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 <= 109class Solution {
public int findKthNumber(int n, int k) {
int curr = 1;
k--;
while (k > 0) {
int step = countSteps(n, curr, curr + 1);
// If the steps are less than or equal to k, we skip this prefix's subtree
if (step <= k) {
// Move to the next prefix and decrease k by the number of steps we skip
curr++;
k -= step;
} else {
// Move to the next level of the tree and decrement k by 1
curr *= 10;
k--;
}
}
return curr;
}
// To count how many numbers exist between prefix1 and prefix2
private int countSteps(int n, long prefix1, long prefix2) {
int steps = 0;
while (prefix1 <= n) {
steps += Math.min(n + 1, prefix2) - prefix1;
prefix1 *= 10;
prefix2 *= 10;
}
return steps;
}
}