#440

Hard

auto_awesomeTrie-driven solution strategy

LeetCode Problem Workspace

K-th Smallest in Lexicographical Order

Find the k-th lexicographically smallest integer in a range using a Trie-based approach.

Problem Statement

Given two integers n and k, return the k-th lexicographically smallest integer in the range [1, n]. The lexicographical order of integers is similar to dictionary order, where numbers are ordered based on their digits, starting from 1. For example, the lexicographical order for numbers up to 13 is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9].

The task requires an efficient algorithm to find the k-th smallest number without generating all the numbers from 1 to n. A direct iteration approach would be too slow, especially for large values of n and k. Instead, a Trie-based strategy provides an optimized solution, leveraging the structure of the numbers and their lexicographical order to solve the problem in logarithmic time.

Examples

Example 1

Input: n = 13, k = 2

Output: 10

The lexicographical order is [1, 10, 11, 12, 13, 2, 3, 4, 5, 6, 7, 8, 9], so the second smallest number is 10.

Example 2

Input: n = 1, k = 1

Output: 1

Example details omitted.

Constraints

1 <= k <= n <= 109

Solution Approach

Trie-based Strategy

This problem can be efficiently solved using a Trie-driven approach, where the numbers are stored in a Trie-like structure. The idea is to navigate through this structure to count how many numbers exist that are lexicographically smaller than the current candidate. Using this counting mechanism, we can find the k-th smallest number without explicitly generating all the numbers.

Count Function

To traverse the Trie, a counting function is used to determine how many numbers in the lexicographical order fall within a given prefix. This function simulates the numbers formed by a specific prefix and counts them by extending the prefix with digits. By progressively refining the search space using this function, we can efficiently locate the k-th smallest number.

Efficient Search Process

Starting from the number 1, the algorithm iteratively tries different digits at each step, counting the number of possible valid numbers under the current prefix. The key is to adjust the current candidate number based on the count, efficiently narrowing down the k-th smallest number in the lexicographical order without generating all numbers in the range.

Complexity Analysis

Metric	Value
Time	O(\log(n)^2)
Space	O(1)

The time complexity of the solution is O(log(n)^2), which comes from traversing the Trie-like structure and repeatedly counting the numbers with a certain prefix. The space complexity is O(1), as the solution does not require additional space proportional to the size of n, aside from a few variables used for counting and tracking the current state of the search.

What Interviewers Usually Probe

Can the candidate describe a method that avoids iterating through all numbers from 1 to n?
Did the candidate implement an efficient counting function that works without generating all possible numbers?
Does the candidate understand how a Trie-driven approach can optimize the search for the k-th smallest number?

Common Pitfalls or Variants

Common pitfalls

Not implementing the Trie structure efficiently, leading to excessive time complexity.
Misunderstanding the counting mechanism and generating all numbers explicitly instead of using the prefix-based counting approach.
Failing to consider edge cases, such as when k is at the upper or lower bound of the range.

Follow-up variants

Solving the problem with a simple iterative approach (not optimized).
Optimizing the approach further to handle extremely large values of n and k by leveraging more advanced Trie techniques.
Adapting the algorithm to handle different ranges or non-integer inputs in a lexicographical order.

FAQ

How does a Trie help solve the K-th Smallest in Lexicographical Order problem?

A Trie allows us to simulate the lexicographical order of numbers and efficiently count how many numbers are smaller than a given candidate, enabling us to find the k-th smallest number without generating all numbers explicitly.

What is the time complexity of the Trie-based approach?

The time complexity is O(log(n)^2), as it involves traversing a Trie-like structure and counting numbers with each prefix.

Why can't we generate all numbers from 1 to n to solve this problem?

Generating all numbers from 1 to n would be too slow, especially when n is very large. Instead, the Trie-based approach allows us to find the k-th smallest number more efficiently.

What is the key idea behind the Trie-based strategy for this problem?

The key idea is to navigate through a Trie-like structure, counting how many numbers are smaller than the current prefix, and adjusting the search to find the k-th smallest number without generating all the numbers.

How can I handle edge cases for small values of k and n?

Edge cases can be handled by ensuring that the counting function correctly accounts for the smallest and largest possible values of k and n, and adjusting the candidate number appropriately.

terminal

Solution

Solution 1: Trie-Based Counting + Greedy Construction

The problem asks for the \$k\$-th smallest number in the range $[1, n]$ when all numbers are sorted in **lexicographical order**. Since $n$ can be as large as $10^9$, we cannot afford to generate and sort all the numbers explicitly. Instead, we adopt a strategy based on **greedy traversal over a conceptual Trie**.

class Solution:
    def findKthNumber(self, n: int, k: int) -> int:
        def count(curr):
            next, cnt = curr + 1, 0
            while curr <= n:
                cnt += min(n - curr + 1, next - curr)
                next, curr = next * 10, curr * 10
            return cnt

        curr = 1
        k -= 1
        while k:
            cnt = count(curr)
            if k >= cnt:
                k -= cnt
                curr += 1
            else:
                k -= 1
                curr *= 10
        return curr

Continue Practicing

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