Find the Difference - Leetcode 389 - Python

0:00 - Read the problem
0:30 - Hashmap Explanation
2:23 - Hashmap Code
3:56 - ASCII Explanation
5:43 - ASCII Code
6:48 - Bit Manipulation Explanation
9:58 - Bit Manipulation Code

leetcode 389

#neetcode #leetcode #python

[gaychin]

Wow, exactly what I needed! Also, for those who don't know, XOR operation is both associative and commutative, therefore, the sequence of characters will not affect the output.

[MP-nyep]

Thank you for the daily problems ! Love the multiple approaches.

[michomapeter]

Bro youre the best, We really appreciate your efforts, Thankyou

[ymncore]

thank you so much for solving the daily challenges !!

[Banaaani]

The last solution is just mind blowing!

[houssainebendhieb]

Thank you for your efforts it's very helpful

[phpostrich]

return chr(sum([ord(ch) for ch in t]) - sum([ord(ch) for ch in s])) # one liner

[anantom]

Did it myself today!!

using the hashing.

char findTheDifference(string s, string t) {

vector<int> mp(27, 0);
char ans = '0';

// we first initialise the vector array with the freq of the characters of string s;
for(auto i=0;i<s.size();i++){
mp[s[i]-'a']++;
}

for(auto i = 0;i<t.size();i++){
mp[t[i]-'a']--;
// if after deleting that character we get a -1 in
//the vector it means that was the unique element.

if(mp[t[i]-'a']==-1){
ans = t[i];
break;
}
}
return ans;
}
S.C. 0(26) ~= O(1). i.e. constant space
T.C. O(N) where N is the size of the string t;

[SASA_maxillo]

my solution:
if s == "":
return t

for i in s:
t = t.replace(i, "", 1)

return t

i think it is easier and more readable 🙃🙃

[rushabhlegion]

I wrote like this

class Solution {
public:
char findTheDifference(string s, string t) {
vector<int>vecS(26, 0);
vector<int>vecT(26, 0);
for(int i=0;i<s.size();i++)
vecS[s[i]-97]++;
for(int i=0;i<t.size();i++)
vecT[t[i]-97]++;
for(int i=0;i<s.size();i++)

return t[i];
return t[t.size()-1];
}
};