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
impl Solution {
    #[allow(dead_code)]
    pub fn minimum_replacement(nums: Vec<i32>) -> i64 {
        if nums.len() == 1 {
            return 0;
        }

        let n = nums.len();
        let mut max = *nums.last().unwrap();
        let mut ret = 0;

        for i in (0..=n - 2).rev() {
            if nums[i] <= max {
                max = nums[i];
                continue;
            }
            // Otherwise make the substitution
            let k = (nums[i] + max - 1) / max;
            ret += (k - 1) as i64;
            // Update the max value, which should be the minimum among the substitution
            max = nums[i] / k;
        }

        ret
    }
}
