LeetCode in Kotlin
440. K-th Smallest in Lexicographical Order
Hard
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
Solution
@Suppress("NAME_SHADOWING")
class Solution {
fun findKthNumber(n: Int, k: Int): Int {
var k = k
var curr = 1
k = k - 1
while (k > 0) {
val steps = calSteps(n, curr.toLong(), curr + 1L)
if (steps <= k) {
curr += 1
k -= steps
} else {
curr *= 10
k -= 1
}
}
return curr
}
// use long in case of overflow
private fun calSteps(n: Int, n1: Long, n2: Long): Int {
var n1 = n1
var n2 = n2
var steps: Long = 0
while (n1 <= n) {
steps += Math.min(n + 1L, n2) - n1
n1 *= 10
n2 *= 10
}
return steps.toInt()
}
}