LeetCode #2343 — MEDIUM

Query Kth Smallest Trimmed Number

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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] = [k_i, trim_i]. For each queries[i], you need to:

Trim each number in nums to its rightmost trim_i digits.
Determine the index of the k_i^th 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 i^th 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 2^nd smallest number is 251 at index 2.
3. Trimmed to the last 2 digits, nums = ["02","73","51","14"]. The 4^th 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 2^nd 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 2^nd 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 <= k_i <= nums.length
1 <= trim_i <= nums[i].length

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array

Example 1

["102","473","251","814"]
[[1,1],[2,3],[4,2],[1,2]]

Example 2

["24","37","96","04"]
[[2,1],[2,2]]

Step 02

Core Insight

What unlocks the optimal approach

Run a simulation to follow the requirement of each query.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2343: Query Kth Smallest Trimmed Number
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;
    }
}

// Accepted solution for LeetCode #2343: Query Kth Smallest Trimmed Number
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
}

# Accepted solution for LeetCode #2343: Query Kth Smallest Trimmed Number
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

// Accepted solution for LeetCode #2343: Query Kth Smallest Trimmed Number
/**
 * [2343] Query Kth Smallest Trimmed Number
 *
 * 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 ki^th 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 i^th 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 2^nd smallest number is 251 at index 2.
 * 3. Trimmed to the last 2 digits, nums = ["02","73","51","14"]. The 4^th 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 2^nd 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 2^nd 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?
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/query-kth-smallest-trimmed-number/
// discuss: https://leetcode.com/problems/query-kth-smallest-trimmed-number/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn smallest_trimmed_numbers(nums: Vec<String>, queries: Vec<Vec<i32>>) -> Vec<i32> {
        vec![]
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2343_example_1() {
        let nums = vec_string!["102", "473", "251", "814"];
        let queries = vec![vec![1, 1], vec![2, 3], vec![4, 2], vec![1, 2]];

        let result = vec![2, 2, 1, 0];

        assert_eq!(Solution::smallest_trimmed_numbers(nums, queries), result);
    }

    #[test]
    #[ignore]
    fn test_2343_example_2() {
        let nums = vec_string!["24", "37", "96", "04"];
        let queries = vec![vec![2, 1], vec![2, 2]];

        let result = vec![3, 0];

        assert_eq!(Solution::smallest_trimmed_numbers(nums, queries), result);
    }
}

// Accepted solution for LeetCode #2343: Query Kth Smallest Trimmed Number
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2343: Query Kth Smallest Trimmed Number
// 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;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(m × n × log n × s)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.