let say number Bottle is 2 and one trial then how many pig we need ?

1 because we are going to feed first bottle to pig 1 if dies then there there is poison in bottle 1 otherwise there is poison in bottle 2.

We can generalize above idea for in binary number.How?

Lets name all bottle from 1 to N .

if in poison bottle I want to know whether in its binary representation first digit is 0 or 1 to do that I will feed pig 1 all bottle in which first digit is one (like 1,3,5,..) if Pig 1 dies then first digit (in binary representation ) is one otherwise 0.

similairy process can be done for other digits

If we have more number of trial than 1 then (lets say 2) we instead of binary number number we can use ternary number.

so this way we will get answer as ceil(log(t+1)(n)); log base t+1 (n)

We can use set or bool array to count character

int count(string & s) {
	set<char>st;
	for (char i : s) {
		st.insert(i);
	}
	return st.size();
}

int count(string & s) {
	int mask = 0;
	for (char i : s) {
		mask |= (1 << (i - 'a'));
	}
	return __builtin_popcount(mask);
}

class Solution {
public:
    int maxProduct(vector<string>& words) {
        vector<int> a(words.size() , 0);
        for(int i = 0;i < words.size();i++){
            for(int j = 0;j < words[i].size();j++){
                a[i] |= 1 << (words[i][j] - 'a');
            }
        }
        int res = 0;
        for(int i = 0;i < words.size();i++){
            for(int j = i + 1;j < words.size();j++){
                if((a[i] & a[j]) == 0){
                    if(words[i].size() * words[j].size() > res){
                        res = words[i].size() * words[j].size();
                    }
                }
            }
        }
        return res;
    }
};

2^count(number of set bits)

class Solution {
public:
    vector<int> findNumOfValidWords(vector<string>& words, vector<string>& puzzles) {
        vector<unordered_map<int,int>>mp(26);
        for(string & i: words){
            int p=0;
            for(char j:i){
                p |= (1<< (j-'a'));
            }
            int vis=0;
            for(char j:i){
                if(!(vis & (1 << (j-'a'))))mp[j-'a'][p]++;
                vis |= (1 << (j-'a'));
            }
        }
        vector<int>ans;
        for(string &i:puzzles){
            int p=0;
            for(char j:i){
                p |= (1<<(j-'a'));
            }
            int k=p;
            int res=0;
            int r=0;
            while(k){
                r++;
                if(mp[i[0]-'a'].count(k))res+=mp[i[0]-'a'][k];
                k=(k-1)&p;
            }
            assert(r==127);
            ans.push_back(res);
        }
        return ans;
    }
};

int g (int n) {
    return n ^ (n >> 1);
}

int rev_g (int g) {
  int n = 0;
  for (; g; g >>= 1)
    n ^= g;
  return n;
}