Leetcode 1657. Determine if Two Strings Are Close

[rabiaaghafoor]

hi I am not understanding the explanation, it seems like reading the code. could you please explain what the intuition is? thank you!

[Phoenix-xigy]

Can you please discuss Leetcode problem 756. Pyramid Transition Matrix .. I am having a hard time understanding it

[ayushgarg]

Sir, please make a video on strings all important questions in single video

[saiei]

class Solution {
public:
bool closeStrings(string word1, string word2) {
//we dont have solve with bruteforce approach
//u can solve this by by seeing if the frequency of the characters in word1 r same as the frequency of charaacters in word2 (its ok if the freq of the specific char is not same) (property2)
//we have to make sure if the chars r same and no new char is present in 1 string and missing in another (both prop1 and pro2 cant distroy a char hence it will alsoways be there in the string irrespective of the operations u do )

//1st we will find the freq of chars
//we also lable 1 if that char is present or not
vector<int>s1(26, 0), s2(26, 0), s11(26, 0), s22(26, 0);
for(char ch:word1)
{
s1[ch-'a']++;
s11[ch-'a']=1;
}
for(char ch:word2)
{
s2[ch-'a']++;
s22[ch-'a']=1;
}
//we have to sort the frequences so that we can compare easily
sort(s1.begin(), s1.end());
sort(s2.begin(), s2.end());

return s1==s2 && s11==s22;
}
};

[parthapratimghose]

There is a better solution in o of n
Why not put all frequencies into two unordered multisets then you can call uset1 quals uset2 it will do it in o1 time and hence overall tc on time

[RitikKumar-bkpj]

Sir please explain a intuition to solve a problem instead of telling a solution directly