package org.pilgrim.leetcode.y2021;

/*Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array.

Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity?

 

Example 1:

Input: nums = [3,0,1]
Output: 2
Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums.
Example 2:

Input: nums = [0,1]
Output: 2
Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums.
Example 3:

Input: nums = [9,6,4,2,3,5,7,0,1]
Output: 8
Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums.
Example 4:

Input: nums = [0]
Output: 1
Explanation: n = 1 since there is 1 number, so all numbers are in the range [0,1]. 1 is the missing number in the range since it does not appear in nums.
 

Constraints:

n == nums.length
1 <= n <= 104
0 <= nums[i] <= n
All the numbers of nums are unique.*/

public class MissingNumber {
    public int missingNumber(int[] nums) {
        if(nums == null){
            return -1;
        }
        
        int n = nums.length;
        
        //sort
        for(int i=0; i<n; i++){
            int k = nums[i];
            
            while(k != i && k != n){
                nums[i] = nums[k];
                nums[k] = k;
                k = nums[i];
            }
        }
        
        for(int i=0; i<n; i++){
            int k = nums[i];
            if(k == n){
                return i;
            }
        }
        
        return n;
    }
}
add MissingNumber