Formatted question description: https://leetcode.ca/all/440.html

440. K-th Smallest in Lexicographical Order

Level

Hard

Description

Given integers n and k, find the lexicographically k-th smallest integer in the range from 1 to n.

Note: 1 ≤ k ≤ n ≤ 10⁹.

Example:

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.

Solution

Starting from 1, let curr be the current prefix, and the next prefix is curr + 1. Count the number of integers with prefix curr. If the number of integers with prefix curr is greater than k, then set curr = curr * 10 and use the updated curr to calculate the next prefix curr + 1. Otherwise, subtract the number of integers from k, and set curr = curr + 1. Repeat the process until k becomes 0. Finally, return curr.

• Java
• C++
• Python

• class Solution {
      public int findKthNumber(int n, int k) {
          int curr = 1;
          k--;
          while (k > 0) {
              int steps = getSteps(n, curr, curr + 1);
              if (steps > k) {
                  curr *= 10;
                  k--;
              } else {
                  curr++;
                  k -= steps;
              }
          }
          return curr;
      }
  
      public int getSteps(int n, long curr, long next) {
          int steps = 0;
          while (curr <= n) {
              steps += Math.min(n + 1, next) - curr;
              curr *= 10;
              next *= 10;
          }
          return steps;
      }
  }
  

• Todo
  

• class Solution(object):
    # naive pre-order traversal on denary tree
    def _findKthNumber(self, n, k):
      """
      :type n: int
      :type k: int
      :rtype: int
      """
  
      def dfs(cur, n):
        if self.k == 0:
          return cur
        self.k -= 1
        if cur == 0:
          for i in range(1, 10):
            if i > n:
              break
            ret = dfs(i, n)
            if ret:
              return ret
        else:
          for i in range(0, 10):
            if cur * 10 + i > n:
              break
            ret = dfs(cur * 10 + i, n)
            if ret:
              return ret
  
      self.k = k
      return dfs(0, n)
  
    # optimized solution
    def findKthNumber(self, n, k):
      def getGap(n, ans):
        gap = 0
        start = ans
        end = start + 1
        while start <= n:
          gap += max(0, min(n + 1, end) - start)
          start *= 10
          end *= 10
        return gap
  
      ans = 1
      k -= 1
      while k > 0:
        gap = getGap(n, ans)
        if gap <= k:
          ans += 1
          k -= gap
        else:
          ans *= 10
          k -= 1
      return ans
  
  

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