Question

1539. Kth Missing Positive Number

中文文档

Description

Given an array arr of positive integers sorted in a strictly increasing order, and an integer k.

Return _the_ kth _positive integer that is missing from this array._

 

Example 1:

Input: arr = [2,3,4,7,11], k = 5
Output: 9
Explanation: The missing positive integers are [1,5,6,8,9,10,12,13,...]. The 5th missing positive integer is 9.

Example 2:

Input: arr = [1,2,3,4], k = 2
Output: 6
Explanation: The missing positive integers are [5,6,7,...]. The 2nd missing positive integer is 6.

 

Constraints:

• 1 <= arr.length <= 1000
• 1 <= arr[i] <= 1000
• 1 <= k <= 1000
• arr[i] < arr[j] for 1 <= i < j <= arr.length

 

Follow up:

Could you solve this problem in less than O(n) complexity?

Solutions

Solution 1

class Solution:
  def findKthPositive(self, arr: List[int], k: int) -> int:
    if arr[0] > k:
      return k
    left, right = 0, len(arr)
    while left < right:
      mid = (left + right) >> 1
      if arr[mid] - mid - 1 >= k:
        right = mid
      else:
        left = mid + 1
    return arr[left - 1] + k - (arr[left - 1] - (left - 1) - 1)

class Solution {
  public int findKthPositive(int[] arr, int k) {
    if (arr[0] > k) {
      return k;
    }
    int left = 0, right = arr.length;
    while (left < right) {
      int mid = (left + right) >> 1;
      if (arr[mid] - mid - 1 >= k) {
        right = mid;
      } else {
        left = mid + 1;
      }
    }
    return arr[left - 1] + k - (arr[left - 1] - (left - 1) - 1);
  }
}

class Solution {
public:
  int findKthPositive(vector<int>& arr, int k) {
    if (arr[0] > k) return k;
    int left = 0, right = arr.size();
    while (left < right) {
      int mid = (left + right) >> 1;
      if (arr[mid] - mid - 1 >= k)
        right = mid;
      else
        left = mid + 1;
    }
    return arr[left - 1] + k - (arr[left - 1] - (left - 1) - 1);
  }
};

func findKthPositive(arr []int, k int) int {
  if arr[0] > k {
    return k
  }
  left, right := 0, len(arr)
  for left < right {
    mid := (left + right) >> 1
    if arr[mid]-mid-1 >= k {
      right = mid
    } else {
      left = mid + 1
    }
  }
  return arr[left-1] + k - (arr[left-1] - (left - 1) - 1)
}