Imagine you’re given an array of numbers, and you need to make it strictly increasing by subtracting prime numbers from each element. Sounds like a math puzzle, right? 🤔 Let’s dive into how we can achieve this efficiently using an optimized approach with prime numbers! 🌟

📝 Problem Statement

Given a 0-indexed integer array nums of length n, you can perform the following operation multiple times:

1. Pick an index i that hasn’t been picked before.
2. Pick a prime number p such that p is strictly less than nums[i].
3. Subtract p from nums[i].

The goal is to make nums a strictly increasing array. Return True if this is possible, otherwise return False.

A Strictly Increasing Array:

An array is strictly increasing if every element is greater than its previous element.

🔍 Examples

Example 1

• Input: nums = [4, 9, 6, 10]
• Output: True
• Explanation:
  1. Select i = 0, choose p = 3. After subtracting, nums becomes [1, 9, 6, 10].
  2. Select i = 1, choose p = 7. After subtracting, nums becomes [1, 2, 6, 10].

  The array is now strictly increasing.

Example 2

• Input: nums = [6, 8, 11, 12]
• Output: True
• Explanation: nums is already strictly increasing, so no operations are needed.

Example 3

• Input: nums = [5, 8, 3]
• Output: False
• Explanation: It’s impossible to make this array strictly increasing by subtracting primes.

🔍 Edge Cases

• Single Element Array: nums = [x] – No operation is required; return True.
• Already Sorted Array: Arrays like [2, 5, 7] – The function should detect this and return True without any modifications.
• All Elements Equal: nums = [7, 7, 7] – It is impossible to make it strictly increasing; return False.
• Large Prime Elements: Arrays with large values may require selecting specific primes for each number to achieve strict ordering.

💡 Approach 1: Basic Solution

The naive approach to solving this problem would involve:

1. Generate All Possible Primes: Create a list of primes that are strictly less than the largest element in nums.
2. Iterate Through Each Element: For each element in nums, try subtracting prime numbers to make it smaller and check if this leads to a strictly increasing array.

However, this brute-force method would be inefficient for larger arrays, as it requires recalculating primes multiple times and does not optimize for minimal operations.

🕰️ Time Complexity of Basic Solution

• Prime Generation: If we generate primes up to the maximum of nums, this can take O(N^2) in the worst case.
• Checking Strict Order: Each element may require an operation, resulting in an additional O(N) complexity.

This leads to an overall complexity of about O(N^2), which may be too slow for large inputs.

💡 Optimized Solution: Using Precomputed Primes and Binary Search 🧑‍💻

The optimized solution leverages a few key strategies:

1. Precompute Primes: We first generate a list of primes up to the largest element in nums using a prime-checking algorithm (like the Sieve of Eratosthenes).
2. Binary Search for Efficiency: For each nums[i], we use binary search to find the largest prime p such that nums[i] - p > nums[i - 1]. This ensures that nums[i] remains larger than nums[i-1], maintaining the strictly increasing property.
3. Early Exit: If we cannot find a prime to satisfy the condition for any nums[i], we immediately return False.

This approach is efficient because:

• Binary Search: Finding the appropriate prime for each element can be done in O(log P) where P is the number of primes, reducing unnecessary checks.
• Single Pass: We process each element once, making this solution very efficient.

🚀 Optimized Solution Code

from bisect import bisect_right
from typing import List

class Solution:
    def primeSubOperation(self, nums: List[int]) -> bool:
        # Generate a list of primes up to the max element in nums
        primes = []
        for i in range(2, max(nums)):
            for j in primes:
                if i % j == 0:
                    break
            else:
                primes.append(i)
        
        # Traverse the list from the second last element down to the first
        n = len(nums)
        for i in range(n - 2, -1, -1):
            if nums[i] < nums[i + 1]:
                continue
            # Use binary search to find the largest prime less than the difference
            j = bisect_right(primes, nums[i] - nums[i + 1])
            if j == len(primes) or primes[j] >= nums[i]:
                return False
            nums[i] -= primes[j]
        return True

🕰️ Time Complexity Analysis

Prime Generation: Generating primes up to max(nums) requires O(N log log N) using the Sieve of Eratosthenes.

Binary Search Operations: For each element in nums, binary search on the list of primes takes O(log P), where P is the number of primes generated.

Total Complexity: The overall time complexity is approximately O(N log log N), which is efficient for the problem constraints.

🧮 Example Walkthrough with Higher Numbers

Let’s take a closer look with an example that has larger values to understand how the solution operates efficiently.

Input: nums = [20, 30, 25, 40]

1. Step 1: Check if nums[2] < nums[3] (i.e., 25 < 40).
   • Yes, this part of the array is already sorted, so we move to the next element.
2. Step 2: Check if nums[1] < nums[2] (i.e., 30 < 25).
   • No, we need to subtract a prime from nums[1].
   • Binary Search: Find the largest prime p such that nums[1] - p > nums[0] and p < nums[1].
   • Select p = 5 (largest prime less than 30 - 25).
   • Subtract 5, so nums becomes [20, 25, 25, 40].
3. Step 3: Check if nums[0] < nums[1] (i.e., 20 < 25).
   • Yes, this part of the array is sorted.

After these adjustments, nums = [20, 25, 25, 40] is strictly increasing, so we return True.

🔚 Conclusion

This problem demonstrates a powerful combination of binary search and precomputed primes to solve a constraint-heavy problem efficiently. By leveraging prime subtraction, we can convert an unsorted array into a strictly increasing sequence, given that the conditions allow. This approach can be valuable in mathematical programming scenarios, emphasizing optimization through binary search and sieve algorithms.

With this in mind, prime subtraction operation provides a unique way to enforce ordering on arrays without traditional sorting, offering a neat mix of number theory and algorithmic efficiency! 🎉