题解 | 小红的双生串

import sys

s = input()
count = 0

s1 = s[:len(s)//2]     # 字符串
s2 = s[len(s)//2:]

s11 = list(set(s1))    # 去重后的列表
s22 = list(set(s2))

i1 = 0
i2 = 0

A = []
B = []

while i1 < len(s11) and i2 < len(s22):
    A.append(s1.count(s11[i1]))
    B.append(s2.count(s22[i2]))
    i1 += 1
    i2 += 1

temp_max_s11 = A[0]
temp_max_s22 = B[0]

temp_max_s11_zifu = s11[0]
temp_max_s22_zifu = s22[0]

j1 = 1
j2 = 1

count_s11 = 0
count_s22 = 0

while j1 < len(s11) and j2 < len(s22):
    if A[j1] > temp_max_s11:
        temp_max_s11 = A[j1]
        temp_max_s11_zifu = s11[j1]
    if B[j2] > temp_max_s22:
        temp_max_s22 = B[j2]
        temp_max_s22_zifu = s22[j2]
    j1 += 1
    j2 += 1

k1 = 0
k2 = 0

while k1 < len(s1) and k2 < len(s2):
    if temp_max_s11_zifu != s1[k1]:
        count_s11 += 1
    if temp_max_s22_zifu != s2[k2]:
        count_s22 += 1
    k1 += 1
    k2 += 1

print(count_s11 + count_s22)

实在不会起名字，自己都被自己写的恶心到了，总之思路就是先把字符串分成两半，分别找到各自出现过次数最多的字符s1，s2,然后把前半部分不是s1的字符全部替换成s1,以及后半部分不是s2的字符全部替换成s2，因为替换数量最多的字符，替换次数就自然最少。