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

<p>You can perform the following operation as many times as you want:</p>

<ul>
	<li>Pick an index <code>i</code> that you haven&rsquo;t picked before, and pick a prime <code>p</code> <strong>strictly less than</strong> <code>nums[i]</code>, then subtract <code>p</code> from <code>nums[i]</code>.</li>
</ul>

<p>Return <em>true if you can make <code>nums</code> a strictly increasing array using the above operation and false otherwise.</em></p>

<p>A <strong>strictly increasing array</strong> is an array whose each element is strictly greater than its preceding element.</p>

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

<pre>
<strong>Input:</strong> nums = [4,9,6,10]
<strong>Output:</strong> true
<strong>Explanation:</strong> In the first operation: Pick i = 0 and p = 3, and then subtract 3 from nums[0], so that nums becomes [1,9,6,10].
In the second operation: i = 1, p = 7, subtract 7 from nums[1], so nums becomes equal to [1,2,6,10].
After the second operation, nums is sorted in strictly increasing order, so the answer is true.</pre>

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

<pre>
<strong>Input:</strong> nums = [6,8,11,12]
<strong>Output:</strong> true
<strong>Explanation: </strong>Initially nums is sorted in strictly increasing order, so we don&#39;t need to make any operations.</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,8,3]
<strong>Output:</strong> false
<strong>Explanation:</strong> It can be proven that there is no way to perform operations to make nums sorted in strictly increasing order, so the answer is false.</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 1000</code></li>
	<li><code><font face="monospace">nums.length == n</font></code></li>
</ul>

from typing import List

class Solution:
    def primeSubOperation(self, nums: List[int]) -> bool:
        # Helper function to generate all primes less than or equal to max_num
        def sieve_of_eratosthenes(max_num):
            is_prime = [True] * (max_num + 1)
            p = 2
            while (p * p <= max_num):
                if (is_prime[p] == True):
                    for i in range(p * p, max_num + 1, p):
                        is_prime[i] = False
                p += 1
            prime_numbers = [p for p in range(2, max_num + 1) if is_prime[p]]
            return prime_numbers
        
        # Generate all primes less than 1000
        primes = sieve_of_eratosthenes(1000)
        
        # Initialize the previous number to be less than the smallest possible number
        prev = 0
        
        for num in nums:
            # Find the largest prime that can be subtracted from num to make it greater than prev
            for prime in reversed(primes):
                if prime < num and num - prime > prev:
                    num -= prime
                    break
            # If no prime was found that satisfies the condition, check if num itself is greater than prev
            if num <= prev:
                return False
            prev = num
        
        return True
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