Prime Pairs With Target Sum

Prime Pairs With Target Sum - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: n = 12

› Output: [[5,7]]

💡 Note: The prime numbers ≤ 12 are: 2, 3, 5, 7, 11. We need pairs that sum to 12: 5 + 7 = 12, and both 5 and 7 are prime, so return [[5,7]].

Example 2 — Multiple Pairs

$ Input: n = 20

› Output: [[3,17],[7,13]]

💡 Note: Prime numbers ≤ 20: 2, 3, 5, 7, 11, 13, 17, 19. Valid pairs summing to 20: 3+17=20 and 7+13=20. Both pairs have prime numbers.

Example 3 — No Valid Pairs

$ Input: n = 4

› Output: []

💡 Note: Only primes ≤ 4 are 2 and 3. We need pairs summing to 4: 2+2=4, but we need x ≤ y and distinct pairs. No valid pairs exist.

Constraints

2 ≤ n ≤ 10⁶

Visualization

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Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is to precompute all prime numbers using the Sieve of Eratosthenes, then check pairs efficiently. Best approach uses sieve preprocessing for O(n log log n) time complexity and O(n) space.

Common Approaches

✓ Brute Force

⏱️ Time: O(n × √n) Space: O(1)

For each possible pair (x, y) where x + y = n, check if both x and y are prime numbers by testing divisibility up to their square roots.

Sieve of Eratosthenes

⏱️ Time: O(n log log n) Space: O(n)

Use the Sieve of Eratosthenes to precompute all prime numbers up to n, then iterate through primes to find valid pairs where both numbers sum to n.

Brute Force — Algorithm Steps

Iterate through all x from 2 to n/2
Calculate y = n - x
Check if both x and y are prime
If both prime and x <= y, add [x, y] to result

Visualization

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Step-by-Step Walkthrough

Test x=2

Check if 2 and n-2 are both prime

Test x=3

Check if 3 and n-3 are both prime

Continue

Test all x up to n/2

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>

bool isPrime(int num) {
    if (num < 2) return false;
    if (num == 2) return true;
    if (num % 2 == 0) return false;
    for (int i = 3; i * i <= num; i += 2) {
        if (num % i == 0) return false;
    }
    return true;
}

int main() {
    int n;
    scanf("%d", &n);
    
    printf("[");
    bool first = true;
    for (int x = 2; x <= n / 2; x++) {
        int y = n - x;
        if (x <= y && isPrime(x) && isPrime(y)) {
            if (!first) printf(",");
            printf("[%d,%d]", x, y);
            first = false;
        }
    }
    printf("]\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n × √n)

For each of n numbers, check primality in O(√n) time

✓ Linear Growth

Space Complexity

O(1)

Only uses variables for iteration and result storage

✓ Linear Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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