<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>

class Solution:
    def primeSubOperation(self, nums: List[int]) -> bool:
        
        def findall_prime(n):
            if n < 2:
                return []
            if n == 2:
                return [2]
            if n == 3:
                return [2, 3]

            res = [2, 3]
            for j in range(4, n + 1):
                for prime in res:
                    if j % prime == 0:
                        break
                else:
                    res.append(j)

            return res

        
        def binary_search(arr, A):
            left, right = 0, len(arr) - 1
            result = None
            while left <= right:
                mid = (left + right) // 2
                
                if arr[mid] < A:
                    result = mid  # arr[mid] is a candidate, store it
                    left = mid + 1  # Continue searching in the right half
                else:
                    right = mid - 1  # Search in the left half

            return result

        all_prime = findall_prime(max(nums))

        prev = 0

        for num in nums[:-1]:
            index = binary_search(all_prime, num - prev)
            if index == None: # didn't find a prime which is strictly smaller than num - prev
                if num <= prev: # directly keep num, but num must be strictly larger than prev, otherwise return False
                    return False
                prev = num
            else:
                prev = num - all_prime[index] # find a prime which is strictly smaller than num - prev

        
        return nums[-1] > prev 
        

206