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3115. Maximum Prime Difference

Description

You are given an integer array nums.

Return an integer that is the maximum distance between the indices of two (not necessarily different) prime numbers in nums.

 

Example 1:

Input: nums = [4,2,9,5,3]

Output: 3

Explanation: nums[1], nums[3], and nums[4] are prime. So the answer is \|4 - 1\| = 3.

Example 2:

Input: nums = [4,8,2,8]

Output: 0

Explanation: nums[2] is prime. Because there is just one prime number, the answer is \|2 - 2\| = 0.

 

Constraints:

  • 1 <= nums.length <= 3 * 105
  • 1 <= nums[i] <= 100
  • The input is generated such that the number of prime numbers in the nums is at least one.

Solutions

Solution 1: Traversal

According to the problem description, we need to find the index $i$ of the first prime number, then find the index $j$ of the last prime number, and return $j - i$ as the answer.

Therefore, we can traverse the array from left to right to find the index $i$ of the first prime number, then traverse the array from right to left to find the index $j$ of the last prime number. The answer is $j - i$.

The time complexity is $O(n \times \sqrt{M})$, where $n$ and $M$ are the length of the array $nums$ and the maximum value in the array, respectively. The space complexity is $O(1)$.

  • class Solution {
    public int maximumPrimeDifference(int[] nums) {
        for (int i = 0;; ++i) {
            if (isPrime(nums[i])) {
                for (int j = nums.length - 1;; --j) {
                    if (isPrime(nums[j])) {
                        return j - i;
                    }
                }
            }
        }
    }

    private boolean isPrime(int x) {
        if (x < 2) {
            return false;
        }
        for (int v = 2; v * v <= x; ++v) {
            if (x % v == 0) {
                return false;
            }
        }
        return true;
    }
}

  • class Solution {
public:
    int maximumPrimeDifference(vector<int>& nums) {
        for (int i = 0;; ++i) {
            if (isPrime(nums[i])) {
                for (int j = nums.size() - 1;; --j) {
                    if (isPrime(nums[j])) {
                        return j - i;
                    }
                }
            }
        }
    }

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

  • class Solution:
    def maximumPrimeDifference(self, nums: List[int]) -> int:
        def is_prime(x: int) -> bool:
            if x < 2:
                return False
            return all(x % i for i in range(2, int(sqrt(x)) + 1))

        for i, x in enumerate(nums):
            if is_prime(x):
                for j in range(len(nums) - 1, i - 1, -1):
                    if is_prime(nums[j]):
                        return j - i


  • func maximumPrimeDifference(nums []int) int {
	for i := 0; ; i++ {
		if isPrime(nums[i]) {
			for j := len(nums) - 1; ; j-- {
				if isPrime(nums[j]) {
					return j - i
				}
			}
		}
	}
}

func isPrime(n int) bool {
	if n < 2 {
		return false
	}
	for i := 2; i <= n/i; i++ {
		if n%i == 0 {
			return false
		}
	}
	return true
}

  • function maximumPrimeDifference(nums: number[]): number {
    const isPrime = (x: number): boolean => {
        if (x < 2) {
            return false;
        }
        for (let i = 2; i <= x / i; i++) {
            if (x % i === 0) {
                return false;
            }
        }
        return true;
    };
    for (let i = 0; ; ++i) {
        if (isPrime(nums[i])) {
            for (let j = nums.length - 1; ; --j) {
                if (isPrime(nums[j])) {
                    return j - i;
                }
            }
        }
    }
}

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