Prime Subtraction Operation

A more efficient approach is to use a greedy strategy. When processing the array from left to right, we make a locally optimal choice at each step. The key insight is that to make it easiest for the subsequent elements to be larger, we should make the current element as small as possible, while still being greater than the previous element.

• Pre-compute all primes up to 1001 using a Sieve.
• Create an auxiliary array, largestPrimeSmallerThan, of size 1002. largestPrimeSmallerThan[k] will store the largest prime strictly less than k. This can be filled in O(M) time after the sieve.
• Initialize a variable prev = 0.
• Iterate through nums from i = 0 to n-1:
  • If nums[i] <= prev, return false.
  • Calculate the difference: diff = nums[i] - prev.
  • Find the largest prime p less than diff using the pre-computed array: p = largestPrimeSmallerThan[diff].
  • If a valid prime p > 0 is found, update the current element's value for the next iteration: current_val = nums[i] - p.
  • Otherwise, we cannot subtract any prime, so current_val = nums[i].
  • Update prev = current_val.
• If the loop completes, return true.

The core idea is to iterate through the nums array from left to right, maintaining the value of the 'previous' element in the modified, strictly increasing sequence. Let's call this prev. Initially, prev can be considered 0.

For each element nums[i], we first check if it's possible to make it larger than prev. If nums[i] is already less than or equal to prev, it's impossible. This is because any subtraction will only make nums[i] smaller, so it can never become greater than prev. In this case, we can immediately return false.

If nums[i] > prev, we have a valid starting point. To be greedy, we want to modify nums[i] to be the smallest possible value that is still greater than prev. This is achieved by subtracting the largest possible prime p from nums[i] such that the result nums[i] - p is still greater than prev.

This condition nums[i] - p > prev is equivalent to p < nums[i] - prev. So, we need to find the largest prime p that is strictly less than the difference nums[i] - prev.

If such a prime exists, we subtract it from nums[i] to get the new value for the current position. If no such prime exists (e.g., if nums[i] - prev is 2 or less), we cannot make nums[i] any smaller, so we leave it as is.

We then update prev to this new, possibly modified value of nums[i] and proceed to the next element.

If we successfully process the entire array, it means a valid sequence can be formed, and we return true.

To efficiently find the 'largest prime less than k', we can pre-compute this information for all numbers up to the maximum possible value.

import java.util.Arrays;

class Solution {
    public boolean primeSubtraction(int[] nums) {
        int maxVal = 1002; // Max possible value for num or diff + buffer
        boolean[] isPrime = new boolean[maxVal];
        Arrays.fill(isPrime, true);
        isPrime[0] = isPrime[1] = false;
        for (int p = 2; p * p < maxVal; p++) {
            if (isPrime[p]) {
                for (int i = p * p; i < maxVal; i += p) {
                    isPrime[i] = false;
                }
            }
        }

        int[] largestPrimeSmallerThan = new int[maxVal];
        // largestPrimeSmallerThan[k] stores the largest prime < k
        for (int i = 3; i < maxVal; i++) {
            if (isPrime[i - 1]) {
                largestPrimeSmallerThan[i] = i - 1;
            } else {
                largestPrimeSmallerThan[i] = largestPrimeSmallerThan[i - 1];
            }
        }

        int prev = 0;
        for (int num : nums) {
            if (num <= prev) {
                return false;
            }
            int diff = num - prev;
            int p = largestPrimeSmallerThan[diff];
            
            if (p > 0) {
                prev = num - p;
            } else {
                prev = num;
            }
        }
        return true;
    }
}