# 440. K-th Smallest in Lexicographical Order

## Description

Given two integers n and k, return the kth 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

## Solutions

• 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;
}
}

• class Solution {
public:
int n;

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

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

• 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


• 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
}