README_EN.md

268. Missing Number

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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.

 

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.

 

Constraints:

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

 

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

Solutions

Python3

class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        return reduce(xor, (i ^ v for i, v in enumerate(nums, 1)))

class Solution:
    def missingNumber(self, nums: List[int]) -> int:
        n = len(nums)
        return (1 + n) * n // 2 - sum(nums)

Java

class Solution {
    public int missingNumber(int[] nums) {
        int n = nums.length;
        int ans = n;
        for (int i = 0; i < n; ++i) {
            ans ^= (i ^ nums[i]);
        }
        return ans;
    }
}

class Solution {
    public int missingNumber(int[] nums) {
        int n = nums.length;
        int ans = n;
        for (int i = 0; i < n; ++i) {
            ans += i - nums[i];
        }
        return ans;
    }
}

C++

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int n = nums.size();
        int ans = n;
        for (int i = 0; i < n; ++i) {
            ans ^= (i ^ nums[i]);
        }
        return ans;
    }
};

class Solution {
public:
    int missingNumber(vector<int>& nums) {
        int n = nums.size();
        return (1 + n) * n / 2 - accumulate(nums.begin(), nums.end(), 0);
    }
};

Go

func missingNumber(nums []int) (ans int) {
	n := len(nums)
	ans = n
	for i, v := range nums {
		ans ^= (i ^ v)
	}
	return
}

func missingNumber(nums []int) (ans int) {
	n := len(nums)
	ans = n
	for i, v := range nums {
		ans += i - v
	}
	return
}

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var missingNumber = function (nums) {
    const n = nums.length;
    let ans = n;
    for (let i = 0; i < n; ++i) {
        ans ^= i ^ nums[i];
    }
    return ans;
};

/**
 * @param {number[]} nums
 * @return {number}
 */
var missingNumber = function (nums) {
    const n = nums.length;
    let ans = n;
    for (let i = 0; i < n; ++i) {
        ans += i - nums[i];
    }
    return ans;
};

PHP

class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function missingNumber($nums) {
        $n = count($nums);
        $sumN = ($n + 1) * $n / 2;
        for ($i = 0; $i < $n; $i++) {
            $sumN -= $nums[$i];
        }
        return $sumN;
    }
}

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