Minimum Replacements to Sort the Array
<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>
Greedy; Array; Math
class Solution {
public:
    long long minimumReplacement(vector<int>& nums) {
        long long ans = 0;
        int n = nums.size();
        int mx = nums[n - 1];
        for (int i = n - 2; i >= 0; --i) {
            if (nums[i] <= mx) {
                mx = nums[i];
                continue;
            }
            int k = (nums[i] + mx - 1) / mx;
            ans += k - 1;
            mx = nums[i] / k;
        }
        return ans;
    }
};