# Split String with Same Distinct Counts - Google Top Interview Questions

### Problem Statement :

You are given a lowercase alphabet string s.

Return the number of ways to split the string into two strings such that the number of distinct characters in each string is the same.

Constraints

1 ≤ n ≤ 100,000 where n is the length of s

Example 1

Input

s = "abaab"

Output

2] = hs.size();
}

Explanation

We can split it by "ab" + "aab" and "aba" + "ab"

### Solution :

Solution in C++ :

int solve(string s) {
vector<int> r(s.size(), 0);

unordered_map<char, int> fre;

for (int j = s.size() - 1; j >= 0; j--) fre[s[j]]++, r[j] = fre.size();

fre.clear();
int res = 0;

for (int i = 0; i + 1 < s.size(); i++) {
fre[s[i]]++;
if (fre.size() == r[i + 1]) res++;
}

return res;
}

Solution in Java :

import java.util.*;

class Solution {
public int solve(String s) {
int n = s.length();
int[] prefix = new int[n];
int[] suffix = new int[n];
HashSet<Character> hs = new HashSet();

for (int i = 0; i < n; i++) {
prefix[i] = hs.size();
}

hs = new HashSet();
for (int i = n - 1; i >= 0; i--) {
suffix[i
int res = 0;
for (int i = 0; i < n - 1; i++) {
if (prefix[i] == suffix[i + 1])
res++;
}
return res;
}
}

Solution in Python :

class Solution:
def solve(self, s):

forward_cnt = [0] * len(s)
seen = set()
temp = 0
for i in range(len(s)):
if s[i] not in seen:
temp += 1
forward_cnt[i] = temp
else:
forward_cnt[i] = forward_cnt[i - 1]

backward_cnt = [0] * len(s)
seen = set()
temp = 0
for i in range(len(s) - 1, -1, -1):
if s[i] not in seen:
temp += 1
backward_cnt[i] = temp
else:
backward_cnt[i] = backward_cnt[i + 1]

ans = 0
for i in range(len(s) - 1):
if forward_cnt[i] == backward_cnt[i + 1]:
ans += 1
return ans

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