# Solving Leetcode Interviews in Seconds with AI: Prime Subtraction Operation


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2601" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a 0-indexed integer array nums of length n. You can perform the following operation as many times as you want:  Pick an index i that you haven’t picked before, and pick a prime p strictly less than nums[i], then subtract p from nums[i].  Return true if you can make nums a strictly increasing array using the above operation and false otherwise. A strictly increasing array is an array whose each element is strictly greater than its preceding element.   Example 1:  Input: nums = [4,9,6,10] Output: true Explanation: 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. Example 2:  Input: nums = [6,8,11,12] Output: true Explanation: Initially nums is sorted in strictly increasing order, so we don't need to make any operations. Example 3:  Input: nums = [5,8,3] Output: false Explanation: 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.   Constraints:  1 <= nums.length <= 1000 1 <= nums[i] <= 1000 nums.length == n  

	# Explanation
	Here's the solution to determine if an array can be made strictly increasing by subtracting primes.

*   **Key Idea:** Iterate through the array from left to right. For each element, determine the minimum value it must have to be strictly greater than the previous element. Then, check if we can reduce the current number to at least that minimum value by subtracting primes. We can achieve this if the difference between the current number and the minimum required value is less than the number itself and greater or equal to zero.

*   **Optimization:** Since the constraint specifies nums[i] <= 1000. We do not need to precompute all the primes. We can generate prime numbers in-line and check if the number can be reduced to at least minimum number required to make the array strictly increasing.

*   **Complexity:** O(n * sqrt(M)), where n is the length of `nums` and M is the maximum value in `nums`. The space complexity is O(1).

	
	# Code
	```python
	def is_strictly_increasing(nums):
    """
    Checks if an array can be made strictly increasing by subtracting primes.

    Args:
        nums: A list of integers.

    Returns:
        True if the array can be made strictly increasing, False otherwise.
    """

    prev = 0
    for i in range(len(nums)):
        required_min = prev + 1
        if nums[i] >= required_min:
            prev = nums[i]
            continue
        
        diff = required_min - nums[i]
        if diff < 0:
            return False

        
        if diff <= nums[i]:
            prev = nums[i]
            
            can_reduce = False
            temp = nums[i]
            
            
            
            n = nums[i]
            val = n
            k = 2
            total_subtracted = 0
            while k * k <= val:
                if val % k:
                    k += 1
                else:
                    total_subtracted += k
                    val //= k
                    k = 2
            if val > 1:
                total_subtracted += val
                
            
            if (nums[i] - diff) <= nums[i] and diff <= total_subtracted:
                can_reduce = True
        
            
            if can_reduce:
                prev = required_min
                continue
        
        return False

    return True
	```