README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Medium	https://github.com/doocs/leetcode/edit/main/solution/2500-2599/2523.Closest%20Prime%20Numbers%20in%20Range/README_EN.md	1649	Weekly Contest 326 Q4	Math Number Theory

2523. Closest Prime Numbers in Range

中文文档

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

 

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Solutions

Solution 1

Python3

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

Java

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;
    }
}

C++

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) {
            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;
            }
        }
        vector<int> ans = {-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;
    }
};

Go

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