Python dictionary, set and counter to check if frequencies can become same

Python dictionary, set and counter to check if frequencies can become same

To check if frequencies of all characters of a string can become the same by just one removal, we can use Python's Counter from the collections module to calculate the frequencies of each character, and then we analyze the frequency counts.

Here's a step-by-step approach:

  1. Count the frequency of each character in the string using Counter.
  2. Get the set of all frequency counts.
  3. If there's only one type of frequency count, return True (all frequencies are already the same).
  4. If there are two types of frequency counts:
    • One of them must be 1 (indicating a single character can be removed), and
    • The other frequency count must be only one higher than the other counts (indicating that removing one occurrence of one character will equalize the frequencies).
  5. Otherwise, return False.

Here is the Python function that does this:

from collections import Counter

def can_frequencies_become_same(s):
    freq = Counter(s)
    freq_counts = Counter(freq.values())

    if len(freq_counts) == 1:
        # All characters already have the same frequency
        return True
    elif len(freq_counts) == 2:
        # Check for the case where one removal can make frequencies the same
        keys = list(freq_counts.keys())
        if 1 in freq_counts.values() and (abs(keys[0] - keys[1]) == 1 or min(keys) == 1):
            return True
    return False

# Examples
print(can_frequencies_become_same('xxxyyzz'))  # Should return True
print(can_frequencies_become_same('xxxxyyzz'))  # Should return False

Explanation of the key part:

  • If there's only one frequency count, all characters already have the same frequency, so the function returns True.
  • If there are exactly two frequency counts, one of the frequencies must be for a single character (i.e., it appears only once in the string), and the other frequency must be only one occurrence higher than the rest. This means you can remove one instance of a character with the higher frequency to make all frequencies the same.

The Counter objects make it easy to implement this logic succinctly and efficiently.

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