Question

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

Solution 1: Bitwise Operation

The XOR operation has the following properties:

• Any number XOR 0 is still the original number, i.e., $x \oplus 0 = x$;
• Any number XOR itself is 0, i.e., $x \oplus x = 0$;

Therefore, we can traverse the array, perform XOR operation between each element and the numbers $[0,..n]$, and the final result will be the missing number.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

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

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

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

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

impl Solution {
  pub fn missing_number(nums: Vec<i32>) -> i32 {
    let n = nums.len() as i32;
    let mut ans = n;
    for (i, v) in nums.iter().enumerate() {
      ans ^= (i as i32) ^ v;
    }
    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;
};

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;
  }
}

Solution 2: Mathematics

We can also solve this problem using mathematics. By calculating the sum of $[0,..n]$, subtracting the sum of all numbers in the array, we can obtain the missing number.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

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

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

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

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

impl Solution {
  pub fn missing_number(nums: Vec<i32>) -> i32 {
    let n = nums.len() as i32;
    let mut ans = n;
    for (i, &v) in nums.iter().enumerate() {
      ans += (i as i32) - v;
    }
    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;
};