Minimum Number of Steps to Make Two Strings Anagram | Simple | Intuitive | Google | Leetcode 1347

[@harshtiwari416]

Bhai bit manipulation ke concepts pe ek marathon video upload kro kyo bahut kam quality videos hai and aapke jaisa koi nhi samjha paa rha hai good work keep it up

[@DurgaShiva7574]

God of concepts 🎉

[@wearevacationuncoverers]

Thanks mik. I stopped the video at 10:16 and coded it up.

[@anuppatankar4294]

Great video 👌

[@DevOpskagyaan]

Simple and clear. Thanks

[@tutuimam3381]

Thanks 👍👍

[@oqant0424]

solved this question on my own ...came here to see the better approach.
done [13.1.24]✅✅

[@gauravbanerjee2898]

Thanks a lot bhaiya ❤❤, I solved this question on my own using the following approach before watching your video, thanks to your previous videos ❤
class Solution {
public:
int minSteps(string s, string t) {
int count[26]={0};
int ans=0;
for(auto ch:s){
count[ch-'a']++;
}
for(auto ch:t){
count[ch-'a']--;
if(count[ch-'a']

[@michaelmaahi7384]

Thank you sir for leetcode problem of the day😊

[@shubhambalodhi6485]

int minSteps(string s, string t) {
int ans=0;
if(s.size()!=t.size())
return 0;
mapmp;
//storing count of each character in s
for(int i=0;i

[@FanIQQuiz]

Done done done.

[@something_went_wrong167]

Today's question was easy

[@thefinalfit]

Too good 👌🏻

[@saurabhKumar-hj6yp]

❤❤

[@S3aw0lf]

Sir if u get time can you pls make a video on this question asked by Bentley Solutions in their coding round.
Question : You are given two arrays A and B each made up of Integers. They represent a grid with N columns and 2 rows where A is the upper row and B is the lower row. Your task is to find the upper left cell to the bottom right cell moving only right and down so that the maximum value over which you pass is as small as possible.
Test cases: A[3,4,6] B[6,5,4] output:5
A[1,2,1,1,1,4] B[1,1,1,3,1,1] output: 2

[@Atasimaji]

Please add the link of related playlist in the discription.

[@codestorywithMIK]

Sure
th-cam.com/play/PLpIkg8OmuX-KRHVXwqSixQC9UE6DsHnWa.html&si=PVo3kOZClG8pU8fU

[@souravjoshi2293]

Thanks mik

[@Whirlwind03]

MIK bhai thanks to your teachings solved it myself using 1 hashmap.
public int minSteps(String s, String t) {
Map map = new HashMap();
for(char ch: s.toCharArray()){
map.put(ch,map.getOrDefault(ch,0) + 1);
}
int res = 0;
for(char ch: t.toCharArray()){
if(map.containsKey(ch) && map.get(ch) > 0){
map.put(ch,map.get(ch) - 1);
}else{
res += 1;
}
}
return res;
}

[@mr_stranger3867]

Hello bhai aaj ke leetcode contest me jo last question tha beautiful indices 2 wala
Usko solution upload krne Wale ho kya aap?

[@codestorywithMIK]

I have already uploaded all.
Last Qn - th-cam.com/video/7Sm0o5lMmfY/w-d-xo.htmlsi=NRb010ZF51LccPeO
3rd Qn Uploaded - 3rd also uploaded - th-cam.com/video/D5yhQ-bqw9E/w-d-xo.htmlsi=BXSsFOq7sGqMEjaC
😇🙏

[@mr_stranger3867]

@@codestorywithMIK Thx bhai♥️

[@bhartendupant8859]

❤

[@indian_gaurav_8648]

8:42 but ye given hai q me string same length ki hai

[@codestorywithMIK]

Yes yes, that’s why such a scenario will never occur

[@crimemastergogo710]

class Solution {
public:
int minSteps(string s, string t) {
unordered_mapmp;
unordered_mapmp2;
for(int i=0;i

[@souravjoshi2293]

Sahi approach lag Raha hai. Maine bhi kuch isi tarah kya tha.

[@crimemastergogo710]

@@souravjoshi2293 can u share ur code pls?

[@dayashankarlakhotia4943]

int minSteps (String s,String t){
int[]cnt=new int[26]; int ans=0;
for(char c:s.toCharArray())
++cnt[c-'a'];
for(char c:t.toCharArray())
--cnt[c-'a'];
for(int i:cnt)
ans+=Math.abs(i);
return ans/2;
}
}