Queue - Print first non-repeating character in given Stream | Using Queue & Map | Java Code

Source Code:thecodingsimpl...
Solution:
- We implement it using Map (to maintain frequency of character) & queue (to maintain first non-repeating character)
- while iterating each character of stream, we update the frequency of that character in map, If frequency is one, then we add in queue as well.
- Now, to get first non-repeating character, we check the from front of queue, if frequency of that character is 1 then we print the value else we remove that element
- If queue is empty, we remove the element.
Time Complexity: O(n)
Space Complexity: O(n)
Do Watch video for more info
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[@mandeep2192]

explained beautifully

[@CodingSimplified]

Thanks. Keep watching.

[@DeepakSaini7900]

Gazab sir
🔥🔥🔥🔥

[@CodingSimplified]

Thanks for your nice feedback. Keep Watching.

[@akshay8675]

I think space complexity if o(1) as the maximum size of queue and hashmap will be 26

[@amorphous4768]

yes bro i think same

[@teetanrobotics5363]

Amazing tutorial

[@CodingSimplified]

Thank you! Cheers!

[@biikaaa.6540]

does it not throw tle? miine did

[@niru1931]

nice

[@CodingSimplified]

Thanks for you feedback. Keep Watching.

[@bitrish34]

🔥

[@CodingSimplified]

Thanks.

[@niru1931]

time complexity??

[@CodingSimplified]

Hi, as we mentioned in video. Time Complexity is O(n)

[@revanthreddy5759]

@@CodingSimplified How is this O(n). We are also looping through the queue right? I think it will be more than O(n). In worst case O(n2) actually.

[@divyngairola6886]

@@revanthreddy5759 at max in queue we only have all the 26 chars so it will not be n^2 but n
and space complexity will be o(1) because queue size at max is 26 and same for map max size is 26