L3. Longest Substring Without Repeating Characters | 2 Pointers and Sliding Window Playlist

[kuldeepdixit]

whenever I am looking for an explanation to a problem and I search it on youtube and if I don't find a solution by striver I get disappointed...Love from Flipkart!

[hashcodez]

Hello Bhaiya!!
From past few weeks i was not able to watch any of your videos.
Was going through depression .

But the moment i joined the journey again with you is a blessing.
You have a different aura ....just looking at you i just simply forget everything and a new spark is ignited within me.

Thank alot Bhaiya for your efforts and helping me!!

[shwetanshusinha]

The new look of the DSA sheet and the whole website is just way too awesome sir.

[pavitrasingh]

I just cant express my gratitude in words towards this man . the way he simplifies every single thing makes everything appear so easy . I am so glad that I found this channel.

[shivamjaiswal]

hi Striver,
Your Graph, DP, binary search playlist are amazing. Please create a playlist for String as well.

[simple_user]

I solved it on my own! Your teaching is amazing bhaiya!!!

[tanvirahammed]

Hi Striver, I'm not sure if this comment will reach you or not, but I just want to say one thing: I've been watching your videos for more than a year now and have covered a huge part of the graph and dynamic programming playlists. Whenever I watch your videos, I really fall in love with your teaching style, and of course, your smile, and sometimes your small jokes. I enjoy your videos, and it feels like you're not just my tutor; it feels like you're someone very close to me, teaching me with fun and pleasure. However, in your recent playlists, I totally miss that. It feels like you're not as friendly anymore; you're just a teacher like any other tutor. Whenever I watch videos from this playlist, sometimes I wonder what happened to you. Why are you so quiet now? Why don't you joke around anymore? After all, I love your work; it's just my opinion.

[rsiszkk]

1 fix in the brute force is that you need to fill the arrays with 0 every time youre moving i. so heres the correctedd brute.public int s) {
int[] arr = new int[256];
int max = 0;
int n = s.length();

for (int i = 0; i < n; i++) {
// Reset the array for each new starting point
Arrays.fill(arr, 0);

for (int j = i; j < n; j++) {
// If character at j has been seen, break the loop
if (arr[s.charAt(j)] == 1) {
break;
}

// Otherwise, add the character to substring
int len = j - i + 1;
max = Math.max(len, max);
// Remember it
arr[s.charAt(j)] = 1;
}
}
return max;
}

[divyanshushukla]

Loved your explaination. Bhaiya instead of using a HashMap we can also use a Set. It will save some more memory.
class Solution {
public int s) {
int i = 0, j = 0, max = 0;
Set<Character> set = new HashSet<>();
while (j < s.length())
{
if (!set.contains(s.charAt(j)))
{
set.add(s.charAt(j));
j++;
max = Math.max(max, set.size());
}
else
{
set.remove(s.charAt(i));
i++;
}
}
return max;
}
}

[AyushEditz-hspf]

i think this is overcomplicating things. The easy code could be:
int n = s.length();
int left = 0;
int right = 0;
int maxLen = 0;
map<char, int> mpp;
int len=0;
while (right < n) {
if (mpp.find( s[right] ) != mpp.end() ) {
left++;
right=left;
mpp.clear();

}
else{
len++;
mpp[s[right]] = right;
maxLen=max(right-left+1, maxLen);
right++;
}

}
return maxLen;

[_sujatachandra]

You are unstoppable... 🙏🙏 You are the best 🙏🙏

[technologicalvivek]

Bhaiya mene isko freq array banakar easy way me kardiya.
class Solution {
public:
int s) {
vector<int> freq(256, 0);

int i = 0, j = 0;
int maxi = 0;
int n = s.length();

while (j < n) {
if (freq[s[j]] == 0) {
freq[s[j]]++;
maxi = max(maxi, j - i + 1);
j++;
} else {
freq[s[i]]--;
i++;
}
}
return maxi;
}
};

[heershah]

with this solution we can make time complexity as O(n) and space complexity O(n)
Set<Character> set=new HashSet<>();

int left=0;
int right=0;
int mxLen=0;
String result = "";
while(right<str.length()) {
{
set.add(str.charAt(right));
if(right-left+1 >mxLen) {
mxLen= right- left +1;
result=str.substring(left, right+1);
}
right++;
}else {

left++;

}
}
return result;
}

[adilkevin]

I think you haven't attached this youtube link in your take youforward site.

[monikayadav-wbpu]

East and west striver bhaiya is best ❤❤

[PrinceKumar-efxf]

00:06 Finding longest substring without repeating characters
02:46 Using 2 Pointers for Substring Generation
04:59 Using hashing to find the longest substring without repeating characters
07:36 Optimizing algorithm using two pointers and sliding window approach
10:02 Understanding two pointer and sliding window approach
12:17 Determine longest substring without repeating characters using hashmap and sliding window
14:45 Updating characters in a sliding window to find longest substring without repeats.
17:06 Sliding window technique for finding longest substring without repeating characters.
19:10 Algorithm explanation and time complexity analysis
21:42 Explanation of sliding window algorithm with two pointer

[ManishKumar-dkhl]

Bhaiya wala code { BY USING MAP NAAM KI ARRAY } :--
class Solution {

public int s) {
int l=0;
int r=0;
int[] map=new int[256];
Arrays.fill(map, -1);
int maxlen=0;
while(r<s.length()){
if(map[s.charAt(r)]==-1){
map[s.charAt(r)]=r; //Now index of map will represent character aur value represent karegi index of string ko.
}
else{
l=Math.max(l, map[s.charAt(r)]+1);
map[s.charAt(r)]=r;
}
maxlen=Math.max(maxlen, r-l+1);
r++;
}
return maxlen;

}
}

HASHMAP WALA CODE :---
int l=0;
int r=0;
HashMap <Character, Integer> mpp=new HashMap<>();
int[] mpp=new int[256];
Arrays.fill(mpp, -1);
int maxlen=0;
while(r<s.length()){
// agar map m value exist nhi krti
mpp.put(s.charAt(r), r);

}
else{
// l=mpp.get(s.charAt(r))+1;
l = Math.max(l, mpp.get(s.charAt(r)) + 1); // iska reason video m hi battaya h
mpp.put(s.charAt(r), r);

}
maxlen=Math.max(maxlen, r-l+1);
r++;

}

ARRAY WALA CODE JYADA FAST HOGA HASHMAP WALE SE🙃

[Divyanshu-qn]

vector< int > mpp(256, -1);
int l=0;
int r=0;
int maxlength=0;
while(r<s.size()) {
if(mpp[s[r]]!=-1) {
if(mpp[s[r]]>=l) l = mpp[s[r]] +1;
}

mpp[s[r]] = r;
maxlength = max(maxlength, r-l+1);
r++;
}
return maxlength;

[theee_]

Very, very smart case where left has to be the rightmost. eg checking in abc, if left is already at 4 and prior presence of b is at 2, we need not update left to 3, instead it should be max of prior instance+1, already present left

[aniketkumar]

int s)
{
if(s.size()==0)return 0;
int i=0;
int j=0;
int maxi=1;
int n=s.size();
unordered_map<char, int>mp;
while(j<n)
{

{
i=mp[s[j]]+1;
}
maxi=max(maxi, j-i+1);
mp[s[j]]=j;
j++;
}
return maxi;

}
finally accepted.
Thankyou Striver😊for your effort.
i like and appreciate you from bottom of my heart.❤