LeetCode-in-Java
3591. Check if Any Element Has Prime Frequency
Easy
You are given an integer array nums.
Return true if the frequency of any element of the array is prime, otherwise, return false.
The frequency of an element x is the number of times it occurs in the array.
A prime number is a natural number greater than 1 with only two factors, 1 and itself.
Example 1:
Input: nums = [1,2,3,4,5,4]
Output: true
Explanation:
4 has a frequency of two, which is a prime number.
Example 2:
Input: nums = [1,2,3,4,5]
Output: false
Explanation:
All elements have a frequency of one.
Example 3:
Input: nums = [2,2,2,4,4]
Output: true
Explanation:
Both 2 and 4 have a prime frequency.
Constraints:
1 <= nums.length <= 1000 <= nums[i] <= 100
Solution
public class Solution {
private boolean isPrime(int n) {
if (n <= 1) {
return false;
}
if (n == 2 || n == 3) {
return true;
}
for (int i = 2; i < n; i++) {
if (n % i == 0) {
return false;
}
}
return true;
}
public boolean checkPrimeFrequency(int[] nums) {
int n = nums.length;
if (n == 1) {
return false;
}
int maxi = Integer.MIN_VALUE;
for (int val : nums) {
maxi = Math.max(val, maxi);
}
int[] hash = new int[maxi + 1];
for (int num : nums) {
hash[num]++;
}
for (int j : hash) {
if (isPrime(j)) {
return true;
}
}
return false;
}
}