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Challenge 116: Permutation Count

curriculum/challenges/english/blocks/daily-coding-challenges-python/69162d64f96574d9bb629f01.md

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--description--

Given a string, return the number of distinct permutations that can be formed from its characters.

  • A permutation is any reordering of the characters in the string.
  • Do not count duplicate permutations.
  • If the string contains repeated characters, repeated arrangements should only be counted once.
  • The string will contain only letters (A-Z, a-z).

For example, given "abb", return 3 because there's three unique ways to arrange the letters: "abb", "bab", and "bba".

--hints--

count_permutations("abb") should return 3.

js
({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_permutations("abb"), 3)`)
}})

count_permutations("abc") should return 6.

js
({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_permutations("abc"), 6)`)
}})

count_permutations("racecar") should return 630.

js
({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_permutations("racecar"), 630)`)
}})

count_permutations("freecodecamp") should return 39916800.

js
({test: () => { runPython(`
from unittest import TestCase
TestCase().assertEqual(count_permutations("freecodecamp"), 39916800)`)
}})

--seed--

--seed-contents--

py
def count_permutations(s):

    return s

--solutions--

py
from math import factorial
from collections import Counter
def count_permutations(s):
    freq = Counter(s)
    n = len(s)

    result = factorial(n)
    for count in freq.values():
        result //= factorial(count)

    return result