1802. Maximum Value at a Given Index in a Bounded Array

public class Solution
{
    public int MaxValue(int n, int index, int maxSum)
    {
        maxSum -= n;

        int l = 0;
        int r = maxSum;

        // Find the first value x s.t. if A[index] = x, then sum(A) >= maxSum.
        while (l < r)
        {
            int m = (l + r) / 2;
            if (GetSum(n, index, m) >= maxSum)
                r = m;
            else
                l = m + 1;
        }

        return GetSum(n, index, l) > maxSum ? l : l + 1;
    }

    // Returns the minimum sum if nums[index] = x.
    private long GetSum(int n, int index, int x)
    {
        long l = Math.Min(index, x - 1);
        long r = Math.Min(n - index, x);
        long lSum = ((x - 1) + (x - 1 - l + 1)) * l / 2;
        long rSum = (x + (x - r + 1)) * r / 2;
        return lSum + rSum;
    }
}