Number of Substrings Containing All Three Characters

//brute force approach: O(N^2)
class Solution {
    public int numberOfSubstrings(String s) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            int hash[] = new int[3];// for a,b and c only
            for (int j = i; j < s.length(); j++) {
                char c = s.charAt(j);
                hash[c-'a']++;
                if(hash[0]>=1 && hash[1] >=1 &&  hash[2]>=1){
                    count++;
                }
            }
        }
        return count;
    }
}
//a little bit of optimization on the brute force approach: but in worst case the time complexity will still be O(n^2)
class Solution {
    public int numberOfSubstrings(String s) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            int hash[] = new int[3];// for a,b and c only
            for (int j = i; j < s.length(); j++) {
                char c = s.charAt(j);
                hash[c-'a']++;
                if(hash[0]>=1 && hash[1] >=1 &&  hash[2]>=1){
                    /*If we have already found a substring that satisfies the condition,
                    then all the other substrings that will come after this (will include this substring as well) will also
                    be valid strings, hence no need to create the rest of the strings we can break out*/
                    count+= s.length()-j;
                    break;
                }
            }
        }
        return count;
    }
}

class Solution {
    public int numberOfSubstrings(String s) {
        //WITH EVERY CHARACTER THERE IS A SUBSTRING THAT ENDS
        int hash[] = new int[3];// it will keep track of last seen indexes of a,b,c 
        Arrays.fill(hash,-1); //initializing the index value of all the a,b and c to -1
        int count =0;
        for(int i =0;i<s.length();i++){
            char c = s.charAt(i);
            hash[c-'a'] = i;
            if(hash[0]!=-1 && hash[1]!=-1 && hash[2]!=-1){
            //we choose min because whichever seems last, all the character before the last seen character will form a substring
                count+= 1 + Integer.min(hash[0],Integer.min(hash[1],hash[2]));// get the last seen index of (a or b or c), since it is 0 based indexing
                //hence if the last index was at index 2 then there will be 3 substrings satisfying the condition, substring starting from 0, 1, and finally 2 itself
            }
        }
        return count;
    }
}