<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code>. In one operation you can replace any element of the array with <strong>any two</strong> elements that <strong>sum</strong> to it.</p>

<ul>
	<li>For example, consider <code>nums = [5,6,7]</code>. In one operation, we can replace <code>nums[1]</code> with <code>2</code> and <code>4</code> and convert <code>nums</code> to <code>[5,2,4,7]</code>.</li>
</ul>

<p>Return <em>the minimum number of operations to make an array that is sorted in <strong>non-decreasing</strong> order</em>.</p>

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

<pre>
<strong>Input:</strong> nums = [3,9,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> Here are the steps to sort the array in non-decreasing order:
- From [3,9,3], replace the 9 with 3 and 6 so the array becomes [3,3,6,3]
- From [3,3,6,3], replace the 6 with 3 and 3 so the array becomes [3,3,3,3,3]
There are 2 steps to sort the array in non-decreasing order. Therefore, we return 2.

</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> 0
<strong>Explanation:</strong> The array is already in non-decreasing order. Therefore, we return 0. 
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minimumReplacement(self, nums: List[int]) -> int:
        n = len(nums)
        ret = 0
        prev = nums[n - 1]
        for ind in range(n - 2, -1, -1):
            num = nums[ind]
            k = ceil(num / prev)

            # (k - 1) is the minimum number of times you'll have to split
            ret += k - 1
            # (num // k) is the maximal number you can create from splitting (k - 1) times
            prev = num // k

        return ret
                    
                    
                
            
        
403