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.
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.😊
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.
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
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; } };
class Solution { public int characterReplacement(String s, int k) { int maxFreq=0, maxLen=0, l=0, r=0; int hash[]=new int[26]; while(rk){ hash[s.charAt(l)-'A']--; //Line Number - 9 l++; } maxLen=Math.max(maxLen,r-l+1); r++; } return maxLen; } } in the Video he has mentioned at the Line number 9 , hash[s.charAt(r)-'A']--; but above solution worked for me in Java, Anyway Thanks for the Logic Explanation
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
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
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
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.
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 .
class Solution { public: int characterReplacement(string str, int l) { int s=0,k=0,longest=0,maxfreq=0;//str consists of only uppercase letters int mpp[26]; while(k
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 😅
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.
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
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
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.
![ILLSLICK X YOUNGOHM - We're The same [Official Video]](http://i.ytimg.com/vi/yUJNkobYR0E/mqdefault.jpg)
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
Thank you brother. I understood it because of you !🙏
Striver - The G.O.A.T of DSA 💥
dude from 2n to n , kudos to you cuz my mind aint gone beyond that!
I like how you go through the naive solution as well and compare the time and space complexities to the optimized solution
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
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.
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 ?
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
Thankyou very much man.I have watched 6 videos and solved this solution optimally by my own.I will learn everything from you 🙂
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
Bro when you explain it becomes so easy to understand.. Thank you so much :)
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;
}
};
19:50 - why no need to decrease the maxFreq when frequency of any char in the map decreases
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
koi python code bhi daaldo
UNDERSTOOD....Thank You So Much for this wonderful video......🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
you are the legend for dsa ......thanks again striver
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;
}
};
#UNDERSTOOD
class Solution {
public int characterReplacement(String s, int k) {
int maxFreq=0, maxLen=0, l=0, r=0;
int hash[]=new int[26];
while(rk){
hash[s.charAt(l)-'A']--; //Line Number - 9
l++;
}
maxLen=Math.max(maxLen,r-l+1);
r++;
}
return maxLen;
}
}
in the Video he has mentioned at the Line number 9 , hash[s.charAt(r)-'A']--; but above solution worked for me in Java, Anyway Thanks for the Logic Explanation
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
19:20 the time complexity should be O( N + N*26) right?
yes
Thanks Striver. I was stuck on what will be the terminating condition for trimming down from left.
Guys here is an simple code :-----
class Solution {
public:
int characterReplacement(string s, int k) {
int l=0,r=0,maxlen=0,maxfreq=0;
int map[26];
fill(map,map+26,0);
int n=s.length();
while(rk){
map[s[l]-'A']--;
l++;
}
if(r-l+1-maxfreq
You are writing it in very very small font can u please write which could be visible properly !
Thanks
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
Bhaia will not the time complexity be N + 26(N) = 27N ??
i agree
keep on making such videos, thanks a lot
Great Content
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
bro rigorously waiting for your heap playlist..
pls pls pls plss upload asap..
Hey peeps , I didn't get "while loop removing to if" can you please elaborate ( it will help alot)
it was quite tough but understood bhaiya
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
can we use use mapping for frequency calculations?
Why I'm not able to understand the time complexity ? There is nested loop but still it is not O(n^2). Why?
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.
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
class Solution {
public:
int characterReplacement(string str, int l) {
int s=0,k=0,longest=0,maxfreq=0;//str consists of only uppercase letters
int mpp[26];
while(k
Awesome brother. Understood.
thanks for this
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 😅
Can u give the link plz?
@@rohang7059 which problem solution link?
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
can anyone explain how assigning mxfreq = 0 and finding maxfreq in further elements will be help?
Where can I find the python codes for this problem?
brilliant
My solution:
int characterReplacement(string s, int k) {
// s consists of only uppercase English letters.
int maxlen = 0, maxf = 0;
int l = 0, r = 0;
unordered_map mpp;
while(r < s.size()){
mpp[s[r]]++;
maxf = max(maxf, mpp[s[r]]);
int changes = (r-l+1) - maxf;
if(changes > k){
char ch = s[l];
mpp[s[l]]--;
if(mpp[s[l]]==0)
mpp.erase(s[l]);
l++;
}
maxlen = max(maxlen, r-l+1);
r++;
}
return maxlen;
}
hacker as always thanks bro
//Bruteforce Approach
class Solution {
public int characterReplacement(String str, int k) {
int maxLength = 0;
for (int start = 0; start < str.length(); start++) {
for (int end = start; end < str.length(); end++) {
String substring = str.substring(start, end + 1);
int numChangesNeeded = substring.length() - getMaxFrequency(substring);
if (numChangesNeeded > k) {
break;
}
maxLength = Math.max(maxLength, end - start + 1);
}
}
return maxLength;
}
private int getMaxFrequency(String str) {
int[] freqArray = new int[26];
int maxFreq = 0;
for (int i = 0; i < str.length(); i++) {
freqArray[str.charAt(i) - 'A']++;
}
for (int frequency : freqArray) {
maxFreq = Math.max(maxFreq, frequency);
}
return maxFreq;
}
}
//Improved Approach
class Solution {
public int characterReplacement(String str, int k) {
int maxLength = 0;
int[] freqArray = new int[26];
for (int start = 0; start < str.length(); start++) {
for (int end = start; end < str.length(); end++) {
for (int range = start; range k) {
break;
}
maxLength = Math.max(maxLength, (end - start + 1));
}
}
return maxLength;
}
private int getMaxFrequency(int[] freqArray) {
int maxFreq = 0;
for (int frequency : freqArray) {
maxFreq = Math.max(maxFreq, frequency);
}
Arrays.fill(freqArray, 0);
return maxFreq;
}
}
//Good Approach
class Solution {
public int characterReplacement(String str, int k) {
int maxLength = 0;
int[] freqArray = new int[26];
for (int start = 0; start < str.length(); start++) {
for (int end = start; end < str.length(); end++) {
freqArray[str.charAt(end) - 'A']++;
int numChangesNeeded = (end - start + 1) - getMaxFrequency(freqArray);
if (numChangesNeeded > k) {
break;
}
maxLength = Math.max(maxLength, (end - start + 1));
}
Arrays.fill(freqArray, 0);
}
return maxLength;
}
private int getMaxFrequency(int[] freqArray) {
int maxFreq = 0;
for (int frequency : freqArray) {
maxFreq = Math.max(maxFreq, frequency);
}
return maxFreq;
}
}
//Better Approach
class Solution {
public int characterReplacement(String str, int k) {
int maxLength = 0, maxFreq = 0, start = 0, end = 0;
int[] freqArray = new int[26];
while (end < str.length()) {
freqArray[str.charAt(end) - 'A']++;
maxFreq = Math.max(maxFreq, freqArray[str.charAt(end) - 'A']);
if ((end - start + 1) - maxFreq k) {
freqArray[str.charAt(start) - 'A']--;
maxFreq = 0;
for (int i = 0; i < 26; i++) {
maxFreq = Math.max(freqArray[i], maxFreq);
}
start++;
}
end++;
}
return maxLength;
}
}
//Further Optimised
class Solution {
public int characterReplacement(String str, int k) {
int maxLength = 0, maxFreq = 0, start = 0, end = 0;
int[] freqArray = new int[26];
while (end < str.length()) {
freqArray[str.charAt(end) - 'A']++;
maxFreq = Math.max(maxFreq, freqArray[str.charAt(end) - 'A']);
if ((end - start + 1) - maxFreq k) {
freqArray[str.charAt(start) - 'A']--;
start++;
}
end++;
}
return maxLength;
}
}
//Optimal Approach
class Solution {
public int characterReplacement(String str, int k) {
int maxLength = 0, maxFreq = 0, start = 0, end = 0;
int[] freqArray = new int[26];
while (end < str.length()) {
freqArray[str.charAt(end) - 'A']++;
maxFreq = Math.max(maxFreq, freqArray[str.charAt(end) - 'A']);
if ((end - start + 1) - maxFreq k) {
freqArray[str.charAt(start) - 'A']--;
start++;
}
end++;
}
return maxLength;
}
}
mera ho gya hai dimag kharab iss question se
understood
Wow good question and nice explanation
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
what data structure is used to store hashmap please anyone tell me?
map or unordered_map in c++
@@editorial9702 is it different from head memory and stack memory
ty sir
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
Great 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.
Understood
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;
}
};
too clean
ugh im not able to understand the maxFreq optimization logic
bhai writing thoda shi se kro , smjh ni aata kya likha h, piche jake dekhna pdta h
28 June 2024 4:49pm
understood++
where is the code in all 3 languages???
do it yourself man
@@lakshsinghania 😂😂
US
God
Striver - The G.O.A.T of DSA 💥
Understood
understood
understood
Understood
understood
understood
understood