两个字符串的不同异或值

Question

You are given two strings of length n, you have to find distinct xor values by rearranging the elements in an arbitrary way of the two strings.

Here is what I tried:

String1 = input()
String2 = input()
s1_ones = String1.count('1')  #Number of 1's in string1
s2_ones = String2.count('1')  #Number of 1's in string2

minimum_overlap = s1_ones+s2_ones-n  #minimum overlapping 1's when performing xor
maximum_overlap = min(s1_ones,s2_ones) #maximum overlapping 1's when performing xor

if(minimum_overlap<0):
    minimum_overlap = 0
ans = 0
for x in range(minimum_overlap,maximum_overlap+1):
    resulting_ones = s1_ones + s2_ones - 2*x   #number of ones in resulting string
    ans+=(nCr(n,resulting_ones)) #nCr is a function which returns number of possible values of resulting String.
print(ans)

I would like to explain nCr(n,r) funtion a little more. Given that you have string of length 5 in which number of ones are say 3. Here, N = 5 String = '11100' nCr(5,3) will give all the possible values of the provided string which is (5!/3!*2!) = 10.
What is wrong with this approach?

Here is the nCr(n,k):

def nCr(n, k): 
    if(k > n - k): 
        k = n - k 
    res = 1
    for i in range(k): 
        res = res * (n - i) 
        res = res / (i + 1) 
    return (res)

Answer 1

I hope, This answers all your questions

import operator as op
from functools import reduce

import operator as op
from functools import reduce

def nCr(n, r):
    r = min(r, n-r)
    numer = reduce(op.mul, range(n, n-r, -1), 1)
    denom = reduce(op.mul, range(1, r+1), 1)
    return numer / denom

String1 = input()
String2 = input()
n=len(String1)
s1_ones = String1.count('1')
s2_ones = String2.count('1') 

minimum_overlap = max(s1_ones,s2_ones)-min(s1_ones,s2_ones)
zero=n-max(s1_ones,s2_ones)
x=min(s1_ones,s2_ones)
ans = 0
while zero>-1 and x>-1:
    zero-=1
    x-=1
    ans+=nCr(n,minimum_overlap)%1000000007
    minimum_overlap+=2
print(int(ans))

now this nCr function is faster than factorial version, but if it still take time to execute, you can use fermat's theorem for nCr%p (also precompute factorial array) which you can find on internet