<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
from bisect import bisect_right

class Solution:
    def primeSubOperation(self, A: List[int]) -> bool:
        # List of all prime numbers up to 1000
        primes = [
            2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 
            107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 
            227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283, 293, 307, 311, 313, 317, 331, 337, 347, 
            353, 359, 367, 373, 379, 383, 389, 397, 401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 
            479, 487, 491, 499, 503, 509, 521, 523, 541, 547, 557, 563, 569, 571, 577, 587, 593, 599, 601, 607, 613, 
            617, 619, 631, 641, 643, 647, 653, 659, 661, 673, 677, 683, 691, 701, 709, 719, 727, 733, 739, 743, 751, 
            757, 761, 769, 773, 787, 797, 809, 811, 821, 823, 827, 829, 839, 853, 857, 859, 863, 877, 881, 883, 887, 
            907, 911, 919, 929, 937, 941, 947, 953, 967, 971, 977, 983, 991, 997
        ]

        # Iterate from the second last element to the first
        for i in range(len(A) - 2, -1, -1):
            # If the current element is already less than the next, continue
            if A[i] < A[i + 1]:
                continue
            
            # Find the smallest prime that can be subtracted from A[i] to make it less than A[i + 1]
            index = bisect_right(primes, A[i] - A[i + 1])
            
            # If no valid prime is found or subtraction results in a non-positive number, return False
            if index == len(primes) or A[i] - primes[index] < 1:
                return False
            
            # Subtract the found prime from the current element
            A[i] -= primes[index]
        
        # If all elements can be adjusted to be strictly increasing, return True
        return True
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