Question

2761. Prime Pairs With Target Sum

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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:
  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

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;
  }
};

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;
}