LeetCode #440 — HARD

K-th Smallest in Lexicographical Order

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given two integers n and k, return the k^th lexicographically smallest integer in the range [1, n].

Example 1:

Input: n = 13, k = 2
Output: 10
Explanation: 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

Constraints:

1 <= k <= n <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given two integers n and k, return the kth lexicographically smallest integer in the range [1, n].

Baseline thinking

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

Pattern signal: Trie

Example 1

13
2

Example 2

1
1

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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

Largest constraint values

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 #440: K-th Smallest in Lexicographical Order
class Solution {
    private int n;

    public int findKthNumber(int n, int k) {
        this.n = n;
        long curr = 1;
        --k;
        while (k > 0) {
            int cnt = count(curr);
            if (k >= cnt) {
                k -= cnt;
                ++curr;
            } else {
                --k;
                curr *= 10;
            }
        }
        return (int) curr;
    }

    public int count(long curr) {
        long next = curr + 1;
        long cnt = 0;
        while (curr <= n) {
            cnt += Math.min(n - curr + 1, next - curr);
            next *= 10;
            curr *= 10;
        }
        return (int) cnt;
    }
}

// Accepted solution for LeetCode #440: K-th Smallest in Lexicographical Order
func findKthNumber(n int, k int) int {
	count := func(curr int) int {
		next := curr + 1
		cnt := 0
		for curr <= n {
			cnt += min(n-curr+1, next-curr)
			next *= 10
			curr *= 10
		}
		return cnt
	}
	curr := 1
	k--
	for k > 0 {
		cnt := count(curr)
		if k >= cnt {
			k -= cnt
			curr++
		} else {
			k--
			curr *= 10
		}
	}
	return curr
}

# Accepted solution for LeetCode #440: K-th Smallest in Lexicographical Order
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

// Accepted solution for LeetCode #440: K-th Smallest in Lexicographical Order
impl Solution {
    pub fn find_kth_number(n: i32, k: i32) -> i32 {
        fn count(mut curr: i64, n: i32) -> i32 {
            let mut next = curr + 1;
            let mut total = 0;
            let n = n as i64;
            while curr <= n {
                total += std::cmp::min(n - curr + 1, next - curr);
                curr *= 10;
                next *= 10;
            }
            total as i32
        }

        let mut curr = 1;
        let mut k = k - 1;

        while k > 0 {
            let cnt = count(curr as i64, n);
            if k >= cnt {
                k -= cnt;
                curr += 1;
            } else {
                k -= 1;
                curr *= 10;
            }
        }

        curr
    }
}

// Accepted solution for LeetCode #440: K-th Smallest in Lexicographical Order
function findKthNumber(n: number, k: number): number {
    function count(curr: number): number {
        let next = curr + 1;
        let cnt = 0;
        while (curr <= n) {
            cnt += Math.min(n - curr + 1, next - curr);
            curr *= 10;
            next *= 10;
        }
        return cnt;
    }

    let curr = 1;
    k--;

    while (k > 0) {
        const cnt = count(curr);
        if (k >= cnt) {
            k -= cnt;
            curr += 1;
        } else {
            k -= 1;
            curr *= 10;
        }
    }

    return curr;
}

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(L)

Space

O(N × L)

Approach Breakdown

HASH SET

O(N × L) time

O(N × L) space

Store all N words in a hash set. Each insert/lookup hashes the entire word of length L, giving O(L) per operation. Prefix queries require checking every stored word against the prefix — O(N × L) per prefix search. Space is O(N × L) for storing all characters.

TRIE

O(L) time

O(N × L) space

Each operation (insert, search, prefix) takes O(L) time where L is the word length — one node visited per character. Total space is bounded by the sum of all stored word lengths. Tries win over hash sets when you need prefix matching: O(L) prefix search vs. checking every stored word.

Shortcut: One node per character → O(L) per operation. Prefix queries are what make tries worthwhile.

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.