ctci/08. Recursion/8.8. Permutations with duplicates.md

8.8. Permutations with duplicates

Write a method to compute all permutations of a string whose characters are not necessarily unique. The list of permutations should not have duplicates.

solution in 357

The easiest solution would be to create a set instead of a list.

• s = anee
• s = an, size=2
  • permutations = an, na
  • total 4
• s = ane, size=3
  • permutations = ean, ane, ena, nae + aen + nea
  • total 6, 3*2
• s = anee, size=4
  • permutations = anee, aeen, aene
  • eaen, eane, eean, eena, enae, enea
  • naee, neae, neea
  • total 12

# approach 1: building from permutations of first n-1 characters
def permutations(s):
    prm = set()
    if len(s) == 0:
        prm.append("")
        return prm
    for word in permutations(s[1:]):
        for i in range(len(word)):
            prm.add(word[:i] + s[0] + word[i:])
    return prm

print(permutations("str"))

Hints

If there are many duplicate letters, it becomes inefficient to check the set each time.

Get the count of each character, abcaac has 3a, 2c and 1b

How does that help? How can I omit doing all the permutations and checking if that permutation is already there?

Pick a starting character. a, b, c. If you start with a, then get all permuations for 2a, 2c and 1b.

Solution

Worst case will be O((n+1)!), but in other cases we want to only create the unique permutations.

Permutations of (a=3, b=1, c=2) is P(2, 1, 2) + P(3, 0, 2) + P(3, 1, 1) and so on recursively for each starting character.

def printPerms(string):
    result = []
    letterCountMap = defaultdict()
    for letter in string:
        letterCountMap[letter] += 1
    
    printPermsInner(letterCountMap, "", len(string), result)
    return result

def printPermsInner(letterCountMap, prefix, remaining, result):
    #base case Permutation has been completed
    if remaining == 0:
        result.append(prefix)
        return
    #try remaining letter for next char, and generate remaining permutations
    for character in letterCountMap:
        count = letterCountMap[character]
        if count > 0:
            letterCountMap[character] -= 1
            printPermsInner(letterCountMap, prefix + character, remaining - 1, result)
            letterCountMap[character] = count

print(printPerms("aaf"))