<p>You are given three positive integers:&nbsp;<code>n</code>, <code>index</code>, and <code>maxSum</code>. You want to construct an array <code>nums</code> (<strong>0-indexed</strong>)<strong> </strong>that satisfies the following conditions:</p>

<ul>
	<li><code>nums.length == n</code></li>
	<li><code>nums[i]</code> is a <strong>positive</strong> integer where <code>0 &lt;= i &lt; n</code>.</li>
	<li><code>abs(nums[i] - nums[i+1]) &lt;= 1</code> where <code>0 &lt;= i &lt; n-1</code>.</li>
	<li>The sum of all the elements of <code>nums</code> does not exceed <code>maxSum</code>.</li>
	<li><code>nums[index]</code> is <strong>maximized</strong>.</li>
</ul>

<p>Return <code>nums[index]</code><em> of the constructed array</em>.</p>

<p>Note that <code>abs(x)</code> equals <code>x</code> if <code>x &gt;= 0</code>, and <code>-x</code> otherwise.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 4, index = 2,  maxSum = 6
<strong>Output:</strong> 2
<strong>Explanation:</strong> nums = [1,2,<u><strong>2</strong></u>,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].
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 6, index = 1,  maxSum = 10
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= maxSum &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt;= index &lt; n</code></li>
</ul>

class Solution:
    def maxValue(self, n: int, index: int, maxSum: int) -> int:
        # Obviously, you want elements to start decreasing from index, i.e. array sort of looks like a mountain with index as the tip.

        # Suppose nums[index] = k. If enough space on one side, will take k elements or k - 1 elements to decrease to 1. The sum will be equal to (k - 1)(k) / 2.
        # What if there's not enough space-- first, what are the conditions?
        # left side: index - 0 < k - 1. sum = [(k - 1)(k) / 2] - [k - 1 - index][k - index]
        #   [0,0,4,3,2,1,0,0]
        # right side: n - 1 - index < k - 1

        maxSum -= n
        l = 0
        r = maxSum
        ans = 1
        while l <= r:
            m = (l + r) // 2
            left_s = self.tri_sum(max(m - index, 0), m - 1)
            right_s = self.tri_sum(max(m - (n - 1 - index), 0), m - 1)
            if left_s + right_s + m <= maxSum:
                ans = m + 1
                l = m + 1
            else:
                r = m - 1
        return ans
    
    def tri_sum(self, l: int, r: int):
        return (l + r) * (r - l + 1) / 2
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