LeetCode 1060. Missing Element in Sorted Array

原题链接在这里：https://leetcode.com/problems/missing-element-in-sorted-array/

题目：

Given a sorted array A of unique numbers, find the K-th missing number starting from the leftmost number of the array.

Example 1:

Input: A = [4,7,9,10], K = 1
Output: 5
Explanation: 
The first missing number is 5.

Example 2:

Input: A = [4,7,9,10], K = 3
Output: 8
Explanation: 
The missing numbers are [5,6,8,...], hence the third missing number is 8.

Example 3:

Input: A = [1,2,4], K = 3
Output: 6
Explanation: 
The missing numbers are [3,5,6,7,...], hence the third missing number is 6.

Note:

1. 1 <= A.length <= 50000
2. 1 <= A[i] <= 1e7
3. 1 <= K <= 1e8

题解：

If the missing numbers count of the whole array < k, then missing number must be after nums[n-1].  res = nums[n-1] + missingCount.

Otherwise, need to find out the starting index to calculate the missing number.

Use binary search to have mid as candidate. 

If missing count < k, then must fall on the right side. l = mid + 1.

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

Space: O(1).

AC Java:

class Solution {
    public int missingElement(int[] nums, int k) {
        int n = nums.length;
        if(nums[n - 1] - nums[0] - (n - 1 - 0) < k){
            return nums[n - 1] + k - missCount(nums, n - 1);
        }
        
        int l = 0;
        int r = n - 1;
        
        while(l < r){
            int mid = l + (r - l) / 2;
            if(missCount(nums, mid) < k){
                l = mid + 1;
            }else{
                r = mid;
            }
        }
        
        return nums[l - 1] + k - missCount(nums, l - 1);
    }
    
    private int missCount(int [] nums, int mid){
        return nums[mid] - nums[0] - mid;
    }
}