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Kth Smallest Serial by Its Tail

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Trains the technique from

LeetCode 2343Query Kth Smallest Trimmed Number

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A batch of serials is given as nums, each a string of digits, and all of them the same length.

Each query queries[j] = [k, trim] asks you to keep only the rightmost trim digits of every serial, then find the k-th smallest of those trimmed serials, counting from 1. Trimmed serials are compared as text, so leading zeros count, and when two are equal the one whose original serial appears earlier in nums is treated as smaller.

Return an array answer where answer[j] is the position in nums of the serial the j-th query picks out. Queries are independent and never change nums.

Examples

Example 1

Input: nums = ["102", "473", "251", "814"], queries = [[1, 1], [2, 3], [4, 2], [1, 2]]
Output: [2, 2, 1, 0]

Keeping one digit gives "2", "3", "1", "4", whose smallest is at position 2. Keeping three digits leaves the serials whole, and the second smallest of those is at position 2 as well.

Example 2

Input: nums = ["119", "219", "319"], queries = [[1, 2], [2, 2], [3, 2]]
Output: [0, 1, 2]

Keeping two digits gives "19" for all three serials, so the tie is settled by the order they appear in.

Example 3

Input: nums = ["10", "01"], queries = [[1, 2], [2, 2]]
Output: [1, 0]

Compared as text, "01" comes before "10", so the first query picks position 1 and the second picks position 0.

Constraints

1 <= nums.length <= 100
1 <= nums[0].length <= 100
nums[i] consists of digits only, and every serial has the same length
1 <= queries.length <= 100
queries[j].length == 2
1 <= queries[j][0] <= 100
1 <= queries[j][1] <= 100
Every query satisfies k <= nums.length and trim <= nums[0].length

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def kth_by_tail(nums: list[str], queries: list[list[int]]) -> list[int]:

Java

public int[] kthByTail(List<String> nums, int[][] queries)

The intended approach

Time O(q * n log n * trim) for q queries over n serials.Space O(n * trim) for the pairs of one query.

The trimmed serials all have the same length, so comparing them as text is the same as comparing them digit by digit from the left, and no numeric conversion is needed. For a query, pair each serial's tail with its position, sort those pairs, and read off the entry the query asks for. Sorting on the pair settles equal tails by position, which is the stated tie-break, and Python's sort is stable so the original order is preserved anyway.

Pythonentry point: kth_by_tail

def kth_by_tail(nums, queries):
    out = []
    for k, trim in queries:
        pairs = sorted((nums[i][-trim:], i) for i in range(len(nums)))
        out.append(pairs[k - 1][1])
    return out

The obvious version

For each query, find the answer by selection rather than sorting: repeatedly walk every serial not yet taken, pick the smallest tail breaking ties by position, and stop after k picks.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_kth_by_tail(nums, queries):
    out = []
    for k, trim in queries:
        taken = [False] * len(nums)
        pick = -1
        for _ in range(k):
            pick = -1
            for i in range(len(nums)):
                if taken[i]:
                    continue
                if pick < 0:
                    pick = i
                    continue
                a = nums[i][len(nums[i]) - trim:]
                b = nums[pick][len(nums[pick]) - trim:]
                if a < b or (a == b and i < pick):
                    pick = i
            taken[pick] = True
        out.append(pick)
    return out

How this problem was checked

Reference solution executed against 10 test cases — 10 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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