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‎solution/3100-3199/3131.Find the Integer Added to Array I/README.md

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# [3131. 找出与数组相加的整数 I](https://leetcode.cn/problems/find-the-integer-added-to-array-i)
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[English Version](/solution/3100-3199/3131.Find%20the%20Integer%20Added%20to%20Array%20I/README_EN.md)
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<!-- tags: -->
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你两个长度相等的数组 <code>nums1</code> 和 <code>nums2</code>。</p>
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<p>数组 <code>nums1</code> 中的每个元素都与变量 <code>x</code> 所表示的整数相加。如果 <code>x</code> 为负数，则表现为元素值的减少。</p>
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<p>在与 <code>x</code> 相加后，<code>nums1</code> 和 <code>nums2</code> <strong>相等</strong> 。当两个数组中包含相同的整数，并且这些整数出现的频次相同时，两个数组 <strong>相等</strong> 。</p>
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<p>返回整数 <code>x</code> 。</p>
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<p>&nbsp;</p>
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<p><strong class="example">示例 1:</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong><span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">nums1 = [2,6,4], nums2 = [9,7,5]</span></p>
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<p><strong>输出：</strong><span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">3</span></p>
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<p><strong>解释：</strong></p>
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<p>与 3 相加后，<code>nums1</code> 和 <code>nums2</code> 相等。</p>
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</div>
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<p><strong class="example">示例 2:</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong><span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">nums1 = [10], nums2 = [5]</span></p>
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<p><strong>输出：</strong><span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">-5</span></p>
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<p><strong>解释：</strong></p>
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<p>与 <code>-5</code> 相加后，<code>nums1</code> 和 <code>nums2</code> 相等。</p>
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</div>
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<p><strong class="example">示例 3:</strong></p>
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<div class="example-block">
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<p><strong>输入：</strong><span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">nums1 = [1,1,1,1], nums2 = [1,1,1,1]</span></p>
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<p><strong>输出：</strong><span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">0</span></p>
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<p><strong>解释：</strong></p>
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<p>与 0 相加后，<code>nums1</code> 和 <code>nums2</code> 相等。</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= nums1.length == nums2.length &lt;= 100</code></li>
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<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 1000</code></li>
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<li>测试用例以这样的方式生成：存在一个整数 <code>x</code>，使得 <code>nums1</code> 中的每个元素都与 <code>x</code> 相加后，<code>nums1</code> 与 <code>nums2</code> 相等。</li>
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</ul>
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## 解法
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### 方法一：求最小值差
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我们可以分别求出两个数组的最小值，然后返回两个最小值的差值即可。
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时间复杂度 $O(n)$，其中 $n$ 为数组的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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```python
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class Solution:
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def addedInteger(self, nums1: List[int], nums2: List[int]) -> int:
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return min(nums2) - min(nums1)
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```
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```java
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class Solution {
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public int addedInteger(int[] nums1, int[] nums2) {
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return Arrays.stream(nums2).min().getAsInt() - Arrays.stream(nums1).min().getAsInt();
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}
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}
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```
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```cpp
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class Solution {
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public:
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int addedInteger(vector<int>& nums1, vector<int>& nums2) {
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return *min_element(nums2.begin(), nums2.end()) - *min_element(nums1.begin(), nums1.end());
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}
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};
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```
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```go
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func addedInteger(nums1 []int, nums2 []int) int {
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return slices.Min(nums2) - slices.Min(nums1)
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}
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```
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```ts
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function addedInteger(nums1: number[], nums2: number[]): number {
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return Math.min(...nums2) - Math.min(...nums1);
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}
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```
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<!-- tabs:end -->
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<!-- end -->

‎solution/3100-3199/3131.Find the Integer Added to Array I/README_EN.md

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# [3131. Find the Integer Added to Array I](https://leetcode.com/problems/find-the-integer-added-to-array-i)
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[中文文档](/solution/3100-3199/3131.Find%20the%20Integer%20Added%20to%20Array%20I/README.md)
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<!-- tags: -->
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## Description
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<p>You are given two arrays of equal length, <code>nums1</code> and <code>nums2</code>.</p>
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<p>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>.</p>
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<p>As a result, <code>nums1</code> becomes <strong>equal</strong> to <code>nums2</code>. Two arrays are considered <strong>equal</strong> when they contain the same integers with the same frequencies.</p>
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<p>Return the integer <code>x</code>.</p>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">nums1 = [2,6,4], nums2 = [9,7,5]</span></p>
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<p><strong>Output:</strong> <span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">3</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The integer added to each element of <code>nums1</code> is 3.</p>
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</div>
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<p><strong class="example">Example 2:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">nums1 = [10], nums2 = [5]</span></p>
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<p><strong>Output:</strong> <span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">-5</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The integer added to each element of <code>nums1</code> is -5.</p>
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</div>
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<p><strong class="example">Example 3:</strong></p>
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<div class="example-block">
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<p><strong>Input:</strong> <span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">nums1 = [1,1,1,1], nums2 = [1,1,1,1]</span></p>
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<p><strong>Output:</strong> <span class="example-io" style="
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font-family: Menlo,sans-serif;
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font-size: 0.85rem;
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">0</span></p>
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<p><strong>Explanation:</strong></p>
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<p>The integer added to each element of <code>nums1</code> is 0.</p>
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</div>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums1.length == nums2.length &lt;= 100</code></li>
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<li><code>0 &lt;= nums1[i], nums2[i] &lt;= 1000</code></li>
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<li>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>.</li>
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</ul>
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## Solutions
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### Solution 1: Calculate Minimum Difference
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We can find the minimum value of each array, then return the difference between the two minimum values.
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The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
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<!-- tabs:start -->
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```python
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class Solution:
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def addedInteger(self, nums1: List[int], nums2: List[int]) -> int:
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return min(nums2) - min(nums1)
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```
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```java
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class Solution {
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public int addedInteger(int[] nums1, int[] nums2) {
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return Arrays.stream(nums2).min().getAsInt() - Arrays.stream(nums1).min().getAsInt();
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}
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}
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```
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```cpp
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class Solution {
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public:
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int addedInteger(vector<int>& nums1, vector<int>& nums2) {
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return *min_element(nums2.begin(), nums2.end()) - *min_element(nums1.begin(), nums1.end());
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}
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};
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```
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```go
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func addedInteger(nums1 []int, nums2 []int) int {
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return slices.Min(nums2) - slices.Min(nums1)
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}
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```
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```ts
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function addedInteger(nums1: number[], nums2: number[]): number {
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return Math.min(...nums2) - Math.min(...nums1);
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}
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```
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<!-- tabs:end -->
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<!-- end -->

‎solution/3100-3199/3131.Find the Integer Added to Array I/Solution.cpp

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class Solution {
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public:
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int addedInteger(vector<int>& nums1, vector<int>& nums2) {
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return *min_element(nums2.begin(), nums2.end()) - *min_element(nums1.begin(), nums1.end());
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}
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};

‎solution/3100-3199/3131.Find the Integer Added to Array I/Solution.go

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func addedInteger(nums1 []int, nums2 []int) int {
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return slices.Min(nums2) - slices.Min(nums1)
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}

‎solution/3100-3199/3131.Find the Integer Added to Array I/Solution.java

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class Solution {
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public int addedInteger(int[] nums1, int[] nums2) {
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return Arrays.stream(nums2).min().getAsInt() - Arrays.stream(nums1).min().getAsInt();
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}
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}

‎solution/3100-3199/3131.Find the Integer Added to Array I/Solution.py

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class Solution:
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def addedInteger(self, nums1: List[int], nums2: List[int]) -> int:
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return min(nums2) - min(nums1)

‎solution/3100-3199/3131.Find the Integer Added to Array I/Solution.ts

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function addedInteger(nums1: number[], nums2: number[]): number {
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return Math.min(...nums2) - Math.min(...nums1);
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}

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