描述
You are given a 0-indexed array of strings nums, where each string is of equal length and consists of only digits. You are also given a 0-indexed 2D integer array queries where queries[i] = [ki, trimi]. For each queries[i], you need to:
- Trim each number in nums to its rightmost trimi digits.
- Determine the index of the kith smallest trimmed number in nums. If two trimmed numbers are equal, the number with the lower index is considered to be smaller.
- Reset each number in nums to its original length.
Return an array answer of the same length as queries, where answer[i] is the answer to the ith query. Note:
- To trim to the rightmost x digits means to keep removing the leftmost digit, until only x digits remain.
- Strings in nums may contain leading zeros.
Example 1:
Input: nums = ["102","473","251","814"], queries = [[1,1],[2,3],[4,2],[1,2]]
Output: [2,2,1,0]
Explanation:
1. After trimming to the last digit, nums = ["2","3","1","4"]. The smallest number is 1 at index 2.
2. Trimmed to the last 3 digits, nums is unchanged. The 2nd smallest number is 251 at index 2.
3. Trimmed to the last 2 digits, nums = ["02","73","51","14"]. The 4th smallest number is 73.
4. Trimmed to the last 2 digits, the smallest number is 2 at index 0.
Note that the trimmed number "02" is evaluated as 2.
Example 2:
Input: nums = ["24","37","96","04"], queries = [[2,1],[2,2]]
Output: [3,0]
Explanation:
1. Trimmed to the last digit, nums = ["4","7","6","4"]. The 2nd smallest number is 4 at index 3.
There are two occurrences of 4, but the one at index 0 is considered smaller than the one at index 3.
2. Trimmed to the last 2 digits, nums is unchanged. The 2nd smallest number is 24.
Note:
1 <= nums.length <= 100
1 <= nums[i].length <= 100
nums[i] consists of only digits.
All nums[i].length are equal.
1 <= queries.length <= 100
queries[i].length == 2
1 <= ki <= nums.length
1 <= trimi <= nums[i].length
解析
根据题意，给定一个 0 索引的字符串 nums 数组，其中每个字符串的长度相等且仅由数字组成。给定一个 0 索引的 2D 整数数组 queries ，其中 queries[i] = [ki, trimi]。 对于每个 queries[i]，您需要做以下的操作：
- 将 nums 中的每个数字修剪到剩下最右边的 trimi 个数字。
- 确定此时 nums 中第 k 个最小数的索引。 如果两个修剪后的数字相等，则认为具有较低索引的数字较小。
- 将 nums 中的每个数字重置为其原始模样。
返回与查询长度相同的数组答案，其中 answer[i] 是第 i 个查询的答案。需要注意的是：
- 修剪到最右边的 x 个数字意味着不断删除最左边的数字，直到只剩下 x 个数字。
- nums 中的字符串可能包含前导零。
这道题很简单，只需要按照题意写代码即可。
时间复杂度为 O(Q*M*NlogN)) ，空间复杂度为 O(N) ，其中 Q 是 queries 的长度，N 时 nums 的长度，M 是 nums[i] 的长度 。
解答
class Solution(object):
def smallestTrimmedNumbers(self, nums, queries):
"""
:type nums: List[str]
:type queries: List[List[int]]
:rtype: List[int]
"""
result = []
for k,t in queries:
tmp = [[n[-t:], i] for i,n in enumerate(nums)]
tmp.sort()
result.append(tmp[k-1][1])
return result
运行结果
251 / 251 test cases passed.
Status: Accepted
Runtime: 1299 ms
Memory Usage: 13.6 MB
原题链接
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