Problem Understanding

Eldrin is tasked with finding a contiguous segment of stones whose sum of magical powers equals the target value $M$. In addition, among all such possible segments, his goal is to choose the one that contains the maximum number of prime-powered stones. If no segment sums to $M$, the solution should return $-1$. The prime numbers indicate stones with special mystical properties, and the problem thus combines two distinct challenges:

• Finding a contiguous segment with a given summation property.
• Counting and maximizing the number of primes in that segment.

The array of stone powers is strictly positive, which hints that the two-pointer or sliding window technique could be efficiently applied.

Solution Analysis

1. Preprocessing Primality:

   • Since each stone has a power represented as a positive integer, we first preprocess each stone's value by determining if it is prime.
   • This preprocessing can be performed using a helper function (e.g., is_prime) for each stone or via a sieve method if the maximum value among stones is high and checking multiple values.

2. Sliding Window Technique:

   • Given that all stone powers are positive, a sliding window (two-pointer) approach is ideal.
   • Initialize pointers: set two indices, left and right, at the start of the list, and maintain two variables: current_sum to store the sum of the current window and current_prime_count for the count of prime stones in the window.
   • Expand the window by moving the right pointer and adding the value (and updating the prime count according to our preprocessed flags) until current_sum is greater than or equal to $M$.
   • If current_sum exceeds $M$, adjust the window by moving the left pointer forward, subtracting the corresponding stone’s value and adjusting the prime count accordingly.
   • When current_sum is exactly $M$, update the answer with the maximum count of prime-powered stones seen so far.
   • Continue this process until the right pointer has processed all stones.

3. Edge Cases:

   • If no contiguous segment summing exactly to $M$ is found, return $-1$.
   • Manage scenarios where the segment could be at the beginning, middle, or end of the array.

Time and Space Complexity Analysis

1. Time Complexity:

   • Preprocessing each stone for primality takes $O(N \times \sqrt{A})$ in the worst case if a naive prime check is used, where $A$ is the value on a stone.
   • The sliding window technique itself runs in $O(N)$ because each stone is processed at most twice (once when the right pointer adds it and once when the left pointer removes it).
   • Hence, the overall time complexity is $O(N \times \sqrt{A})$ if individual prime checking is used. If a more efficient sieve is applied when appropriate, the preprocessing can be reduced accordingly.

2. Space Complexity:

   • An auxiliary array for the prime status of each stone requires $O(N)$ space.
   • Additional variables are used for pointers and count tracking, which require $O(1)$ extra space.
   • Therefore, the total space complexity is $O(N)$.