Check whether the frequencies of all the characters in a string are prime or not in Python

Sometimes we need to check whether the frequencies of all characters in a string are prime numbers. A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself.

So, if the input is like s = "apuuppa", then the output will be True as there are two 'a's, three 'p's and two 'u's. Since 2 and 3 are prime numbers, all character frequencies are prime.

Algorithm

To solve this problem, we will follow these steps ?

• Create a frequency map containing all characters and their counts
• For each character frequency, check if it's a prime number
• If any frequency is not prime, return False
• If all frequencies are prime, return True

Example

Let us see the implementation to get better understanding ?

from collections import defaultdict

def isPrime(num):
    if num > 1:
        for i in range(2, int(num**0.5) + 1):
            if num % i == 0:
                return False
        return True
    return False

def solve(s):
    freq = defaultdict(int)
    
    # Count frequency of each character
    for char in s:
        freq[char] += 1
    
    # Check if all frequencies are prime
    for char in freq:
        if not isPrime(freq[char]):
            return False
    
    return True

s = "apuuppa"
print(f"String: {s}")
print(f"Result: {solve(s)}")

# Let's see the frequencies
freq = defaultdict(int)
for char in s:
    freq[char] += 1

print("\nCharacter frequencies:")
for char, count in freq.items():
    print(f"'{char}': {count} (Prime: {isPrime(count)})")

String: apuuppa
Result: True

Character frequencies:
'a': 2 (Prime: True)
'p': 3 (Prime: True)
'u': 2 (Prime: True)

Testing with Another Example

Let's test with a string where not all frequencies are prime ?

def isPrime(num):
    if num > 1:
        for i in range(2, int(num**0.5) + 1):
            if num % i == 0:
                return False
        return True
    return False

def solve(s):
    freq = {}
    
    # Count frequency of each character
    for char in s:
        freq[char] = freq.get(char, 0) + 1
    
    # Check if all frequencies are prime
    for char in freq:
        if not isPrime(freq[char]):
            return False
    
    return True

test_string = "aaaa"
print(f"String: {test_string}")
print(f"Result: {solve(test_string)}")

# Show frequencies
freq = {}
for char in test_string:
    freq[char] = freq.get(char, 0) + 1

print("\nCharacter frequencies:")
for char, count in freq.items():
    print(f"'{char}': {count} (Prime: {isPrime(count)})")

String: aaaa
Result: False

Character frequencies:
'a': 4 (Prime: False)

Key Points

• We optimize the prime checking function by only checking divisors up to ?num
• Character frequency counting can be done using defaultdict or regular dictionary
• Numbers 1 and below are not considered prime
• The function returns False as soon as any non-prime frequency is found

Conclusion

This solution efficiently checks if all character frequencies in a string are prime numbers. The time complexity is O(n + k?m) where n is string length, k is unique characters, and m is the maximum frequency.