L8. Longest Repeating Character Replacement | 2 Pointers and Sliding Window Playlist
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- čas přidán 20. 08. 2024
- Notes/Codes/Problem links under step 10 of A2Z DSA Course: takeuforward.o...
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striver bro please upload the greedy playlist asap .. ur lectures are literally dominating all dsa premium courses.. far more better than anyone
it's been already uploaded
bantu bhai ko ram ram
dude from 2n to n , kudos to you cuz my mind aint gone beyond that!
Striver - The G.O.A.T of DSA 💥
lagta hai jindgi tutorial dekte dekte nikal jayigi
kasam s ...sala karne betho to ye kese krenge...bas yahi chlta h
sahi kaha bhaiya😂koi qn sliding window me slove nahi ho raha he
@@anandunique1472 wahi sabse jyda aata hain 🤣
mt dekho tutorial.. khud se code likho fir socho kya glt h. Key Tip: Dry Run your code with pen and paper. At last, kch smjh na aae to tutorial dekho approach smjhne k liye bs.
@@rishujeetrai5780 mein to aise kr raha tha pehle video solution dekho then khud try karo
Optimal:
class Solution {
public:
int characterReplacement(string s, int k) {
int n = s.size();
int left = 0, right = 0, maxFreq = 0, maxLen = 0;
int hash[26] = {0};
while(right maxFreq){
maxFreq = hash[s[right]-'A'];
};
if((right-left+1)-maxFreq > k){ //trimming the left portion
hash[s[left]-'A']--;
left++;
}
maxLen = max(maxLen, right-left+1);
right++;
}
return maxLen;
}
};
Though this code runs gets accepted in LC, but i think something is missing: if((right-left+1)-maxFreq > k){ //trimming the left portion
hash[s[left]-'A']--;
left++;
} what if my maxFreq was from i part and also contributing, here i need to recalculate the max freq. I may be wrong but this is what i think.
@@beinginnit Yeah,I see that,but it neetcode channel also he explained in this way to optimize more to O(1) space like this,please have a look at it and explain
Last 5 minutes are very precious, it is a must watch.
That trick could be used in other questions as well for optimisation.
That he has been mentioning in his previous videos as well.
Thankyou very much man.I have watched 6 videos and solved this solution optimally by my own.I will learn everything from you 🙂
First solution witout optimization:
class Solution {
public:
int characterReplacement(string s, int k) {
//FML
int l = 0, r = 0, ans = 0, maxf = 0;
unordered_map umap;
while(r < s.length()){
char c = s[r];
if(umap.find(c) == umap.end()) umap.insert({c, 1});
else umap.find(c)->second++;
maxf = max(maxf, umap.find(c)->second);
while(r - l + 1 - maxf > k){
umap.find(s[l])->second--;
for(auto x : umap){
maxf = max(maxf, x.second);
}
l++;
}
ans = max(ans, r - l + 1);
r++;
}
return ans;
}
};
Thank you brother. I understood it because of you !🙏
Bro when you explain it becomes so easy to understand.. Thank you so much :)
After understanding the logic and without seeing the solution now i am able to code it myself.
Thanks striver for making us understand this algorithm.😊
Ok let's understand how much you really know the depth. give me pattern of all substring in ABC string with k=1.
@@kaushit {"A", "B", "C", "AB", "BC"} hi hoga na ?
Java Code for java language forks
class Solution {
public int characterReplacement(String s, int k) {
int n=s.length();
int maxlen=0,maxf=0;
int i=0,j=0;
Mapmap=new HashMap();
while(jk){
char ch2=s.charAt(i);
map.put(ch2,map.get(ch2)-1); //decrement the frequency
i++;
}
// if((j-i+1)+maxf
Trick is we dont want to minimize our max freq variable to get better result, ie. if we get result for max freq 3 we will continue to look for max freq greater than 3 if it exists and ignore freq less than equal 3. This will eliminate need of freq map scan and also remove while loop if we want to look for better max length.
thanks bhaiya,now i got it.
Thanks a lot bro
you are the legend for dsa ......thanks again striver
Bhaiya I am very much Greedy for your Greedy Playlist So when you are uploading that ?? 🙌🙌🤑🤑
yes we want greedy playlist bhaiyyaa
same bro plz give greedy playlist after this much needed
UNDERSTOOD....Thank You So Much for this wonderful video......🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
it was quite tough but understood bhaiya
understood
keep on making such videos, thanks a lot
Great Content
C++ Leetcode solution:
class Solution {
public:
int characterReplacement(string s, int k) {
int n = s.size();
int l = 0;
int r = 0;
int ans = 0;
int hash[26]={0};
int maxFre = 0;
while(rmaxFre){
maxFre = hash[s[r]-'A'];
}
if((r-l+1)-maxFre>k) {
hash[s[l]-'A']--;
l++;
}
ans = max(ans, r-l+1);
r++;
}
return ans;
}
};
Did all the questions of sliding windows by myself with the optimal solutions. Just leaving the fruit in basket one.
Watch the video to know all the possible ways of solving that particular problem ;>
Easy Peasy Lemon Squeezy 😅
Please upload the code.. And looking forward for Heaps playlist please. Nowhere in youtube heaps problems are covered.. Please do it 🙏🏼🙏🏼🙏🏼🙏🏼 . It's being asked in interviews many times
You are writing it in very very small font can u please write which could be visible properly !
Thanks
public int characterReplacement(String s,int k){
int[]freq=new int[26];
char[]ans=s.toCharArray();
int n=ans.length,l=0,max=0;
for(int r=0;r
Great bro !
what data structure is used to store hashmap please anyone tell me?
Hey peeps , I didn't get "while loop removing to if" can you please elaborate ( it will help alot)
sir what a video this is just amazing 🙏🙌🙌🔻
O((n + n)) logic is failing on AABABBB, we cannot because of the while loop directly jump to a point where some ans is missed out, if we write if instead of while (like in logic 3) then it will work.
Awesome brother. Understood.
Striver bhaiya, I didnt understand why didnt we update the maxf when we are incrementing l. because the max trip will also change right when we change shorten the length of the subarray from front?
you're talking about his most optimal approach, i presume. In that case he didn't modify the maxfreq variable under the inner while loop because we don't need a maxfreq variable less than the current maxfreq variable regardless of the trimmed down substring we have right now. that's because, say, the current length of the substring is lesser but we still have the maxfreq value pertaining to the previous valid substring; in this case we simply would not require to have to check the length and change the consequent letters, we'd just ignore it. And to optimize it further, instead of looping through the letters till we get the valid window of (current_length - max_freq) > k, we could replace it with 'if' to check the condition only once because we can again simply ignore the additional trouble of having to loop through for each invalid checks which striver explains why in the other videos of his
@@sparksfly4421 thanks bhai
Even if you don't update the maxfreq in while loop of left then also it is working smoothly.
Although thanks for wonderfull series striver .
code - class Solution {
public int characterReplacement(String s, int k) {
int[] hash=new int[26];
int maxlen=Integer.MIN_VALUE;
int maxfreq=Integer.MIN_VALUE;
int l=0;
int n=s.length();
for(int r=0;rk){
char cl=s.charAt(l);
hash[cl-'A']--;
l++;
}
maxlen=Math.max(maxlen,r-l+1);
}
return maxlen;
}
}
Why would this work? Also once let's say maxfreq is 3 because of 'A', and then if we remove 'A' from the window how do we update maxfreq from 3 to 2 assuming no other char having freq 3?
@@fractionofinfinity842actually we don't need to update the maxf value to the lesser value . Becz if u update it to lesser value like 2 , again it will come under > k while condition .
use if (r-l+1-maxf>k) rather than while
Bhaia will not the time complexity be N + 26(N) = 27N ??
i agree
can we use use mapping for frequency calculations?
hacker as always thanks bro
int window3(string& s,int k){ //O(n) most optimal pure O(n) logic
int i=0;
int j=0;
int len=0;
int max_freq=0;
vectorfreq(26,0);
while(j < n){
freq[s[j] - 'A']++;
max_freq = max(max_freq,freq[s[j] - 'A']);
if((j-i+1) - max_freq > k){
freq[s[i]-'A']--;
i++;
// max_freq = 0;//use less
//max_freq = *max_element(freq.begin(),freq.end());//it will make no sense to calculate max freq here in if part we will get max in next iteration with line: max_freq = max(max_freq,freq[s[j] - 'A']);
}else{
len = max(len,j-i+1);
}
j++;
}
return len;
}
this is most optimal logic, max_freq = 0, logic not clear to me and it is even not working with while version of this function, but this one is intuitive and clear.
Hi anna....!!!
I am super excited with your sliding window protocol series. I understood every video till now in this series except this video with the difference between only if condition and inner while loop and its inner for loop. Could you please elaborate it in depth with some specific examples. So, that I can understand the thought process behind it. Please please anna...😊😊😊
Trick is we dont want to minimize our max freq variable to get better result, ie. if we get result for max freq 3 we will continue to look for max freq greater than 3 if it exists and ignore freq less than equal 3. This will eliminate need of freq map scan and also remove while loop if we want to look for better max length.
@@nirmalgurjar8181 aaah i get it now ,thanks i got the 26 for loop part , but not the inner while loop part
class Solution {
public:
int characterReplacement(string s, int k) {
int n = s.size();
int l=0,r = 0,maxf = 0,maxlen = 0;
mapmpp;
while(rk)
{
mpp[s[l]]--;
l = l+1;
}
maxlen = max(maxlen,r-l+1);
r++;
}
return maxlen;
}
};
19:20 the time complexity should be O( N + N*26) right?
yes
ty sir
Where i can find sudo code?
Is the naive or brute force code work properly for all test cases ? IF YES CAN ANYONE FIND ERROR IN MY CODE because It does not satisfy test case 2 of leetcode
int[] arr = new int[26];
int maxc = 0;
int maxl = 0;
for(int i = 0; i< s.length();i++){
for(int j = i ; j < s.length() ; j++){
arr[s.charAt(j) - 'A']++;
maxc = Math.max(maxc,arr[s.charAt(j)-'A']);
if( (j - i + 1 ) - maxc
int[] arr = new int[26]; should be after first loop bcz in next iteration is should be updated to 0
I didn't understand , why we are doing maxf=0 in the else part. Can someone please explain ?
I guess he did it initially as we was going to check again for the max freq using that 0-26 for loop its not in else everything is in if I guess
ugh im not able to understand the maxFreq optimization logic
why we are doing " -'A' " in hash[S[r]-'A' ????
S[r]-'A' will give a integer value which we will use for indexing in our hash map
For example if s[r]=B
So, 'B'-'A'(i.e 66-65(ASCII VALUES OF B AND A) =1)
So hash[1] will increase
mera ho gya hai dimag kharab iss question se
understood++
28 June 2024 4:49pm
US
where is the code in all 3 languages???
do it yourself man
@@lakshsinghania 😂😂
bhai writing thoda shi se kro , smjh ni aata kya likha h, piche jake dekhna pdta h
God
understood
understood
understood
understood