Problem statement

The set S originally contains numbers from 1 to n. But unfortunately, due to the data error, one of the numbers in the set got duplicated to another number in the set, which results in repetition of one number and loss of another number.

Given an array nums representing the data status of this set after the error. Your task is to firstly find the number occurs twice and then find the number that is missing. Return them in the form of an array.

Example 1:

Input: nums = [1,2,2,4]
Output: [2,3]

Note:

1. The given array size will in the range [2, 10000].
2. The given array‘s numbers won‘t have any order.

Solution one:

This problem is similar with .

Since all the value in the array ranges from [1, n]. They are all positive. Each time when we loop the number, we can use the value as index to set the corresponding position value as negative.

And the number is the duplicated one if we use the index to get the value is negative.  Find the duplicated.

Loop the array again, if the value is positive, the index is the missing number.

Time complexity is O(2 * n), space complexity is O(1).

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        vector<int> error;
        for(auto idx : nums){
            if(nums[abs(idx) - 1] > 0){
                nums[abs(idx) - 1] = -nums[abs(idx) - 1];
            } else {
                error.push_back(abs(idx));
            }
        }
        for(int i = 0; i < nums.size(); i++){
            if(nums[i] > 0){
                error.push_back(i + 1);
            }
        }
        return error;
    }
};

Solution two: Set + sum calculation.

We use a set to find the duplicated number.

There is a math equation to calculate the sum, duplicated number and the missing number.

The original sum is 1 + 2 + 3 + ... + n   = sum1;

The missing number array is sum2 = sum1 + duplicated - missing

We just loop once, putting the number in set to find the duplicated, and sum all the numbers to calculate the missing value at the end of loop.

Time complexity is O(n), space complexity is O(n).

class Solution {
public:
    vector<int> findErrorNums(vector<int>& nums) {
        set<int> s;
        int dup = 0;
        int sum = (nums.size() + 1) * nums.size() / 2;
        for(auto num : nums){
            if(s.find(num) != s.end()){
                dup = num;
            }
            sum -= num;
            s.insert(num);
        }
        return {dup, sum + dup};
    }
};