Valid Anagram | 2 Approaches | Leetcode-242 | UBER

[@DevOpskagyaan]

Hats off to your consistency. I got to learn a lot from you . I wish i had got this channel in early 2023. But it’s ok.
2024 I AM READY

[@adarshjha5126]

Consistency++❤❤

[@drishtid4943]

I am definitely a big fan of your teaching.
building intuitions non stop bcs of your help. Thanks MIK

[@aadarshpandit2929]

Thank you sir aapke solution video dekh dekh ke main daily question practice karta hu. I am a total beginner but 3 days tak ki consistency hain daily question mein

[@codestorywithMIK]

❤️❤️🙏🙏

[@FanIQQuiz]

Thank you , you have improved by DSA and logical skills during this year 2023.

[@DeepakGupta_er]

2:00 how someone can sort the string in O(1) space complexity i.e. without converting String into char array?

[@souravjoshi2293]

I got the Annual badge. Thanks to you MIK sir

[@oqant0424]

I got the Annual badge. Thanks Bhaiya 💖

[@dayashankarlakhotia4943]

class Solution {
public boolean isAnagram (String s,String t){
if(s.length()!=t.length ()) return false;
int[]cnt=new int [128];
for(char c:s.toCharArray ())
++cnt[c];
for(char c:t.toCharArray())
if(--cnt[c]

[@tutuimam3381]

Thanks a lot

[@Ankit_Maurya_12]

first approach come to my mind is XOR, tried some cases passed some not can anyone explain why??

[@codeandtalk6]

received 3 badges yesterday ❤❤❤

[@codestorywithMIK]

Awesome 🔥
Congratulations

[@saurabhKumar-hj6yp]

❤❤

[@Strawcontamination]

pls upload geeksforgeeks potd solutions also

[@jagadeeshp1163]

i think we dont need 2 for loops we can do it in one pass
def isAnagram(self, s: str, t: str) -> bool:
if len(s)!=len(t):
return False
freq=[0 for i in range(26)]
for i in range(len(s)):
freq[ord(s[i])-ord('a')]+=1
freq[ord(t[i])-ord('a')]-=1
for i in range(26):
if freq[i]>0:
return False
return True

[@wearevacationuncoverers]

Yes. I also though the same.

[@codestorywithMIK]

Loved this. Thanks for sharing ❤

[@yadav.nitin03]

Yo🎉

[@utkarshsahay9908]

❤

[@WiFiWaLaaa]

// java code ;)
class Solution {
public boolean isAnagram(String s, String t) {

char[] s1 = s.toCharArray();
char[] t1 = t.toCharArray();
Arrays.sort(s1);
Arrays.sort(t1);
return Arrays.equals(s1, t1);
}
}

[@anushkathakur6531]

I got this question in suggestions.... 438. Find All Anagrams in a String....do you mind explaining this?

[@codestorywithMIK]

Let me check soon

[@Ramneet04]

Why in December it is giving so easy question its like half a month all easy questions except 1 was medium,but was too easy too.

[@oqant0424]

Solved this on my own and optimised it too🥹

[@iamvtyagii9540]

❤❤❤❤