Check if Any Element Has Prime Frequency

Check if Any Element Has Prime Frequency - Problem

Array Hash Table Math Counting Number Theory 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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [4,6,4,6,4]

› Output: true

💡 Note: Element 4 appears 3 times and 3 is prime. Element 6 appears 2 times and 2 is prime. Since both frequencies are prime, return true.

Example 2 — No Prime Frequencies

$ Input: nums = [1,1,1,1]

› Output: false

💡 Note: Element 1 appears 4 times. 4 is not prime (divisible by 1, 2, and 4), so return false.

Example 3 — Single Element

$ Input: nums = [5]

› Output: false

💡 Note: Element 5 appears 1 time. 1 is not prime (prime numbers must be greater than 1), so return false.

Constraints

1 ≤ nums.length ≤ 1000
-1000 ≤ nums[i] ≤ 1000

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is to count element frequencies efficiently using a hash map, then check each frequency for primality with optimized testing (only check divisors up to square root). Best approach is the frequency count method. Time: O(n + k√f), Space: O(n)

Common Approaches

✓ Optimized

⏱️ Time: N/A Space: N/A

Brute Force

⏱️ Time: O(n² × √f) Space: O(1)

For each element, count its frequency by scanning the entire array. Then for each frequency, check if it's prime by testing divisibility from 2 to frequency-1.

Frequency Count with Hash Map

⏱️ Time: O(n + k√f) Space: O(n)

First pass through array builds a frequency map. Second pass checks each frequency for primality using optimized prime testing (only check divisors up to square root).

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>

bool isPrime(int n) {
    if (n <= 1) {
        return false;
    }
    if (n <= 3) {
        return true;
    }
    if (n % 2 == 0 || n % 3 == 0) {
        return false;
    }
    
    int i = 5;
    while (i * i <= n) {
        if (n % i == 0 || n % (i + 2) == 0) {
            return false;
        }
        i += 6;
    }
    return true;
}

bool solution(int* nums, int numsSize) {
    if (numsSize == 0) {
        return false;
    }
    
    // Count frequencies using simple array approach
    int values[10000];
    int freqs[10000];
    int uniqueCount = 0;
    
    for (int i = 0; i < numsSize; i++) {
        int found = -1;
        for (int j = 0; j < uniqueCount; j++) {
            if (values[j] == nums[i]) {
                found = j;
                break;
            }
        }
        
        if (found >= 0) {
            freqs[found]++;
        } else {
            values[uniqueCount] = nums[i];
            freqs[uniqueCount] = 1;
            uniqueCount++;
        }
    }
    
    // Check if any frequency is prime
    for (int i = 0; i < uniqueCount; i++) {
        if (isPrime(freqs[i])) {
            return true;
        }
    }
    return false;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array manually
    int nums[1000];
    int numsSize = 0;
    
    char* ptr = line;
    while (*ptr && *ptr != '[') ptr++; // Find start bracket
    if (*ptr == '[') ptr++;
    
    while (*ptr && *ptr != ']' && numsSize < 1000) {
        if (*ptr == '-' || (*ptr >= '0' && *ptr <= '9')) {
            nums[numsSize] = strtol(ptr, &ptr, 10);
            numsSize++;
        } else {
            ptr++;
        }
    }
    
    bool result = solution(nums, numsSize);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

✓ Linear Space

8.5K Views

Medium Frequency

~12 min Avg. Time

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Ln 1, Col 1

Smart Actions

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