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1539. Kth Missing Positive Number

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

Binary search.

  • 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);
    }
};

  • 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)


  • 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)
}

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