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:
nums to its rightmost trimi digits.kith smallest trimmed number in nums. If two trimmed numbers are equal, the number with the lower index is considered to be smaller.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:
x digits means to keep removing the leftmost digit, until only x digits remain.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 <= 1001 <= nums[i].length <= 100nums[i] consists of only digits.nums[i].length are equal.1 <= queries.length <= 100queries[i].length == 21 <= ki <= nums.length1 <= 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?
Problem Overview: You are given an array of equal-length numeric strings. Each query asks you to trim every number to its last trim digits, then return the index of the k-th smallest trimmed value. If two trimmed numbers are equal, the smaller original index wins. Each query works independently, so the challenge is efficiently comparing many trimmed substrings.
Approach 1: Simple Sorting Approach (O(q · n log n) time, O(n) space)
The direct approach processes every query independently. For a query [k, trim], iterate through the array and extract the last trim digits from each string using substring operations. Store pairs of (trimmed_value, index). Sort this list lexicographically by the trimmed value, and break ties using the index. The answer for the query is the index at position k-1 after sorting.
This method relies on standard sorting and works well when the number of queries or numbers is small. The downside is repeated work: the same trims may be recomputed many times. If there are q queries and n numbers, sorting for each query leads to O(q · n log n) time. Still, the logic is straightforward and easy to implement in any language.
Approach 2: Optimized Bucket Sort Method (Radix Style) (O(m · n) time, O(n) space)
A faster solution observes that trimming always keeps the suffix of the number. Instead of recomputing trims for every query, process digits from right to left using a radix-style technique. Maintain an ordered list of indices representing numbers sorted by their last t digits. For each additional digit position, perform a stable bucket/counting sort based on the next digit to the left.
Because digits range from 0–9, each step distributes indices into 10 buckets, preserving previous order. After processing t digits, the list represents numbers sorted by their last t digits. Store this ordering so queries asking for that trim length can directly access the k-1 position. This technique is essentially a specialized radix sort applied to numeric strings.
The preprocessing runs for at most the number length m, and each pass processes all n numbers, giving O(m · n) time. Query answers become constant time lookups. The idea combines properties of arrays and stable bucket sorting to avoid repeated comparisons.
Recommended for interviews: Start by describing the simple sorting approach. It demonstrates understanding of trimming logic, lexicographic comparison, and tie-breaking with indices. Then explain the radix-style bucket optimization. Interviewers typically expect the optimized idea because it avoids redundant sorting and shows familiarity with digit-based sorting techniques.
The simple approach involves trimming the numbers based on each query and then sorting them to find the k-th smallest element. We will trim each number, pair it with its original index, sort the list of pairs, and select the k-th smallest element by considering the first k elements of the sorted list.
This C solution works by first trimming the numbers based on the given number of digits specified in each query, then sorting the resultant numbers while keeping track of their original positions. We use the qsort function for sorting the trimmed numbers with their original indices, ensuring any ties during sorting are resolved using the original indices.
Time Complexity: O(Q * N * log(N)), where Q is the number of queries, and N is the number of numbers in the input list.
Space Complexity: O(N), which is needed to store the trimmed numbers for comparison and the original indices.
For this approach, we leverage the Radix Sort when nums is restricted by the constraints given the uniformity and limited set size, specifically using the properties of counting sort, given that precision can be limited to 10 digits (0-9). We'll create buckets to sort the numbers efficiently for each trim and query.
In this C example, we use radix sort, which is a non-comparison-based sorting algorithm that groups numbers (or strings) based on digits. The solution applies a counting sort for each digit position, starting from the least significant digit. Although radix sort typically operates efficiently with integer representations, constraints on string lengths allow this adaptation.
Time Complexity: O(D * (N + B)), where D is the maximum number of digits in nums, N is the number of strings, and B is the maximum value within a single bucket.
Space Complexity: O(N), for storing temporary ordering.
According to the problem description, we can simulate the cropping process, then sort the cropped strings, and finally find the corresponding number based on the index.
The time complexity is O(m times n times log n times s), and the space complexity is O(n). Here, m and n are the lengths of nums and queries respectively, and s is the length of the string nums[i].
| Approach | Complexity |
|---|---|
| Simple Sorting Approach | Time Complexity: O(Q * N * log(N)), where Q is the number of queries, and N is the number of numbers in the input list. |
| Optimized Bucket Sort Method | Time Complexity: O(D * (N + B)), where D is the maximum number of digits in nums, N is the number of strings, and B is the maximum value within a single bucket. |
| Simulation | — |
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Simple Sorting | O(q · n log n) | O(n) | Best for quick implementation or when constraints are small |
| Bucket Sort (Radix Style) | O(m · n) | O(n) | Preferred for large inputs or many queries; avoids repeated sorting |
Leetcode Weekly contest 302 - Medium - Query Kth Smallest Trimmed Number • Prakhar Agrawal • 1,131 views views
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