Problem statement

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 <= 10^5.

• 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.

Solution: Marking the appearances

#include <vector>
#include <iostream>
using namespace std;
vector<int> findDisappearedNumbers(vector<int>& nums) {        
    vector<bool> exist(nums.size() + 1, false);        
    for (auto& i : nums) {
        exist[i] = true;
    }
    vector<int> result;
    for (int i = 1; i <= nums.size(); i++) {
        if (!exist[i]) {
            result.push_back(i);
        }
    }
    return result;
}
void print(vector<int>& nums) {
    cout << "[";
    for (auto& n : nums) {
        cout << n << ",";
    }
    cout << "]\n";
}
int main() {
    vector<int> nums = {4,3,2,7,8,2,3,1};
    auto result = findDisappearedNumbers(nums);
    print(result);
    nums = {1,1};
    result = findDisappearedNumbers(nums);
    print(result);
}

Output:
[5,6,]
[2,]

Complexity

• Runtime: O(N), where N = nums.length.

• Extra space: O(1) (vector<bool> is optimized for space efficiency).

References

• https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/

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