<p>You are given a <strong>0-indexed</strong> array <code>nums</code> comprising of <code>n</code> non-negative integers.</p>

<p>In one operation, you must:</p>

<ul>
	<li>Choose an integer <code>i</code> such that <code>1 &lt;= i &lt; n</code> and <code>nums[i] &gt; 0</code>.</li>
	<li>Decrease <code>nums[i]</code> by 1.</li>
	<li>Increase <code>nums[i - 1]</code> by 1.</li>
</ul>

<p>Return<em> the <strong>minimum</strong> possible value of the <strong>maximum</strong> integer of </em><code>nums</code><em> after performing <strong>any</strong> number of operations</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,7,1,6]
<strong>Output:</strong> 5
<strong>Explanation:</strong>
One set of optimal operations is as follows:
1. Choose i = 1, and nums becomes [4,6,1,6].
2. Choose i = 3, and nums becomes [4,6,2,5].
3. Choose i = 1, and nums becomes [5,5,2,5].
The maximum integer of nums is 5. It can be shown that the maximum number cannot be less than 5.
Therefore, we return 5.
</pre>

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

<pre>
<strong>Input:</strong> nums = [10,1]
<strong>Output:</strong> 10
<strong>Explanation:</strong>
It is optimal to leave nums as is, and since 10 is the maximum value, we return 10.
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

'''
- So it seems that I need to act on all indices that are the max first, and I can act on them as much as I want until nums[i - 1] == nums[i], then it's not worth it
- I'm increasing the index behind me by 1 

I can also binary search for the solution -> cause I can determine in O(n) time how to get values in the array to be <= x

Pseudocode
- Find the max, binary search between 0 and max - O(log(max))
- Determine if it is possible to get the max of the array to be x
    - Iterate backwards, if arr[i] >= max, decrease it as much as you can, it works if arr[0] <= x



'''

class Solution:
    def minimizeArrayValue(self, nums: List[int]) -> int:
        low = 0
        high = max(nums)
        
 
       
        while low < high:
            mid = (low + high) // 2

            if self.possible(nums.copy(), mid):
                high = mid
            else:
                low = mid + 1
        return low
    
    def possible(self, nums, x):
        for i in range(len(nums) - 1, 0, -1):
            diff = nums[i] - x
            
            if diff > 0:
                nums[i - 1] += diff
        return nums[0] <= x
1312