<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code>.</p>

<p>A <strong>split</strong> at an index <code>i</code> where <code>0 &lt;= i &lt;= n - 2</code> is called <strong>valid</strong> if the product of the first <code>i + 1</code> elements and the product of the remaining elements are coprime.</p>

<ul>
	<li>For example, if <code>nums = [2, 3, 3]</code>, then a split at the index <code>i = 0</code> is valid because <code>2</code> and <code>9</code> are coprime, while a split at the index <code>i = 1</code> is not valid because <code>6</code> and <code>3</code> are not coprime. A split at the index <code>i = 2</code> is not valid because <code>i == n - 1</code>.</li>
</ul>

<p>Return <em>the smallest index </em><code>i</code><em> at which the array can be split validly or </em><code>-1</code><em> if there is no such split</em>.</p>

<p>Two values <code>val1</code> and <code>val2</code> are coprime if <code>gcd(val1, val2) == 1</code> where <code>gcd(val1, val2)</code> is the greatest common divisor of <code>val1</code> and <code>val2</code>.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/12/14/second.PNG" style="width: 450px; height: 211px;" />
<pre>
<strong>Input:</strong> nums = [4,7,8,15,3,5]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The table above shows the values of the product of the first i + 1 elements, the remaining elements, and their gcd at each index i.
The only valid split is at index 2.
</pre>

<p><strong>Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/12/14/capture.PNG" style="width: 450px; height: 215px;" />
<pre>
<strong>Input:</strong> nums = [4,7,15,8,3,5]
<strong>Output:</strong> -1
<strong>Explanation:</strong> The table above shows the values of the product of the first i + 1 elements, the remaining elements, and their gcd at each index i.
There is no valid split.
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
</ul>

# generate prime numbers and all the factorizations
M = 1_000_000
firstPrime = [0 for _ in range(M+1)]
primes = []

for i in range(2, M+1):
    if firstPrime[i] == 0:
        primes.append(i)
    for prime in primes:
        if i * prime > M:
            break
        firstPrime[i * prime] = prime

        # break, because the next numbers will have firstPrime[i]
        # as the smallest prime factor
        if prime == firstPrime[i]:
            break

class Solution:
    def findValidSplit(self, nums: List[int]) -> int:
        if len(nums) < 2:
            return -1
        n = len(nums)

        totalProd = 1
        for num in nums:
            totalProd *= num
        
        # keep track of the prime factors already used
        leftDivs = Counter()
        rightDivs = Counter()

        @cache
        def decompose(x: int):
            cnt = Counter()
            while x != 1:
                if firstPrime[x] == 0:
                    cnt[x] += 1
                    break
                cnt[firstPrime[x]] += 1
                x = x // firstPrime[x]
            return cnt
    
        for num in nums:
            rightDivs += decompose(num)
        uniqueFactors = len(rightDivs)
        
        for i, num in enumerate(nums[:-1]):
            current = decompose(num)

            for key in current:
                leftDivs[key] += current[key]
                rightDivs[key] -= current[key]
                if rightDivs[key] == 0:
                    rightDivs.pop(key)

            if len(leftDivs) + len(rightDivs) == uniqueFactors:
                decompose.cache_clear()
                return i
        return -1
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