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feat: add solutions to lc problem: No.1802
No.1802.Maximum Value at a Given Index in a Bounded Array
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‎solution/1800-1899/1802.Maximum Value at a Given Index in a Bounded Array/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：二分查找**
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根据题目描述，如果我们确定了 $nums[index]$ 的值为 $x$，此时我们可以找到一个最小的数组总和。也就是说，在 $index$ 左侧的数组元素从 $x-1$ 一直递减到 $1$，如果还有剩余的元素，那么剩余的元素都为 $1$；同理，在 $index$ 及右侧的数组元素从 $x$ 一直递减到 $1$，如果还有剩余的元素，那么剩余的元素都为 $1$。
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这样我们就可以计算出数组的总和，如果总和小于等于 $maxSum$，那么此时的 $x$ 是合法的。随着 $x$ 的增大，数组的总和也会增大，因此我们可以使用二分查找的方法，找到一个最大的且符合条件的 $x$。
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为了方便计算数组左侧、右侧的元素之和，我们定义一个函数 $sum(x, cnt)$，表示一共有 $cnt$ 个元素，且最大值为 $x$ 的数组的总和。函数 $sum(x, cnt)$ 可以分为两种情况：
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- 如果 $x \geq cnt$，那么数组的总和为 $\frac{(x + x - cnt + 1) \times cnt}{2}$
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- 如果 $x \lt cnt$，那么数组的总和为 $\frac{(x + 1) \times x}{2} + cnt - x$
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接下来，定义二分的左边界 $left = 1$，右边界 $right = maxSum$，然后二分查找 $nums[index]$ 的值 $mid$，如果 $sum(mid - 1, index) + sum(mid, n - index) \leq maxSum$，那么此时的 $mid$ 是合法的，我们可以将 $left$ 更新为 $mid$，否则我们将 $right$ 更新为 $mid - 1$。
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最后将 $left$ 作为答案返回即可。
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时间复杂度 $O(\log M)$，空间复杂度 $O(1)$。其中 $M=maxSum$。
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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 maxValue(self, n: int, index: int, maxSum: int) -> int:
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def sum(x, cnt):
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return (x + x - cnt + 1) * cnt // 2 if x >= cnt else (x + 1) * x // 2 + cnt - x
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left, right = 1, maxSum
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while left < right:
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mid = (left + right + 1) >> 1
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if sum(mid - 1, index) + sum(mid, n - index) <= maxSum:
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left = mid
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else:
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right = mid - 1
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return left
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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 int maxValue(int n, int index, int maxSum) {
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int left = 1, right = maxSum;
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while (left < right) {
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int mid = (left + right + 1) >>> 1;
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if (sum(mid - 1, index) + sum(mid, n - index) <= maxSum) {
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left = mid;
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} else {
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right = mid - 1;
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}
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}
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return left;
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}
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private long sum(long x, int cnt) {
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return x >= cnt ? (x + x - cnt + 1) * cnt / 2 : (x + 1) * x / 2 + cnt - x;
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}
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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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int maxValue(int n, int index, int maxSum) {
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auto sum = [](long x, int cnt) {
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return x >= cnt ? (x + x - cnt + 1) * cnt / 2 : (x + 1) * x / 2 + cnt - x;
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};
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int left = 1, right = maxSum;
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while (left < right) {
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int mid = (left + right + 1) >> 1;
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if (sum(mid - 1, index) + sum(mid, n - index) <= maxSum) {
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left = mid;
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} else {
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right = mid - 1;
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}
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}
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return left;
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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 maxValue(n int, index int, maxSum int) int {
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sum := func(x, cnt int) int {
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if x >= cnt {
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return (x + x - cnt + 1) * cnt / 2
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}
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return (x+1)*x/2 + cnt - x
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}
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return sort.Search(maxSum, func(x int) bool {
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x++
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return sum(x-1, index)+sum(x, n-index) > maxSum
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})
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}
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```
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### **...**

‎solution/1800-1899/1802.Maximum Value at a Given Index in a Bounded Array/README_EN.md

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### **Python3**
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```python
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class Solution:
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def maxValue(self, n: int, index: int, maxSum: int) -> int:
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def sum(x, cnt):
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return (x + x - cnt + 1) * cnt // 2 if x >= cnt else (x + 1) * x // 2 + cnt - x
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left, right = 1, maxSum
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while left < right:
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mid = (left + right + 1) >> 1
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if sum(mid - 1, index) + sum(mid, n - index) <= maxSum:
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left = mid
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else:
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right = mid - 1
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return left
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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 int maxValue(int n, int index, int maxSum) {
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int left = 1, right = maxSum;
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while (left < right) {
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int mid = (left + right + 1) >>> 1;
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if (sum(mid - 1, index) + sum(mid, n - index) <= maxSum) {
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left = mid;
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} else {
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right = mid - 1;
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}
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}
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return left;
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}
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private long sum(long x, int cnt) {
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return x >= cnt ? (x + x - cnt + 1) * cnt / 2 : (x + 1) * x / 2 + cnt - x;
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}
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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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int maxValue(int n, int index, int maxSum) {
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auto sum = [](long x, int cnt) {
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return x >= cnt ? (x + x - cnt + 1) * cnt / 2 : (x + 1) * x / 2 + cnt - x;
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};
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int left = 1, right = maxSum;
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while (left < right) {
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int mid = (left + right + 1) >> 1;
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if (sum(mid - 1, index) + sum(mid, n - index) <= maxSum) {
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left = mid;
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} else {
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right = mid - 1;
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}
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}
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return left;
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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 maxValue(n int, index int, maxSum int) int {
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sum := func(x, cnt int) int {
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if x >= cnt {
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return (x + x - cnt + 1) * cnt / 2
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}
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return (x+1)*x/2 + cnt - x
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}
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return sort.Search(maxSum, func(x int) bool {
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x++
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return sum(x-1, index)+sum(x, n-index) > maxSum
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})
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}
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```
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### **...**

‎solution/1800-1899/1802.Maximum Value at a Given Index in a Bounded Array/Solution.cpp

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class Solution {
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public:
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int maxValue(int n, int index, int maxSum) {
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auto sum = [](long x, int cnt) {
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return x >= cnt ? (x + x - cnt + 1) * cnt / 2 : (x + 1) * x / 2 + cnt - x;
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};
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int left = 1, right = maxSum;
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while (left < right) {
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int mid = (left + right + 1) >> 1;
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if (sum(mid - 1, index) + sum(mid, n - index) <= maxSum) {
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left = mid;
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} else {
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right = mid - 1;
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}
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}
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return left;
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}
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};

‎solution/1800-1899/1802.Maximum Value at a Given Index in a Bounded Array/Solution.go

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func maxValue(n int, index int, maxSum int) int {
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sum := func(x, cnt int) int {
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if x >= cnt {
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return (x + x - cnt + 1) * cnt / 2
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}
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return (x+1)*x/2 + cnt - x
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}
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return sort.Search(maxSum, func(x int) bool {
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x++
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return sum(x-1, index)+sum(x, n-index) > maxSum
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})
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}

‎solution/1800-1899/1802.Maximum Value at a Given Index in a Bounded Array/Solution.java

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class Solution {
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public int maxValue(int n, int index, int maxSum) {
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int left = 1, right = maxSum;
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while (left < right) {
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int mid = (left + right + 1) >>> 1;
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if (sum(mid - 1, index) + sum(mid, n - index) <= maxSum) {
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left = mid;
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} else {
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right = mid - 1;
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}
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}
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return left;
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}
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private long sum(long x, int cnt) {
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return x >= cnt ? (x + x - cnt + 1) * cnt / 2 : (x + 1) * x / 2 + cnt - x;
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}
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}

‎solution/1800-1899/1802.Maximum Value at a Given Index in a Bounded Array/Solution.py

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class Solution:
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def maxValue(self, n: int, index: int, maxSum: int) -> int:
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def sum(x, cnt):
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return (x + x - cnt + 1) * cnt // 2 if x >= cnt else (x + 1) * x // 2 + cnt - x
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left, right = 1, maxSum
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while left < right:
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mid = (left + right + 1) >> 1
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if sum(mid - 1, index) + sum(mid, n - index) <= maxSum:
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left = mid
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else:
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right = mid - 1
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return left

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