Closest Prime Numbers in Range

I just used the Sieve of Eratosthenes algorithm to pre-generate a list of all the primes. Then, I can simply just iterate through [left, right] to find the 2 primes with the minimum difference.

code

class Solution {
public:
    vector<int> closestPrimes(int left, int right) {
        vector<int> primes (right+1, 1);
        primes[1] = 0; // 1 is not a prime

        // Sieve of Eratosthenes algorithm:
        // 1. start from smallest prime number (in this case is 2)
        // 2. set all its multiple's value to 0
        // 3. find the next smallest prime number (e.g. 3)
        // 4. repeat step 2.
        // 5. repeat step 3 and 4 until the end
        int i = 2;
        for (int i = 2; i*i <= right; ++i){
            if (primes[i] == 1) {
                for (int j = i*i; j <= right; j += i) {
                    primes[j] = 0;
                }
            }
        }

        vector<int> ans(2, -1);
        int smallest_difference = INT32_MAX;
        int previous_prime = -1;

        // iterate through every integer in [left, right]
        // checking if it is prime
        for (int i = left; i <= right; ++i) {
            if (primes[i] == 1) {
                // update value only if current difference
                // is smaller than the smallest_difference
                if (i-previous_prime < smallest_difference) {
                    smallest_difference = i-previous_prime;
                    ans[0] = previous_prime;
                    ans[1] = i;
                }
                // always set the previous_prime afterwards
                previous_prime = i;
            }
            // we know that the difference cannot be smaller than 2
            // (only exception is [2, 3])
            // so we can break early if we have found it already
            if (smallest_difference <= 2) {
                break;
            }
        }
        // if exists less than 2 primes in the set
        // return {-1, -1}
        if (ans[0] == -1) return {-1, -1};
        // else return ans
        return ans;
    }
};

complexity

• Time: O(right * log(log(right)))
• Space: O(right)

I did not realise this at the time, but a brute force solution gets 0 ms run-time on Leetcode, only if you break when smallest_difference <= 2. This is because the upper bound on the distance between twin primes is 1452 when you consider the limits in-place for this question (1 <= left <= right <= 10^6) (which can be verified through simulation).

Therefore, this means that a brute force solution is not all that bad as you only have to iterate through a maximum of 1452 integers and check if they are prime or not before you can guarantee a twin prime.

Code:

class Solution {
public:
    bool isPrime(int n) {
        if (n == 1) return false;
        for (int i = 2; i < n; ++i) {
            if (i*i>n){
                break;
            }
            if (n%i == 0) {
                return false;
            }
        }
        return true;
    }
    vector<int> closestPrimes(int left, int right) {
        vector<int> ans(2, -1);
        int prev = -1;
        int minDiff = INT32_MAX;

        for (int i = left; i <= right; ++i) {
            if (i%2==0 and i != 2){
                continue;
            }
            if (isPrime(i)) {
                if (i-prev < minDiff) {
                    minDiff = i-prev;
                    ans[0] = prev;
                    ans[1] = i;
                }
                prev = i;
            }
            if (minDiff <= 2){
                break;
            }
        }
        if (ans[0] == -1) return {-1, -1};
        return ans;
    }
};

Extra for experts:

complexity

• Time: O(min(1452, right-left) * sqrt(right))
• Space: O(1)

learnings

• while loops are so slow (use for loops whenever you can)
• the second solution is stupid