Maximum Value at a Given Index in a Bounded Array

class Solution {
    private long calculateSum(int idx, int val, int size) {
        long total = 0;

        if (val > idx) {
            total += (long)(val + val - idx) * (idx + 1) / 2;
        } else {
            total += (long)(val + 1) * val / 2 + idx - val + 1;
        };

        if (val >= size - idx) {
            total += (long)(val + val - size + 1 + idx) * (size - idx) / 2;
        } else {
            total += (long)(val + 1) * val / 2 + size - idx - val;
        }   
        
        return total - val;
        
    }
    public int findMaxValue(int size, int idx, int maxSum) {
        int start = 1, end = maxSum;
        while (start < end) {
            int mid = (start + end + 1) / 2;
            if (calculateSum(idx, mid, size) <= maxSum) {
                start = mid;
            } else {
                end = mid - 1;
            }
        }
        
        return start;
    }
}

This Java solution finds the maximum value x that can be placed at a specific index in an array of given size such that the sum of all elements in the array does not exceed a provided maximum sum. The approach utilizes binary search to efficiently determine the optimal value of x.

• Begin by setting the initial search bounds start at 1 (the minimum possible value of x) and end at maxSum (the upper constraint).
• Use a while loop to continue the search as long as start is less than end.
• Calculate the midpoint, mid, between start and end to test if an array with the middle element mid at the given index can have a sum less than or equal to maxSum.
• calculateSum is a helper function that computes the potential array sum, given the current middle value, the size of the array, and the target index. It differentiates situations where the proposed middle value exceeds either side of the index and adjusts the total accordingly.
• If the calculated sum from calculateSum is within the acceptable range (<= maxSum), the search space is narrowed to the upper half (start = mid); otherwise, it is narrowed to the lower half (end = mid - 1).
• The binary search concludes when start equals end, and start will depict the maximum value that meets the problem conditions.

By leveraging a binary search algorithm, this implementation ensures that the solution is optimized for large input sizes, providing a logarithmic time complexity relative to the constraints.

class Solution:
    def calculateSum(self, idx: int, val: int, len: int) -> int:
        result = 0
        if val > idx:
            result += (val + val - idx) * (idx + 1) // 2
        else:
            result += (val + 1) * val // 2 + idx - val + 1

        if val >= len - idx:
            result += (val + val - len + 1 + idx) * (len - idx) // 2
        else:
            result += (val + 1) * val // 2 + len - idx - val
        
        return result - val
    
    def findMaxValue(self, length: int, position: int, totalSum: int) -> int:
        low, high = 1, totalSum
        while low < high:
            mid = (low + high + 1) // 2
            if self.calculateSum(position, mid, length) <= totalSum:
                low = mid
            else:
                high = mid - 1
        
        return low

This problem focuses on finding the maximum value that can be placed at a specified index in an array, subject to certain constraints. The array must meet a total sum requirement, and each element must fall within a bounded range.

The solution consists of two primary methods in the Solution class coded in Python:

• calculateSum(idx, val, len): This computes the sum of the array if the element at index idx has a value val, while assuming that this value val is the peak and the array values taper down towards the start and the end of the array.

• findMaxValue(length, position, totalSum): This method uses binary search to find the maximum value that can be placed at a specific index (position) in an array of a given length such that the total sum of the array does not exceed totalSum. The binary search is implemented by adjusting the low and high bounds based on the calculated sum using the aforementioned calculateSum method.

Below is an outline of the algorithmic strategy:

1. Implement the helper function calculateSum, which calculates the sum of a hypothetical array distribution, based on the peak value being at a given index.
2. Define findMaxValue which leverages binary search. Start with possible values of the element at the position ranging from 1 to totalSum.
   • For each value (mid), check if the total sum with mid as the value at position can be supported without exceeding totalSum.
   • Adjust the binary search bounds (low and high) depending on whether the current arrangement's sum is less than or exceeds totalSum.
   • Continue until the optimal (maximum) value is located by narrowing down the search range.
3. Return the maximum suitable value that fits the sum and position criteria without violating the constraints.

This approach ensures that you effectively find the maximum potential value for any given position in the array within the specified sum limits, using efficient binary search and sum calculation methodologies.