LeetCode 1539. Kth Missing Positive Number

原题链接在这里：https://leetcode.com/problems/kth-missing-positive-number/description/

题目：

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?

题解：

Assume x is the final result, from 1 to x, there are m numbers not missing.

Try to find the m position.

Use the binary search, arr[mid] - mid - 1 is the missing numbers up to index mid.

if it is smaller than k, than m must be on the right, l = mid + 1. 

else, than m could on the left including current m. r = m.

Note: initialize r as arr.length.

Time Complexity: O(logn). n = arr.length.

Space: O(1).

AC Java:

 1 class Solution {
 2     public int findKthPositive(int[] arr, int k) {
 3         int l = 0;
 4         int r = arr.length;
 5         while(l < r){
 6             int mid = l + (r - l) / 2;
 7             if(arr[mid] - mid  - 1 < k){
 8                 l = mid + 1;
 9             }else{
10                 r = mid;
11             }
12         }
13 
14         return l + k;
15     }
16 }