Longest Substring Without Repeating Characters | Leetcode 3 | Sliding window

Time Complexity : O(n)
Space Complexity : O(1)


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Longest Substring Without Repeating Characters solution
Longest Substring Without Repeating Characters Leetcode
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Longest Substring Without Repeating Characters Java
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[jasper]

Thanks. Here is the Java code if someone needs it -

public int maxSubString(String s) {
int[] count = new int[256];
int l = 0;
int r = 0;
int ans = 0;

while (r < s.length()) {
char currChar = s.charAt(r);
count[currChar]++;

while (count[currChar] > 1) {
count[s.charAt(l)]--;
l++;
}

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

[azhargayawi]

after wasting approx 5 hours on you tube i found the genuine logic here, thank you mam. greaat....

[cs_soldier]

This can help someone :)



class Solution {
public:
int s) {
//creating a set to store the list
unordered_set<char> ans;
int value=0;
int i=0, j=0;
while(i<s.length() && j<s.length()){
if(ans.find(s[j])==ans.end())
//if the character is not present in the list then we insert the char into the list and update the right pointer.
{
ans.insert(s[j++]);
//we update the length of the current longest substring
value=max(value, j-i);
}else{
//if the character is found for the second time then we stop the process and increase the left pointer by 1 by erasing the left most character.
ans.erase(s[i++]);
}
}
//returning the length of the longest substring
return value;
}
};

[venkatvasamsetti]

I have already done this problem using hashmap.but without sliding window.thank you for this optimization of code

[RajeshS-nj]

well explained solution, in particular I like your making use of blackboard to go over approach to solution first, helps us a lot. 😀

[harshvardhan]

for java coders, there will be slight changes:-


public int s) {
int[] arr = new int[128];
int n = s.length();
int len = 0;
for(int i = 0, j = 0; j < n; j++){
int c = s.charAt(j);
arr[c]++;
if(arr[c] > 1){
while(arr[c] > 1){
arr[s.charAt(i)]--;
i++;
}
}
len = Math.max(len, j - i + 1);
}
return len;
}

[FootB_shorts]

After wasting a lot of time i found A helpful video .
thanks❤❤

[BADASSBLADE]

Solve Today's Leetcode Challenge(Erasure Problem, ) by using this approach, thankyou keep up the good work

[luckilygamer]

int s)
{
unordered_map<char, int> umap;
int n = s.length();
int ans=0;
int start=0;

for (int i = 0; i < n; i++) {

// update the start index to the next character after the last occurrence
start = max(start, umap[s[i]] + 1);
}
umap[s[i]] = i;
ans=max(ans, i-start+1);
}
return ans;
}

[arqamazmi]

Nice Explanation
Try this also.
Longest Repeating Character Replacement

[coding]

Good explanation but I guess the space complexity mentioned in the description should be O(k), where k are the unique characters and not O(1).

[Darshan-prk]

hi aushi, using two while loop may increase time complexity else we can use hash map for storing index of characters

[abhishekprasad]

Di should i buy coding ninja dsa course or study from free resource . pls response

[jagmohangautam]

it is giving runtime error for string containing space

[sumitpal]

Java code is not working in testcase "pwwkew". can any one help me.

public int s) {

int count[] = new int[256];

int maxLen = 0, i=0, j=0;

while(j < s.length()){
int index = s.charAt(j);
count[index]++;

while(count[index] > 1){
count[index]--;
i++;
}

maxLen = Math.max(maxLen, ( j-i)+1);

j++;
}

return maxLen;
}

[priyankakataria]

Time complexity worst case will be O(n^2) right? Coz two while loops

[vinaykulkarni]

int s) {
unordered_map<char, int>mp;
int i=0, j=0, ans=0;
int n=s.length();
while(j<n){
mp[s[j]]++;
while(mp[s[j]]>1){
mp[s[i]]--;
i++;
}
ans=max(ans, j-i+1);
j++;
}
return ans;
}
};

[abhisheknavhal]

The solution for the test case 'pwwkew' is not giving the correct answer.

[vamsiv]

This code gives wrong answer for aabccbb

[NikhilKumar-brnj]

Didi if possible pls give java code also