Minimum Number of Operations to Make All Array Elements Equal to 1
<p>You are given a <strong>0-indexed</strong>&nbsp;array <code>nums</code> consisiting of <strong>positive</strong> integers. You can do the following operation on the array <strong>any</strong> number of times:</p>

<ul>
	<li>Select an index <code>i</code> such that <code>0 &lt;= i &lt; n - 1</code> and replace either of&nbsp;<code>nums[i]</code> or <code>nums[i+1]</code> with their gcd value.</li>
</ul>

<p>Return <em>the <strong>minimum</strong> number of operations to make all elements of </em><code>nums</code><em> equal to </em><code>1</code>. If it is impossible, return <code>-1</code>.</p>

<p>The gcd of two integers is the greatest common divisor of the two integers.</p>

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

<pre>
<strong>Input:</strong> nums = [2,6,3,4]
<strong>Output:</strong> 4
<strong>Explanation:</strong> We can do the following operations:
- Choose index i = 2 and replace nums[2] with gcd(3,4) = 1. Now we have nums = [2,6,1,4].
- Choose index i = 1 and replace nums[1] with gcd(6,1) = 1. Now we have nums = [2,1,1,4].
- Choose index i = 0 and replace nums[0] with gcd(2,1) = 1. Now we have nums = [1,1,1,4].
- Choose index i = 2 and replace nums[3] with gcd(1,4) = 1. Now we have nums = [1,1,1,1].
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,10,6,14]
<strong>Output:</strong> -1
<strong>Explanation:</strong> It can be shown that it is impossible to make all the elements equal to 1.
</pre>

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

<ul>
	<li><code>2 &lt;= nums.length &lt;= 50</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
</ul>
Array; Math; Number Theory
class Solution {
public:
    int minOperations(vector<int>& nums) {
        int n = nums.size();
        int cnt = 0;
        for (int x : nums) {
            if (x == 1) {
                ++cnt;
            }
        }
        if (cnt) {
            return n - cnt;
        }
        int mi = n + 1;
        for (int i = 0; i < n; ++i) {
            int g = 0;
            for (int j = i; j < n; ++j) {
                g = gcd(g, nums[j]);
                if (g == 1) {
                    mi = min(mi, j - i + 1);
                }
            }
        }
        return mi > n ? -1 : n - 1 + mi - 1;
    }
};