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.
Solution
class Solution {
public int missingNumber(int[] nums) {
int n = nums.length;
int sum = 0;
for (int num: nums) {
sum += num;
}
return ((n + 1) * n) / 2 - sum;
}
// 0 1 2 3 4 5 6 7 8 9
// 10 * 9 / 2 = 45
}
Solution 2022-01-30
class Solution {
public int missingNumber(int[] nums) {
int sum = 0;
for (int num: nums) {
sum += num;
}
int n = nums.length;
return n * (n + 1) / 2 - sum;
}
}