You are given three positive integers: `n`

, `index`

, and `maxSum`

. You want to construct an array `nums`

(**0-indexed**)** **that satisfies the following conditions:

`nums.length == n`

`nums[i]`

is a**positive**integer where`0 <= i < n`

.`abs(nums[i] - nums[i+1]) <= 1`

where`0 <= i < n-1`

.- The sum of all the elements of
`nums`

does not exceed`maxSum`

. `nums[index]`

is**maximized**.

Return `nums[index]`

* of the constructed array*.

Note that `abs(x)`

equals `x`

if `x >= 0`

, and `-x`

otherwise.

**Example 1:**

Input:n = 4, index = 2, maxSum = 6Output:2Explanation:nums = [1,2,,1] is one array that satisfies all the conditions. There are no arrays that satisfy all the conditions and have nums[2] == 3, so 2 is the maximum nums[2].2

**Example 2:**

Input:n = 6, index = 1, maxSum = 10Output:3

**Constraints:**

`1 <= n <= maxSum <= 10`

^{9}`0 <= index < n`

class Solution {
public int maxValue(int n, int index, int maxSum) {
}
}