Longest Substring Without Repeating Characters | Live Coding with Explanation | Leetcode - 3

[@AlgorithmsMadeEasy]

We hope you all are enjoying our videos!!! Don't forget to leave a comment!!! Please like the video to support us!!!
Questions you might like:
✅✅✅[ Tree Data Structure ] : ruclips.net/p/PLJtzaiEpVo2zx-rCqLMmcFEpZw1UpGWls
✅✅✅[ Graphs Data Structure ] : ruclips.net/p/PLJtzaiEpVo2xg89cZzZCHqX03a1Vb6w7C
✅✅✅[ December Leetcoding Challenge ] : ruclips.net/p/PLJtzaiEpVo2xo8OdPZxrpybGR8FmzZpCA
✅✅✅[ November Leetcoding Challenge ] : ruclips.net/p/PLJtzaiEpVo2yMYz5RPH6pfB0wNnwWsK7e
✅✅✅[ August Leetcoding Challenge ] : ruclips.net/p/PLJtzaiEpVo2xu4h0gYQzvOMboclK_pZMe
✅✅✅July Leetcoding challenges: ruclips.net/p/PLJtzaiEpVo2wrUwkvexbC-vbUqVIy7qC-
✅✅✅June Leetcoding challenges: ruclips.net/p/PLJtzaiEpVo2xIfpptnCvUtKrUcod2zAKG
✅✅✅May Leetcoding challenges: ruclips.net/p/PLJtzaiEpVo2wRmUCq96zsUwOVD6p66K9e
✅✅✅Cracking the Coding Interview - Unique String: ruclips.net/p/PLJtzaiEpVo2xXf4LZb3y_BopOnLC1L4mE
Struggling in a question??
Leave in a comment and we will make a video!!!🙂🙂🙂

[@crunchworking3601]

it was quite helpful

[@AlgorithmsMadeEasy]

Thank you!!

[@thugken]

"gbcjkancb" in this case, the output shows "jkanc" not "jkancb" because 'b' is already in the map. don't you need to update map?

[@atularya7124]

really hard to understand please provide more explanation

[@EntEduRealm]

ruclips.net/video/vHZjMkrSc2M/видео.html

[@cricsinghvlogz]

nice

[@AlgorithmsMadeEasy]

Thanks Pawanpreet!! 🙂

[@Progresser436]

Ek no.🙂

[@AlgorithmsMadeEasy]

👍👍

[@pallavithukral3469]

Please code in c++ language instead of Java

[@lrchoudhary6875]

C++ Code :
class Solution {
public:
// Sliding Window
// TC : O(n)
// SC :O(min(m,n)); m - length of charset
int lengthOfLongestSubstring(string s) {
int n = s.size();
int maxLen = 0;
// Store last seen index of each character
unordered_map charIndex;
int start = 0;
for (int end = 0; end < n; end++) {
if (charIndex.find(s[end]) != charIndex.end()) {
start = charIndex[s[end]];
maxLen = max(maxLen, end - start);
}
charIndex[s[end]] = end;
maxLen = max(maxLen, end - start);
}
return maxLen;
}
};