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2761. Prime Pairs With Target Sum

Description

You are given an integer n. We say that two integers x and y form a prime number pair if:

  • 1 <= x <= y <= n
  • x + y == n
  • x and y are prime numbers

Return the 2D sorted list of prime number pairs [xi, yi]. The list should be sorted in increasing order of xi. If there are no prime number pairs at all, return an empty array.

Note: A prime number is a natural number greater than 1 with only two factors, itself and 1.

 

Example 1:

Input: n = 10
Output: [[3,7],[5,5]]
Explanation: In this example, there are two prime pairs that satisfy the criteria. 
These pairs are [3,7] and [5,5], and we return them in the sorted order as described in the problem statement.

Example 2:

Input: n = 2
Output: []
Explanation: We can show that there is no prime number pair that gives a sum of 2, so we return an empty array.

 

Constraints:

  • 1 <= n <= 106

Solutions

Solution 1: Preprocessing + Enumeration

First, we pre-process all the prime numbers within the range of $n$, and record them in the array $primes$, where $primes[i]$ is true if $i$ is a prime number.

Next, we enumerate $x$ in the range of $[2, \frac{n}{2}]$. In this case, $y = n - x$. If both $primes[x]$ and $primes[y]$ are true, then $(x, y)$ is a pair of prime numbers, which is added to the answer.

After the enumeration is complete, we return the answer.

The time complexity is $O(n \log \log n)$ and the space complexity is $O(n)$, where $n$ is the number given in the problem.

  • class Solution {
    public List<List<Integer>> findPrimePairs(int n) {
        boolean[] primes = new boolean[n];
        Arrays.fill(primes, true);
        for (int i = 2; i < n; ++i) {
            if (primes[i]) {
                for (int j = i + i; j < n; j += i) {
                    primes[j] = false;
                }
            }
        }
        List<List<Integer>> ans = new ArrayList<>();
        for (int x = 2; x <= n / 2; ++x) {
            int y = n - x;
            if (primes[x] && primes[y]) {
                ans.add(List.of(x, y));
            }
        }
        return ans;
    }
}

  • class Solution {
public:
    vector<vector<int>> findPrimePairs(int n) {
        bool primes[n];
        memset(primes, true, sizeof(primes));
        for (int i = 2; i < n; ++i) {
            if (primes[i]) {
                for (int j = i + i; j < n; j += i) {
                    primes[j] = false;
                }
            }
        }
        vector<vector<int>> ans;
        for (int x = 2; x <= n / 2; ++x) {
            int y = n - x;
            if (primes[x] && primes[y]) {
                ans.push_back({x, y});
            }
        }
        return ans;
    }
};

  • class Solution:
    def findPrimePairs(self, n: int) -> List[List[int]]:
        primes = [True] * n
        for i in range(2, n):
            if primes[i]:
                for j in range(i + i, n, i):
                    primes[j] = False
        ans = []
        for x in range(2, n // 2 + 1):
            y = n - x
            if primes[x] and primes[y]:
                ans.append([x, y])
        return ans


  • func findPrimePairs(n int) (ans [][]int) {
	primes := make([]bool, n)
	for i := range primes {
		primes[i] = true
	}
	for i := 2; i < n; i++ {
		if primes[i] {
			for j := i + i; j < n; j += i {
				primes[j] = false
			}
		}
	}
	for x := 2; x <= n/2; x++ {
		y := n - x
		if primes[x] && primes[y] {
			ans = append(ans, []int{x, y})
		}
	}
	return
}

  • function findPrimePairs(n: number): number[][] {
    const primes: boolean[] = new Array(n).fill(true);
    for (let i = 2; i < n; ++i) {
        if (primes[i]) {
            for (let j = i + i; j < n; j += i) {
                primes[j] = false;
            }
        }
    }
    const ans: number[][] = [];
    for (let x = 2; x <= n / 2; ++x) {
        const y = n - x;
        if (primes[x] && primes[y]) {
            ans.push([x, y]);
        }
    }
    return ans;
}

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