Description#
You have a set of integers s, which originally contains all the numbers from 1 to n. Unfortunately, due to some error, one of the numbers in s got duplicated to another number in the set, which results in repetition of one number and loss of another number.
You are given an integer array nums representing the data status of this set after the error.
Find the number that occurs twice and the number that is missing and return them in the form of an array.
Example 1:
Input: nums = [1,2,2,4]
Output: [2,3]
Example 2:
Input: nums = [1,1]
Output: [1,2]
Constraints:
2 <= nums.length <= 1041 <= nums[i] <= 104
Solutions#
Solution 1: Mathematics#
We denote $s_1$ as the sum of all numbers from $[1,..n]$, $s_2$ as the sum of the numbers in the array $nums$ after removing duplicates, and $s$ as the sum of the numbers in the array $nums$.
Then $s - s_2$ is the duplicate number, and $s_1 - s_2$ is the missing number.
The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $nums$. Extra space is needed to store the array after de-duplication.
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| class Solution:
def findErrorNums(self, nums: List[int]) -> List[int]:
n = len(nums)
s1 = (1 + n) * n // 2
s2 = sum(set(nums))
s = sum(nums)
return [s - s2, s1 - s2]
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| class Solution {
public int[] findErrorNums(int[] nums) {
int n = nums.length;
int s1 = (1 + n) * n / 2;
int s2 = 0;
Set<Integer> set = new HashSet<>();
int s = 0;
for (int x : nums) {
if (set.add(x)) {
s2 += x;
}
s += x;
}
return new int[] {s - s2, s1 - s2};
}
}
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| class Solution {
public:
vector<int> findErrorNums(vector<int>& nums) {
int n = nums.size();
int s1 = (1 + n) * n / 2;
int s2 = 0;
unordered_set<int> set(nums.begin(), nums.end());
for (int x : set) {
s2 += x;
}
int s = accumulate(nums.begin(), nums.end(), 0);
return {s - s2, s1 - s2};
}
};
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| func findErrorNums(nums []int) []int {
n := len(nums)
s1 := (1 + n) * n / 2
s2, s := 0, 0
set := map[int]bool{}
for _, x := range nums {
if !set[x] {
set[x] = true
s2 += x
}
s += x
}
return []int{s - s2, s1 - s2}
}
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| function findErrorNums(nums: number[]): number[] {
const n = nums.length;
const s1 = (n * (n + 1)) >> 1;
const s2 = [...new Set(nums)].reduce((a, b) => a + b);
const s = nums.reduce((a, b) => a + b);
return [s - s2, s1 - s2];
}
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| use std::collections::HashSet;
impl Solution {
pub fn find_error_nums(nums: Vec<i32>) -> Vec<i32> {
let n = nums.len() as i32;
let s1 = ((1 + n) * n) / 2;
let s2 = nums.iter().cloned().collect::<HashSet<i32>>().iter().sum::<i32>();
let s: i32 = nums.iter().sum();
vec![s - s2, s1 - s2]
}
}
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Solution 2: Hash Table#
We can also use a more intuitive method, using a hash table $cnt$ to count the occurrence of each number in the array $nums$.
Next, iterate through $x \in [1, n]$, if $cnt[x] = 2$, then $x$ is the duplicate number, if $cnt[x] = 0$, then $x$ is the missing number.
The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the array $nums$.
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| class Solution:
def findErrorNums(self, nums: List[int]) -> List[int]:
cnt = Counter(nums)
n = len(nums)
ans = [0] * 2
for x in range(1, n + 1):
if cnt[x] == 2:
ans[0] = x
if cnt[x] == 0:
ans[1] = x
return ans
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| class Solution {
public int[] findErrorNums(int[] nums) {
int n = nums.length;
Map<Integer, Integer> cnt = new HashMap<>(n);
for (int x : nums) {
cnt.merge(x, 1, Integer::sum);
}
int[] ans = new int[2];
for (int x = 1; x <= n; ++x) {
int t = cnt.getOrDefault(x, 0);
if (t == 2) {
ans[0] = x;
} else if (t == 0) {
ans[1] = x;
}
}
return ans;
}
}
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| class Solution {
public:
vector<int> findErrorNums(vector<int>& nums) {
int n = nums.size();
unordered_map<int, int> cnt;
for (int x : nums) {
++cnt[x];
}
vector<int> ans(2);
for (int x = 1; x <= n; ++x) {
if (cnt[x] == 2) {
ans[0] = x;
} else if (cnt[x] == 0) {
ans[1] = x;
}
}
return ans;
}
};
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| func findErrorNums(nums []int) []int {
n := len(nums)
cnt := map[int]int{}
for _, x := range nums {
cnt[x]++
}
ans := make([]int, 2)
for x := 1; x <= n; x++ {
if cnt[x] == 2 {
ans[0] = x
} else if cnt[x] == 0 {
ans[1] = x
}
}
return ans
}
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| function findErrorNums(nums: number[]): number[] {
const n = nums.length;
const cnt: Map<number, number> = new Map();
for (const x of nums) {
cnt.set(x, (cnt.get(x) || 0) + 1);
}
const ans: number[] = new Array(2).fill(0);
for (let x = 1; x <= n; ++x) {
const t = cnt.get(x) || 0;
if (t === 2) {
ans[0] = x;
} else if (t === 0) {
ans[1] = x;
}
}
return ans;
}
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| impl Solution {
pub fn find_error_nums(nums: Vec<i32>) -> Vec<i32> {
let mut xs = 0;
for (i, x) in nums.iter().enumerate() {
xs ^= ((i + 1) as i32) ^ x;
}
let mut a = 0;
let lb = xs & -xs;
for (i, x) in nums.iter().enumerate() {
if (((i + 1) as i32) & lb) != 0 {
a ^= (i + 1) as i32;
}
if (*x & lb) != 0 {
a ^= *x;
}
}
let b = xs ^ a;
for x in nums.iter() {
if *x == a {
return vec![a, b];
}
}
vec![b, a]
}
}
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Solution 3: Bit Operation#
According to the properties of the XOR operation, for integer $x$, we have $x \oplus x = 0$ and $x \oplus 0 = x$. Therefore, if we perform the XOR operation on all elements in the array $nums$ and all numbers $i \in [1, n]$, we can eliminate the numbers that appear twice, leaving only the XOR result of the missing number and the duplicate number, i.e., $xs = a \oplus b$.
Since these two numbers are not equal, there must be at least one bit in the XOR result that is $1$. We find the lowest bit of $1$ in the XOR result through the $lowbit$ operation, and then divide all numbers in the array $nums$ and all numbers $i \in [1, n]$ into two groups according to whether this bit is $1$. In this way, the two numbers are divided into different groups. The XOR result of one group of numbers is $a$, and the XOR result of the other group is $b$. These two numbers are the answers we are looking for.
Next, we only need to determine which of $a$ and $b$ is the duplicate number and which is the missing number. Therefore, iterate through the array $nums$, for the traversed number $x$, if $x=a$, then $a$ is the duplicate number, return $[a, b]$, otherwise, at the end of the iteration, return $[b, a]$.
The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$, only using a constant size of extra space.
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| class Solution:
def findErrorNums(self, nums: List[int]) -> List[int]:
xs = 0
for i, x in enumerate(nums, 1):
xs ^= i ^ x
a = 0
lb = xs & -xs
for i, x in enumerate(nums, 1):
if i & lb:
a ^= i
if x & lb:
a ^= x
b = xs ^ a
for x in nums:
if x == a:
return [a, b]
return [b, a]
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| class Solution {
public int[] findErrorNums(int[] nums) {
int n = nums.length;
int xs = 0;
for (int i = 1; i <= n; ++i) {
xs ^= i ^ nums[i - 1];
}
int lb = xs & -xs;
int a = 0;
for (int i = 1; i <= n; ++i) {
if ((i & lb) > 0) {
a ^= i;
}
if ((nums[i - 1] & lb) > 0) {
a ^= nums[i - 1];
}
}
int b = xs ^ a;
for (int i = 0; i < n; ++i) {
if (nums[i] == a) {
return new int[] {a, b};
}
}
return new int[] {b, a};
}
}
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| class Solution {
public:
vector<int> findErrorNums(vector<int>& nums) {
int n = nums.size();
int xs = 0;
for (int i = 1; i <= n; ++i) {
xs ^= i ^ nums[i - 1];
}
int lb = xs & -xs;
int a = 0;
for (int i = 1; i <= n; ++i) {
if (i & lb) {
a ^= i;
}
if (nums[i - 1] & lb) {
a ^= nums[i - 1];
}
}
int b = xs ^ a;
for (int i = 0; i < n; ++i) {
if (nums[i] == a) {
return {a, b};
}
}
return {b, a};
}
};
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| func findErrorNums(nums []int) []int {
xs := 0
for i, x := range nums {
xs ^= x ^ (i + 1)
}
lb := xs & -xs
a := 0
for i, x := range nums {
if (i+1)&lb != 0 {
a ^= (i + 1)
}
if x&lb != 0 {
a ^= x
}
}
b := xs ^ a
for _, x := range nums {
if x == a {
return []int{a, b}
}
}
return []int{b, a}
}
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| function findErrorNums(nums: number[]): number[] {
const n = nums.length;
let xs = 0;
for (let i = 1; i <= n; ++i) {
xs ^= i ^ nums[i - 1];
}
const lb = xs & -xs;
let a = 0;
for (let i = 1; i <= n; ++i) {
if (i & lb) {
a ^= i;
}
if (nums[i - 1] & lb) {
a ^= nums[i - 1];
}
}
const b = xs ^ a;
return nums.includes(a) ? [a, b] : [b, a];
}
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