题解 | #abb#

n = int(input())
s = input()
#回溯
# result = []
# path = []
# def back(s):
    
    
#     if len(path) == 3 and path[0]!=path[1] and path[1] == path[2]:
#         # print(path)
#         result.append(path.copy())
#         return
#     if len(path)>=3:
#         return

#     for i in range(len(s)):
#         if path == [] or path[0] != s[i]:
#             path.append(s[i])
#             back(s[i+1:])
#             path.pop()

# back(s)
# print(len(result))

#数学方法
D = {}
for i in s:
    if i not in D:
        D[i] = 1
    else:
        D[i] += 1

result = 0
for i in range(len(s)):
    D[s[i]] -= 1 #扫描一个，向后走一个，把走过的在字典中减一
    for j in D:
        if j!=s[i]: #确保后面的和第一个是不同的字母
            result += (D[j] * (D[j] -1) ) //2  
print(result)

回溯法过了8/11，效率较低

数学用后缀和的方式进行，全部通过