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feat: add solutions to lc problem: No.2219
No.2219.Maximum Sum Score of Array
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# [2219. Maximum Sum Score of Array](https://leetcode-cn.com/problems/maximum-sum-score-of-array)
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[English Version](/solution/2200-2299/2219.Maximum%20Sum%20Score%20of%20Array/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code>.</p>
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<p>The <strong>sum </strong><strong>score</strong> of <code>nums</code> at an index <code>i</code> where <code>0 &lt;= i &lt; n</code> is the <strong>maximum</strong> of:</p>
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<ul>
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<li>The sum of the <strong>first</strong> <code>i + 1</code> elements of <code>nums</code>.</li>
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<li>The sum of the <strong>last</strong> <code>n - i</code> elements of <code>nums</code>.</li>
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</ul>
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<p>Return <em>the <strong>maximum</strong> <strong>sum </strong><strong>score</strong> of </em><code>nums</code><em> at any index.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [4,3,-2,5]
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<strong>Output:</strong> 10
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<strong>Explanation:</strong>
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The sum score at index 0 is max(4, 4 + 3 + -2 + 5) = max(4, 10) = 10.
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The sum score at index 1 is max(4 + 3, 3 + -2 + 5) = max(7, 6) = 7.
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The sum score at index 2 is max(4 + 3 + -2, -2 + 5) = max(5, 3) = 5.
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The sum score at index 3 is max(4 + 3 + -2 + 5, 5) = max(10, 5) = 10.
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The maximum sum score of nums is 10.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [-3,-5]
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<strong>Output:</strong> -3
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<strong>Explanation:</strong>
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The sum score at index 0 is max(-3, -3 + -5) = max(-3, -8) = -3.
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The sum score at index 1 is max(-3 + -5, -5) = max(-8, -5) = -5.
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The maximum sum score of nums is -3.
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</pre>
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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>n == nums.length</code></li>
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<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
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<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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前缀和。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maximumSumScore(self, nums: List[int]) -> int:
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s = [0] + list(accumulate(nums))
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return max(max(s[i + 1], s[-1] - s[i]) for i in range(len(nums)))
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public long maximumSumScore(int[] nums) {
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int n = nums.length;
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long[] s = new long[n + 1];
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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long ans = Long.MIN_VALUE;
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for (int i = 0; i < n; ++i) {
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ans = Math.max(ans, Math.max(s[i + 1], s[n] - s[i]));
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}
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return ans;
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}
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}
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```
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### **TypeScript**
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```ts
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function maximumSumScore(nums: number[]): number {
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const n = nums.length;
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let s = new Array(n + 1).fill(0);
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for (let i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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let ans = -Infinity;
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for (let i = 0; i < n; ++i) {
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ans = Math.max(ans, Math.max(s[i + 1], s[n] - s[i]));
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}
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return ans;
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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long long maximumSumScore(vector<int>& nums) {
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int n = nums.size();
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vector<long long> s(n + 1);
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for (int i = 0; i < n; ++i) s[i + 1] = s[i] + nums[i];
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long long ans = INT_MIN;
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for (int i = 0; i < n; ++i) ans = max(ans, max(s[i + 1], s[n] - s[i]));
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maximumSumScore(nums []int) int64 {
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n := len(nums)
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s := make([]int64, n+1)
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for i, v := range nums {
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s[i+1] = s[i] + int64(v)
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}
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var ans int64 = math.MinInt64
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for i := 0; i < n; i++ {
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ans = max(ans, max(s[i+1], s[n]-s[i]))
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}
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return ans
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}
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func max(a, b int64) int64 {
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if a > b {
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return a
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}
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return b
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number[]} nums
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* @return {number}
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*/
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var maximumSumScore = function (nums) {
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const n = nums.length;
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let s = new Array(n + 1).fill(0);
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for (let i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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let ans = -Infinity;
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for (let i = 0; i < n; ++i) {
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ans = Math.max(ans, Math.max(s[i + 1], s[n] - s[i]));
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}
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return ans;
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};
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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# [2219. Maximum Sum Score of Array](https://leetcode.com/problems/maximum-sum-score-of-array)
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[中文文档](/solution/2200-2299/2219.Maximum%20Sum%20Score%20of%20Array/README.md)
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## Description
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<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code>.</p>
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<p>The <strong>sum </strong><strong>score</strong> of <code>nums</code> at an index <code>i</code> where <code>0 &lt;= i &lt; n</code> is the <strong>maximum</strong> of:</p>
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<ul>
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<li>The sum of the <strong>first</strong> <code>i + 1</code> elements of <code>nums</code>.</li>
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<li>The sum of the <strong>last</strong> <code>n - i</code> elements of <code>nums</code>.</li>
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</ul>
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<p>Return <em>the <strong>maximum</strong> <strong>sum </strong><strong>score</strong> of </em><code>nums</code><em> at any index.</em></p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [4,3,-2,5]
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<strong>Output:</strong> 10
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<strong>Explanation:</strong>
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The sum score at index 0 is max(4, 4 + 3 + -2 + 5) = max(4, 10) = 10.
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The sum score at index 1 is max(4 + 3, 3 + -2 + 5) = max(7, 6) = 7.
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The sum score at index 2 is max(4 + 3 + -2, -2 + 5) = max(5, 3) = 5.
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The sum score at index 3 is max(4 + 3 + -2 + 5, 5) = max(10, 5) = 10.
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The maximum sum score of nums is 10.
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [-3,-5]
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<strong>Output:</strong> -3
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<strong>Explanation:</strong>
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The sum score at index 0 is max(-3, -3 + -5) = max(-3, -8) = -3.
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The sum score at index 1 is max(-3 + -5, -5) = max(-8, -5) = -5.
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The maximum sum score of nums is -3.
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</pre>
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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>n == nums.length</code></li>
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<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
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<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def maximumSumScore(self, nums: List[int]) -> int:
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s = [0] + list(accumulate(nums))
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return max(max(s[i + 1], s[-1] - s[i]) for i in range(len(nums)))
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```
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### **Java**
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```java
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class Solution {
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public long maximumSumScore(int[] nums) {
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int n = nums.length;
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long[] s = new long[n + 1];
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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long ans = Long.MIN_VALUE;
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for (int i = 0; i < n; ++i) {
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ans = Math.max(ans, Math.max(s[i + 1], s[n] - s[i]));
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}
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return ans;
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}
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}
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```
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### **TypeScript**
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```ts
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function maximumSumScore(nums: number[]): number {
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const n = nums.length;
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let s = new Array(n + 1).fill(0);
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for (let i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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let ans = -Infinity;
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for (let i = 0; i < n; ++i) {
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ans = Math.max(ans, Math.max(s[i + 1], s[n] - s[i]));
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}
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return ans;
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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long long maximumSumScore(vector<int>& nums) {
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int n = nums.size();
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vector<long long> s(n + 1);
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for (int i = 0; i < n; ++i) s[i + 1] = s[i] + nums[i];
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long long ans = INT_MIN;
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for (int i = 0; i < n; ++i) ans = max(ans, max(s[i + 1], s[n] - s[i]));
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return ans;
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}
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};
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```
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### **Go**
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```go
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func maximumSumScore(nums []int) int64 {
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n := len(nums)
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s := make([]int64, n+1)
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for i, v := range nums {
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s[i+1] = s[i] + int64(v)
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}
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var ans int64 = math.MinInt64
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for i := 0; i < n; i++ {
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ans = max(ans, max(s[i+1], s[n]-s[i]))
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}
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return ans
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}
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func max(a, b int64) int64 {
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if a > b {
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return a
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}
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return b
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}
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```
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### **JavaScript**
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```js
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/**
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* @param {number[]} nums
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* @return {number}
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*/
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var maximumSumScore = function (nums) {
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const n = nums.length;
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let s = new Array(n + 1).fill(0);
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for (let i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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let ans = -Infinity;
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for (let i = 0; i < n; ++i) {
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ans = Math.max(ans, Math.max(s[i + 1], s[n] - s[i]));
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}
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return ans;
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};
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```
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### **...**
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```
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```
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<!-- tabs:end -->
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class Solution {
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public:
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long long maximumSumScore(vector<int>& nums) {
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int n = nums.size();
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vector<long long> s(n + 1);
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for (int i = 0; i < n; ++i) s[i + 1] = s[i] + nums[i];
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long long ans = INT_MIN;
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for (int i = 0; i < n; ++i) ans = max(ans, max(s[i + 1], s[n] - s[i]));
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return ans;
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}
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};
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func maximumSumScore(nums []int) int64 {
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n := len(nums)
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s := make([]int64, n+1)
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for i, v := range nums {
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s[i+1] = s[i] + int64(v)
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}
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var ans int64 = math.MinInt64
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for i := 0; i < n; i++ {
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ans = max(ans, max(s[i+1], s[n]-s[i]))
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}
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return ans
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}
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func max(a, b int64) int64 {
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if a > b {
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return a
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}
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return b
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}
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class Solution {
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public long maximumSumScore(int[] nums) {
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int n = nums.length;
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long[] s = new long[n + 1];
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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long ans = Long.MIN_VALUE;
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for (int i = 0; i < n; ++i) {
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ans = Math.max(ans, Math.max(s[i + 1], s[n] - s[i]));
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}
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return ans;
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}
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}

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