**Unique Characters of Every Substring - Microsoft Top Interview Questions**

### Problem Statement :

Given a lowercase alphabet string s, return the sum of the number of characters that are unique in every substring of s. Mod the result by 10 ** 9 + 7. Constraints n ≤ 100,000 where n is the length of s Example 1 Input s = "aab" Output 6 Explanation Here are the substrings and their counts: "a" - 1 "a" - 1 "b" - 1 "aa" - 0 (since "a" is not unique) "ab" - 2 "aab" - 1 ("since "a" is not unique)

### Solution :

` ````
Solution in C++ :
int solve(string s) {
vector<int> ind(26, -1);
vector<int> left(s.size(), 0), right(s.size(), 0);
for (int i = 0; i < s.size(); i++) {
left[i] = ind[s[i] - 'a'];
ind[s[i] - 'a'] = i;
}
fill(ind.begin(), ind.end(), s.size());
for (int i = s.size() - 1; i >= 0; i--) {
right[i] = ind[s[i] - 'a'];
ind[s[i] - 'a'] = i;
}
int mod = 1e9 + 7;
long res = 0;
for (int i = 0; i < s.size(); i++) {
res = (res + ((i - left[i]) * (right[i] - i)) % mod) % mod;
}
return res;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
public int solve(String s) {
long MOD = 1000000007L;
int N = s.length();
ArrayList<Integer>[] indices = new ArrayList[26];
for (int i = 0; i < 26; i++) {
indices[i] = new ArrayList<Integer>();
indices[i].add(-1);
}
for (int i = 0; i < N; i++) {
indices[s.charAt(i) - 'a'].add(i);
}
int[] cur = new int[26];
for (int i = 0; i < 26; i++) {
cur[i] = 1;
indices[i].add(N);
}
long ans = 0L;
for (char c : s.toCharArray()) {
int n = c - 'a';
long a = indices[n].get(cur[n] + 1) - indices[n].get(cur[n]);
long b = indices[n].get(cur[n]) - indices[n].get(cur[n] - 1);
ans += a * b;
ans %= MOD;
cur[n] += 1;
}
return (int) (ans);
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, s):
total = 0
found = defaultdict(lambda: [-1])
for c, char in enumerate(s):
found[char].append(c)
if len(found[char]) == 2:
continue
current = found[char]
total += (current[-1] - current[-2]) * (current[-2] - current[-3])
end = len(s)
for item in found:
current = found[item]
total += (end - current[-1]) * (current[-1] - current[-2])
return total % (10 ** 9 + 7)
```

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