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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

• Java
• C++
• Python
• Go

• 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
  }
  

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