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440. K-th Smallest in Lexicographical Order 👍

  • Time: $O(\log^2 n)$
  • Space: $O(1)$
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class Solution {
 public:
  int findKthNumber(int n, int k) {
    long ans = 1;

    for (int i = 1; i < k;) {
      const long gap = getGap(ans, ans + 1, n);
      if (i + gap <= k) {
        i += gap;
        ++ans;
      } else {
        ++i;
        ans *= 10;
      }
    }

    return ans;
  }

 private:
  long getGap(long a, long b, long n) {
    long gap = 0;
    while (a <= n) {
      gap += min(n + 1, b) - a;
      a *= 10;
      b *= 10;
    }
    return gap;
  };
};

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class Solution {
  public int findKthNumber(int n, int k) {
    long ans = 1;

    for (int i = 1; i < k;) {
      final long gap = getGap(ans, ans + 1, n);
      if (i + gap <= k) {
        i += gap;
        ++ans;
      } else {
        ++i;
        ans *= 10;
      }
    }

    return (int) ans;
  }

  private long getGap(long a, long b, long n) {
    long gap = 0;
    while (a <= n) {
      gap += Math.min(n + 1, b) - a;
      a *= 10;
      b *= 10;
    }
    return gap;
  }
}

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class Solution:
  def findKthNumber(self, n: int, k: int) -> int:
    ans = 1

    i = 1
    while i < k:
      gap = self._getGap(ans, ans + 1, n)
      if i + gap <= k:
        i += gap
        ans += 1
      else:
        i += 1
        ans *= 10

    return ans

  def _getGap(self, a: int, b: int, n: int) -> int:
    gap = 0
    while a <= n:
      gap += min(n + 1, b) - a
      a *= 10
      b *= 10
    return gap