268. Missing Number

Description

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.

Constraints

• n == nums.length

• 1 <= n <= 104

• 0 <= nums[i] <= n

• All the numbers of nums are unique.

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

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.

Solutions

Solution 1

/**
 * Time complexity : O(NlogN)
 * Space complexity : O(1)
 */

class Solution {
    public int missingNumber(int[] nums) {
        Arrays.sort(nums);

        // Ensure that n is at the last index
        if (nums[nums.length-1] != nums.length) {
            return nums.length;
        }
        // Ensure that 0 is at the first index
        else if (nums[0] != 0) {
            return 0;
        }

        // If we get here, then the missing number is on the range (0, n)
        for (int i = 1; i < nums.length; i++) {
            int expectedNum = nums[i-1] + 1;
            if (nums[i] != expectedNum) {
                return expectedNum;
            }
        }

        // Array was not missing any numbers
        return -1;
    }
}

Solution 2

/**
 * Time complexity : O(n)
 * Space complexity : O(n)
 */
 
 class Solution {
    public int missingNumber(int[] nums) {
        Set<Integer> numSet = new HashSet<Integer>();
        for (int num : nums) numSet.add(num);

        int expectedNumCount = nums.length + 1;
        for (int number = 0; number < expectedNumCount; number++) {
            if (!numSet.contains(number)) {
                return number;
            }
        }
        return -1;
    }
}

Solution 3

/**
 * Time complexity : O(n)
 * Space complexity : O(1)
 */
 
 class Solution {
    public int missingNumber(int[] nums) {
        int expectedSum = nums.length*(nums.length + 1)/2;
        int actualSum = 0;
        for (int num : nums) actualSum += num;
        return expectedSum - actualSum;
    }
}

Follow up

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