Given three positive integers, n, index, and maxSum, output the nums[index] by constructing an array of nums with the length of n, which satisfies the following conditions:

• The length of the array nums is equal to n.

• Each element nums[i] is a positive integer, where $1$$\leq$i $\lt$n.

• The absolute difference between two consecutive elements, nums[i] and nums[i+1], is at most $1$.

• The sum of all elements in nums does not exceed maxSum.

• The element at nums[index] contains the maximum value.

Constraints:

• $1$$\leq$n $\leq$maxSum $\leq$$10^9$

• $0$$\leq$index $\lt$n

The solution to maximizing the value at a given index in a bounded array involves using a binary search approach combined with mathematical calculations of arithmetic sequences. To achieve this, we need to ensure that the values on both sides of nums[index] decrease by at most $1$ until reaching $1$ (we cannot go beyond 1 because the elements should be positive), minimizing the overall sum of the array.

The algorithm to solve this problem is as follows:

Initialization

• Initialize two variables, left and right, with the values $1$ and maxSum, respectively. We can assign a minimum value of $1$ to an element, and the maximum value of maxSum.

• Start a loop and Iterate until left is less than right, and calculate the sum of left and right elements as listed below:

Calculate mid

• Calculate mid as $mid = (left + right + 1) / 2$.

Determine values on the left side

• If mid is greater than index, there will be a sequence starting from mid to the $mid - index$(right to left). This sequence will form a series $[mid - index, mid - index + 1, mid - 1, ..., mid]$, which is an arithmetic series. This arithmetic series is calculated as $(mid + mid - index) * (index + 1) / 2.$ ...