README_EN.md

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2343. Query Kth Smallest Trimmed Number

中文文档

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

Solution 1

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}