# (c++) 387 - First Unique Character in a String

Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.

### Brute force

##### Input: s = “leetcode”
• I take a look at l, I check if does repeat I go next. In this case it doesnt so it is the solution. We only care about the first one which doesnt repeat.
• I take a look at a, I check if does repeat I go next. In this case it repeat thefore I skip. Next value is also a, as before, so we should delete all the instances of a to be sure.

• Now the input is s = “ccddleetcode”, we repeat the previous step
• Now the input is s = “ddleetcode”, we repeat the previous step
• Now the input is s = “leetcode”, we repeat the previous step
• Now L doesnt repeat so it is the solution, its position in the original input was 4.

This is a $O(n^2)$

### Bottlenecks

Searching the repeated values makes the solution slow. Ideally we could make something like this

x = (frecuency, first index) “a = (3,0) c = (3,2) d = (3,5) l = (1,6) - exit e t o d “

Maybe I could a map with frequencies going once through the array, then I go again check the letters and its frequecy. This would be O(n)!

hash map of input - O(n) “a = 3 c = 3 d = 3 l = 1 e = 3 t = 1 o = 1”

check the array again - O(n)

a - 3, skip, a skip c skip … l return index

### Code

int firstUniqChar(string s) {
unordered_map<char,int> hs{};

for(const auto &i:s) hs[i]++; // O(n)

for(int i=0; i < s.size(); i++)
if(hs.at(s[i])==1) return i; //n x O(1)

return -1;
}


I need to go through the array at least once to check if the first values repeats or not. The best possible runtime is a O(n) order

## Code 2.0 without maps

We fix an array with every letter, in theory the lookup time is faster in this case than a map.

int firstUniqChar(string s) {
std::vector<int> v; hs(26);

for(const auto &i:s) hs[i]++; // O(n)

for(int i=0; i < s.size(); i++)
if(hs[s[i]]==1) return i; //n x O(1) - here is faster

return -1;
}

Date: June 11, 2022
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