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Formatted question description: https://leetcode.ca/all/2343.html

# 2343. Query Kth Smallest Trimmed Number

## Description

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.


Constraints:

• 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

Follow up: Could you use the Radix Sort Algorithm to solve this problem? What will be the complexity of that solution?

## Solutions

• class Solution {
public int[] smallestTrimmedNumbers(String[] nums, int[][] queries) {
int n = nums.length;
int m = queries.length;
int[] ans = new int[m];
String[][] t = new String[n][2];
for (int i = 0; i < m; ++i) {
int k = queries[i][0], trim = queries[i][1];
for (int j = 0; j < n; ++j) {
t[j] = new String[] {nums[j].substring(nums[j].length() - trim), String.valueOf(j)};
}
Arrays.sort(t, (a, b) -> {
int x = a[0].compareTo(b[0]);
return x == 0 ? Long.compare(Integer.valueOf(a[1]), Integer.valueOf(b[1])) : x;
});
ans[i] = Integer.valueOf(t[k - 1][1]);
}
return ans;
}
}

• class Solution {
public:
vector<int> smallestTrimmedNumbers(vector<string>& nums, vector<vector<int>>& queries) {
int n = nums.size();
vector<pair<string, int>> t(n);
vector<int> ans;
for (auto& q : queries) {
int k = q[0], trim = q[1];
for (int j = 0; j < n; ++j) {
t[j] = {nums[j].substr(nums[j].size() - trim), j};
}
sort(t.begin(), t.end());
ans.push_back(t[k - 1].second);
}
return ans;
}
};

• class Solution:
def smallestTrimmedNumbers(
self, nums: List[str], queries: List[List[int]]
) -> List[int]:
ans = []
for k, trim in queries:
t = sorted((v[-trim:], i) for i, v in enumerate(nums))
ans.append(t[k - 1][1])
return ans


• func smallestTrimmedNumbers(nums []string, queries [][]int) []int {
type pair struct {
s string
i int
}
ans := make([]int, len(queries))
t := make([]pair, len(nums))
for i, q := range queries {
for j, s := range nums {
t[j] = pair{s[len(s)-q[1]:], j}
}
sort.Slice(t, func(i, j int) bool { a, b := t[i], t[j]; return a.s < b.s || a.s == b.s && a.i < b.i })
ans[i] = t[q[0]-1].i
}
return ans
}