仅通过一次交换，如何找到字符串的所有可能排列？

Question

The complexity should be O(n) where n is length of string. Ex: For 'abc', answer is 'bac', 'cba', 'acb'. 'bca' and 'cab' should not be in the list as two swaps are required to convert it to 'abc'.

I have made a O(n ² ) algorithm but it is very slow.

def f(s):
    temp=list(s)
    l=[]
    for i in range(len(s)):
        for j in range(len(s)):
            temp=list(s)
            temp[i],temp[j]=temp[j],temp[i]
            l.append("".join(str(i) for i in temp))
    print set(l)

Answer 1

The number of possible outcomes for a string (with distinct characters) of length n is nC2 = n * (n-1) / 2 , since we can choose two letters at any two indices and swap them.

Hence, if you plan to print all of the outcomes, the complexity will be O(n^2) and a O(n) solution is not possible.

For duplicate characters, the reasoning becomes more complex. Suppose there is exactly one duplicate character repeated k times. Then there are nk swaps that will be identical. So if there is exactly one character repeated, and it is repeated k times, the number of possibilities is nC2 - (nk) . This can be extended to more repeated characters using the inclusion-exclusion principle.

Answer 2

Use itertools.combinations to find all possible combination of two char index, then swapped it.

# -*- coding: utf-8 -*-
import itertools


def f(s):
    result = []
    # This will produce: [0, 1], [0, 2], [1, 2]
    for idx1, idx2 in itertools.combinations(range(len(s)), 2):
        swapped_s = list(s)
        swapped_s[idx1], swapped_s[idx2] = swapped_s[idx2], swapped_s[idx1]
        result.append(''.join(swapped_s))
    return result

if __name__ == '__main__':
    print f('abc')

The above code will give you the correct result:

['bac', 'cba', 'acb']

Hope it helps!