Optimized Solution for Finding Prime Pairs Using Sieve of Eratosthenes
# IntuitionWhen given an integer (n), we are tasked with finding all unique pairs of prime numbers such that their sum is equal to (n). A brute force approach w
### # Intuition
When given an integer \(n\), we are tasked with finding all unique pairs of prime numbers such that their sum is equal to \(n\). A brute force approach where we check each number in the range \([2, n/2]\) for primality and then check if \(n - i\) is prime would be inefficient, especially for larger values of \(n\). Instead, we can utilize **Sieve of Eratosthenes** to efficiently find all prime numbers up to \(n\). Then, we can iterate through the primes to identify pairs whose sum equals \(n\).

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### # Approach
1. **Sieve of Eratosthenes**: First, we generate a list of prime numbers up to \(n\) using the Sieve of Eratosthenes. This allows us to determine whether a number is prime in constant time \(O(1)\).
2. **Prime Pair Search**: After constructing the list of primes, we iterate through the numbers from \(2\) to \(n/2\). For each number \(i\), we check if both \(i\) and \(n - i\) are prime. If they are, we add the pair \((i, n - i)\) to the result list.
3. **Avoid Duplicates**: To ensure that each pair is only added once, we mark the primes we've already used as non-prime once we've added the pair to the result.

This approach efficiently finds all the prime pairs in \(O(nloglogn)\) time due to the sieve, and then the pair-finding process runs in \(O(n/2)\), which is linear in nature.

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### # Complexity

- **Time complexity**:  
  The time complexity for generating the prime numbers using the Sieve of Eratosthenes is \(O(nloglogn)\). The iteration over the range \([2, n/2]\) to check the prime pairs takes \(O(n)\) time. Thus, the overall time complexity is \(O(nloglogn) \).

- **Space complexity**:  
  The space complexity is \(O(n)\), as we store a boolean array to track which numbers up to \(n\) are prime. The space for storing the result list is \(O(n)\) in the worst case.


# Code
```java []
class Solution {

    // Sieve of Eratosthenes
    private boolean[] f(int n) {
        boolean[] isPrime = new boolean[n + 1];
        Arrays.fill(isPrime, true);
        isPrime[0] = false;
        isPrime[1] = false;
        for(int p = 2; p * p <= n; p++) {
            if(isPrime[p]) {
                for(int i = p * p; i <= n; i += p) {
                    isPrime[i] = false;
                }
            }
        }
        return isPrime;
    }

    public List<List<Integer>> findPrimePairs(int n) {
        boolean[] isPrime = this.f(n);
        List<List<Integer>> answer = new ArrayList<>();
        for(int i = 2; i <= n / 2; i++) {
            if(isPrime[i] && isPrime[n - i]) {
                answer.add(Arrays.asList(i, n - i));
                isPrime[i] = false;
                isPrime[n - i] = false;
            }
        }
        return answer;
    }
}
```
['Math', 'Number Theory', 'Java']