Approach: Mark by sign / seen set

def findDisappearedNumbers(nums):
    seen = set(nums)
    return [x for x in range(1, len(nums)+1) if x not in seen]

Complexity: O(n) time, O(n) space.

Full Solution — all languages

class Solution:
    def findDisappearedNumbers(self, nums):
        seen = set(nums)
        return [x for x in range(1, len(nums) + 1) if x not in seen]

import java.util.*;
class Solution {
    public int[] findDisappearedNumbers(int[] nums) {
        int n = nums.length;
        boolean[] seen = new boolean[n + 1];
        for (int x : nums) seen[x] = true;
        List<Integer> res = new ArrayList<>();
        for (int x = 1; x <= n; x++) if (!seen[x]) res.add(x);
        int[] r = new int[res.size()];
        for (int i = 0; i < r.length; i++) r[i] = res.get(i);
        return r;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: vector<int> findDisappearedNumbers(vector<int>& nums) {
        int n = (int)nums.size();
        vector<bool> seen(n + 1, false);
        for (int x : nums) seen[x] = true;
        vector<int> r;
        for (int x = 1; x <= n; x++) if (!seen[x]) r.push_back(x);
        return r;
    }
};

/**
 * @param {number[]} nums
 * @return {number[]}
 */ var findDisappearedNumbers = function(nums) {
    const n = nums.length, seen = new Array(n + 1).fill(false);
    for (const x of nums) seen[x] = true;
    const r = [];
    for (let x = 1; x <= n; x++) if (!seen[x]) r.push(x);
    return r;
};

Find All Numbers Disappeared in an Array

Example 1

Example 2

Constraints