# LC 3131. Find the Integer Added to Array I

### [Problem link](https://leetcode.com/problems/find-the-integer-added-to-array-i/)

###### tags: `leedcode` `easy` `c++`

You are given two arrays of equal length, <code>nums1</code> and <code>nums2</code>.

Each element in <code>nums1</code> has been increased (or decreased in the case of negative) by an integer, represented by the variable <code>x</code>.

As a result, <code>nums1</code> becomes **equal** to <code>nums2</code>. Two arrays are considered **equal** when they contain the same integers with the same frequencies.

Return the integer <code>x</code>.

**Example 1:**

Input: nums1 = [2,6,4], nums2 = [9,7,5]
Output: 3

Explanation:
The integer added to each element of nums1 is 3.

**Example 2:**

Input: nums1 = [10], nums2 = [5]
Output: -5

Explanation:
The integer added to each element of nums1 is -5.

**Example 3:**

Input: nums1 = [1,1,1,1], nums2 = [1,1,1,1]
Output: 0

Explanation:
The integer added to each element of nums1 is 0.

**Constraints:**

- `1 <= nums1.length == nums2.length <= 100`
- `0 <= nums1[i], nums2[i] <= 1000`
- The test cases are generated in a way that there is an integer <code>x</code> such that <code>nums1</code> can become equal to <code>nums2</code> by adding <code>x</code> to each element of <code>nums1</code>.

## Solution 1
#### C++

class Solution {
public:
    int addedInteger(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        return nums2[0] - nums1[0];
    }
};

>### Complexity
>n = max(nums1.length, nums2.length)
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O(nlogn) | O(1) |

## Note
x