<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 getSum(self, index: int, value: int, n: int) -> int:
        count = 0
        
        # On index's left:
        # If value > index, there are index + 1 numbers in the arithmetic sequence:
        # [value - index, ..., value - 1, value].
        # Otherwise, there are value numbers in the arithmetic sequence:
        # [1, 2, ..., value - 1, value], plus a sequence of length (index - value + 1) of 1s. 
        if value > index:
            count += (value + value - index) * (index + 1) // 2
        else:
            count += (value + 1) * value // 2 + index - value + 1
        
        # On index's right:
        # Last Element=Starting Value−(Number of Terms−1)
        # If value >= n - index, there are n - index numbers in the arithmetic sequence:
        # [value, value - 1, ..., value - n + 1 + index].
        # Otherwise, there are value numbers in the arithmetic sequence:
        # [value, value - 1, ..., 1], plus a sequence of length (n - index - value) of 1s. 
        if value >= n - index:
            count += (value + value - n + 1 + index) * (n - index) // 2
        else:
            count += (value + 1) * value // 2 + n - index - value
        
        return count - value # bc double counting value
    
    def maxValue(self, n: int, index: int, maxSum: int) -> int:
        left, right = 1, maxSum
        while left < right:
            mid = (left + right + 1) // 2
            if self.getSum(index, mid, n) <= maxSum:
                left = mid
            else:
                right = mid - 1
        
        return left
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