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2523. Closest Prime Numbers in Range

Description

Given two positive integers left and right, find the two integers num1 and num2 such that:

  • left <= num1 < num2 <= right .
  • num1 and num2 are both prime numbers.
  • num2 - num1 is the minimum amongst all other pairs satisfying the above conditions.

Return the positive integer array ans = [num1, num2]. If there are multiple pairs satisfying these conditions, return the one with the minimum num1 value or [-1, -1] if such numbers do not exist.

A number greater than 1 is called prime if it is only divisible by 1 and itself.

 

Example 1:

Input: left = 10, right = 19
Output: [11,13]
Explanation: The prime numbers between 10 and 19 are 11, 13, 17, and 19.
The closest gap between any pair is 2, which can be achieved by [11,13] or [17,19].
Since 11 is smaller than 17, we return the first pair.

Example 2:

Input: left = 4, right = 6
Output: [-1,-1]
Explanation: There exists only one prime number in the given range, so the conditions cannot be satisfied.

 

Constraints:

  • 1 <= left <= right <= 106

 

Solutions

Solution 1

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class Solution:
    def closestPrimes(self, left: int, right: int) -> List[int]:
        cnt = 0
        st = [False] * (right + 1)
        prime = [0] * (right + 1)
        for i in range(2, right + 1):
            if not st[i]:
                prime[cnt] = i
                cnt += 1
            j = 0
            while prime[j] <= right // i:
                st[prime[j] * i] = 1
                if i % prime[j] == 0:
                    break
                j += 1
        p = [v for v in prime[:cnt] if left <= v <= right]
        mi = inf
        ans = [-1, -1]
        for a, b in pairwise(p):
            if (d := b - a) < mi:
                mi = d
                ans = [a, b]
        return ans

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class Solution {
    public int[] closestPrimes(int left, int right) {
        int cnt = 0;
        boolean[] st = new boolean[right + 1];
        int[] prime = new int[right + 1];
        for (int i = 2; i <= right; ++i) {
            if (!st[i]) {
                prime[cnt++] = i;
            }
            for (int j = 0; prime[j] <= right / i; ++j) {
                st[prime[j] * i] = true;
                if (i % prime[j] == 0) {
                    break;
                }
            }
        }
        int i = -1, j = -1;
        for (int k = 0; k < cnt; ++k) {
            if (prime[k] >= left && prime[k] <= right) {
                if (i == -1) {
                    i = k;
                }
                j = k;
            }
        }
        int[] ans = new int[] {-1, -1};
        if (i == j || i == -1) {
            return ans;
        }
        int mi = 1 << 30;
        for (int k = i; k < j; ++k) {
            int d = prime[k + 1] - prime[k];
            if (d < mi) {
                mi = d;
                ans[0] = prime[k];
                ans[1] = prime[k + 1];
            }
        }
        return ans;
    }
}

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class Solution {
public:
    vector<int> closestPrimes(int left, int right) {
        int cnt = 0;
        bool st[right + 1];
        memset(st, 0, sizeof st);
        int prime[right + 1];
        for (int i = 2; i <= right; ++i) {