LeetCode #3591 — EASY

Check if Any Element Has Prime Frequency

Build confidence with an intuition-first walkthrough focused on array fundamentals.

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The Problem

Problem Statement

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 <= 100
0 <= nums[i] <= 100

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Math

Example 1

[1,2,3,4,5,4]

Example 2

[1,2,3,4,5]

Example 3

[2,2,2,4,4]

Step 02

Core Insight

What unlocks the optimal approach

Implement a function to check if a number is prime
Find the frequencies of all the distinct elements in the array (using data structures like map in C++), check if there is any element with prime frequency

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3591: Check if Any Element Has Prime Frequency
import java.util.*;

class Solution {
    public boolean checkPrimeFrequency(int[] nums) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int x : nums) {
            cnt.merge(x, 1, Integer::sum);
        }

        for (int x : cnt.values()) {
            if (isPrime(x)) {
                return true;
            }
        }
        return false;
    }

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

// Accepted solution for LeetCode #3591: Check if Any Element Has Prime Frequency
func checkPrimeFrequency(nums []int) bool {
	cnt := make(map[int]int)
	for _, x := range nums {
		cnt[x]++
	}
	for _, x := range cnt {
		if isPrime(x) {
			return true
		}
	}
	return false
}

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

# Accepted solution for LeetCode #3591: Check if Any Element Has Prime Frequency
class Solution:
    def checkPrimeFrequency(self, nums: List[int]) -> bool:
        def is_prime(x: int) -> bool:
            if x < 2:
                return False
            return all(x % i for i in range(2, int(sqrt(x)) + 1))

        cnt = Counter(nums)
        return any(is_prime(x) for x in cnt.values())

// Accepted solution for LeetCode #3591: Check if Any Element Has Prime Frequency
fn check_prime_frequency(nums: Vec<i32>) -> bool {
    use std::collections::HashMap;

    let mut primes = [true; 101];
    primes[0] = false;
    primes[1] = false;

    for i in 2..=100 {
        if !primes[i] {
            continue;
        }
        primes[i] = true;
        let mut v = i * 2;
        while v <= 100 {
            primes[v] = false;
            v += i;
        }
    }

    let mut h = HashMap::new();
    for num in nums {
        *h.entry(num).or_insert(0) += 1;
    }

    for &v in h.values() {
        if primes[v as usize] {
            return true;
        }
    }

    false
}

fn main() {
    let nums = vec![3,0,3,6,3,3];
    let ret = check_prime_frequency(nums);
    println!("ret={ret}");
}

#[test]
fn test() {
    {
        let nums = vec![1, 2, 3, 4, 5, 4];
        assert!(check_prime_frequency(nums));
    }
    {
        let nums = vec![1, 2, 3, 4, 5];
        assert!(!check_prime_frequency(nums));
    }
    {
        let nums = vec![2, 2, 2, 4, 4];
        assert!(check_prime_frequency(nums));
    }
    {
        let nums = vec![3, 0, 3, 6, 3, 3];
        assert!(!check_prime_frequency(nums));
    }
}

// Accepted solution for LeetCode #3591: Check if Any Element Has Prime Frequency
function checkPrimeFrequency(nums: number[]): boolean {
    const cnt: Record<number, number> = {};
    for (const x of nums) {
        cnt[x] = (cnt[x] || 0) + 1;
    }
    for (const x of Object.values(cnt)) {
        if (isPrime(x)) {
            return true;
        }
    }
    return false;
}

function isPrime(x: number): boolean {
    if (x < 2) {
        return false;
    }
    for (let i = 2; i * i <= x; i++) {
        if (x % i === 0) {
            return false;
        }
    }
    return true;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × sqrtM)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.