# 448. Find All Numbers Disappeared in an Array

## Description

Given an array nums of n integers where nums[i] is in the range [1, n], return an array of all the integers in the range [1, n] that do not appear in nums.

Example 1:

Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]


Example 2:

Input: nums = [1,1]
Output: [2]


Constraints:

• n == nums.length
• 1 <= n <= 105
• 1 <= nums[i] <= n

Follow up: Could you do it without extra space and in O(n) runtime? You may assume the returned list does not count as extra space.

## Solutions

• class Solution {
public List<Integer> findDisappearedNumbers(int[] nums) {
int n = nums.length;
for (int x : nums) {
int i = Math.abs(x) - 1;
if (nums[i] > 0) {
nums[i] *= -1;
}
}
List<Integer> ans = new ArrayList<>();
for (int i = 0; i < n; i++) {
if (nums[i] > 0) {
}
}
return ans;
}
}

• class Solution {
public:
vector<int> findDisappearedNumbers(vector<int>& nums) {
int n = nums.size();
for (int& x : nums) {
int i = abs(x) - 1;
if (nums[i] > 0) {
nums[i] = -nums[i];
}
}
vector<int> ans;
for (int i = 0; i < n; ++i) {
if (nums[i] > 0) {
ans.push_back(i + 1);
}
}
return ans;
}
};

• class Solution:
def findDisappearedNumbers(self, nums: List[int]) -> List[int]:
for x in nums:
i = abs(x) - 1
if nums[i] > 0:
nums[i] *= -1
return [i + 1 for i in range(len(nums)) if nums[i] > 0]


• func findDisappearedNumbers(nums []int) (ans []int) {
n := len(nums)
for _, x := range nums {
i := abs(x) - 1
if nums[i] > 0 {
nums[i] = -nums[i]
}
}
for i := 0; i < n; i++ {
if nums[i] > 0 {
ans = append(ans, i+1)
}
}
return
}

func abs(x int) int {
if x < 0 {
return -x
}
return x
}

• function findDisappearedNumbers(nums: number[]): number[] {
const n = nums.length;
for (const x of nums) {
const i = Math.abs(x) - 1;
if (nums[i] > 0) {
nums[i] *= -1;
}
}
const ans: number[] = [];
for (let i = 0; i < n; ++i) {
if (nums[i] > 0) {
ans.push(i + 1);
}
}
return ans;
}