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.length1 <= n <= 1040 <= 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).
PythonJavaC++GoTypeScriptRustJavaScriptPHP
class Solution:
def missingNumber(self, nums: List[int]) -> int:
return reduce(xor, (i ^ v for i, v in enumerate(nums, 1)))(code-box)
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;
}
}(code-box)
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;
}
};(code-box)
func missingNumber(nums []int) (ans int) {
n := len(nums)
ans = n
for i, v := range nums {
ans ^= (i ^ v)
}
return
}(code-box)
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;
}(code-box)
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
}
}(code-box)
/**
* @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;
};(code-box)
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;
}
}(code-box)
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).
PythonJavaC++GoTypeScriptRustJavaScript
class Solution:
def missingNumber(self, nums: List[int]) -> int:
n = len(nums)
return (1 + n) * n // 2 - sum(nums)(code-box)
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;
}
}(code-box)
class Solution {
public:
int missingNumber(vector<int>& nums) {
int n = nums.size();
return (1 + n) * n / 2 - accumulate(nums.begin(), nums.end(), 0);
}
};(code-box)
func missingNumber(nums []int) (ans int) {
n := len(nums)
ans = n
for i, v := range nums {
ans += i - v
}
return
}(code-box)
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;
}(code-box)
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
}
}(code-box)
/**
* @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;
};(code-box)
