] * 10
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tmp = [("", 0)] * len(nums)
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for num, _ in nums:
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count[ord(num[-trim]) - 48] += 1
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for i in range(1, 10):
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count[i] += count[i - 1]
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for num, i in nums[::-1]:
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count[ord(num[-trim]) - 48] -= 1
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tmp[count[ord(num[-trim]) - 48]] = (num, i)
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for i, k in trims.get(trim, []):
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answer[i] = tmp[k][1]
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nums = tmp
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return answer

‎README.md

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[2337][2337l]|[Move Pieces to Obtain a String][2337] |![rs]
11371137
[2341][2341l]|[Maximum Number of Pairs in Array][2341] |![rs]
11381138
[2342][2342l]|[Max Sum of a Pair With Equal Sum of Digits][2342] |![rs]
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[2343][2343l]|[Query Kth Smallest Trimmed Number][2343] |![py]
11391140
[2347][2347l]|[Best Poker Hand][2347] |![py]
11401141
[2348][2348l]|[Number of Zero-Filled Subarrays][2348] |![rs]
11411142
[2349][2349l]|[Design a Number Container System][2349] |![rs]
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[2337]:Problemset/2337-Move%20Pieces%20to%20Obtain%20a%20String/README.md#2337-move-pieces-to-obtain-a-string
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[2341]:Problemset/2341-Maximum%20Number%20of%20Pairs%20in%20Array/README.md#2341-maximum-number-of-pairs-in-array
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[2342]:Problemset/2342-Max%20Sum%20of%20a%20Pair%20With%20Equal%20Sum%20of%20Digits/README.md#2342-max-sum-of-a-pair-with-equal-sum-of-digits
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[2343]:Problemset/2343-Query%20Kth%20Smallest%20Trimmed%20Number/README.md#2343-query-kth-smallest-trimmed-number
23862388
[2347]:Problemset/2347-Best%20Poker%20Hand/README.md#2347-best-poker-hand
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[2348]:Problemset/2348-Number%20of%20Zero-Filled%20Subarrays/README.md#2348-number-of-zero-filled-subarrays
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[2349]:Problemset/2349-Design%20a%20Number%20Container%20System/README.md#2349-design-a-number-container-system
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[2337l]:https://leetcode.com/problems/move-pieces-to-obtain-a-string/
36343636
[2341l]:https://leetcode.com/problems/maximum-number-of-pairs-in-array/
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[2342l]:https://leetcode.com/problems/max-sum-of-a-pair-with-equal-sum-of-digits/
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[2343l]:https://leetcode.com/problems/query-kth-smallest-trimmed-number/
36363639
[2347l]:https://leetcode.com/problems/best-poker-hand/
36373640
[2348l]:https://leetcode.com/problems/number-of-zero-filled-subarrays/
36383641
[2349l]:https://leetcode.com/problems/design-a-number-container-system/

‎README_CN.md

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[2337][2337l]|[移动片段得到字符串][2337] |![rs]
11371137
[2341][2341l]|[数组能形成多少数对][2341] |![rs]
11381138
[2342][2342l]|[数位和相等数对的最大和][2342] |![rs]
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[2343][2343l]|[裁剪数字后查询第 K 小的数字][2343] |![py]
11391140
[2347][2347l]|[最好的扑克手牌][2347] |![py]
11401141
[2348][2348l]|[全 0 子数组的数目][2348] |![rs]
11411142
[2349][2349l]|[设计数字容器系统][2349] |![rs]
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23832384
[2337]:Problemset/2337-Move%20Pieces%20to%20Obtain%20a%20String/README_CN.md#2337-移动片段得到字符串
23842385
[2341]:Problemset/2341-Maximum%20Number%20of%20Pairs%20in%20Array/README_CN.md#2341-数组能形成多少数对
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[2342]:Problemset/2342-Max%20Sum%20of%20a%20Pair%20With%20Equal%20Sum%20of%20Digits/README_CN.md#2342-数位和相等数对的最大和
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[2343]:Problemset/2343-Query%20Kth%20Smallest%20Trimmed%20Number/README_CN.md#2343-裁剪数字后查询第-k-小的数字
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[2347]:Problemset/2347-Best%20Poker%20Hand/README_CN.md#2347-最好的扑克手牌
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[2348]:Problemset/2348-Number%20of%20Zero-Filled%20Subarrays/README_CN.md#2348-全-0-子数组的数目
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[2349]:Problemset/2349-Design%20a%20Number%20Container%20System/README_CN.md#2349-设计数字容器系统
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[2337l]:https://leetcode.cn/problems/move-pieces-to-obtain-a-string/
36343636
[2341l]:https://leetcode.cn/problems/maximum-number-of-pairs-in-array/
36353637
[2342l]:https://leetcode.cn/problems/max-sum-of-a-pair-with-equal-sum-of-digits/
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[2343l]:https://leetcode.cn/problems/query-kth-smallest-trimmed-number/
36363639
[2347l]:https://leetcode.cn/problems/best-poker-hand/
36373640
[2348l]:https://leetcode.cn/problems/number-of-zero-filled-subarrays/
36383641
[2349l]:https://leetcode.cn/problems/design-a-number-container-system/

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