131. Palindrome Partitioning | Recursion | Backtracking

131. Palindrome Partitioning | Recursion | Backtracking | 2 Approaches

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Күн бұрын

Пікірлер: 13

@ARYANMITTAL 6 ай бұрын

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@anonymousftw6386 6 ай бұрын

Instead of writing palindrome You can use stl if(equal(s.begin(),s.end(),s.rbegin()))

@mayankshakya9200 6 ай бұрын

Bhaiya aapne apne time pe dsa kaha se seekha tha which course???

@hydrocy.9165 5 ай бұрын

wallahi im finished

@anonanon1305 6 ай бұрын

Leercode 1718

@vladpvl 6 ай бұрын

Hello brother, I have an issue. I know a lot of theory, but can't use it in practise. When I read excercise I understand how to do it, but I cant write code. Maybe you can help me with advice

@arijeetchakraborty117 6 ай бұрын

okay, now if you know how to do it, means you understand the question properly and you can at least figure out one approach to solve it, be it a very efficient approach or not that optimal way. I can suggest you from my personal experience that if you are facing these kind of problem, you should first break down the code in small modules. Try the very easy parts first and then move on to the harder parts. And feel free to take help from anywhere you can get until you can just understand what is happening. This technique helped me a lot.

@vladpvl 6 ай бұрын

@@arijeetchakraborty117 thank you!!! I appreciate your advice

@pratikpandey6680 6 ай бұрын

Why are we doing pop_back() ? Can you please tell😊

@theslimeoxidizer3308 6 ай бұрын

so that everytime we get empty vector for storing another palindrome after returning from the calls.

@WiFiWaLaaa 6 ай бұрын

it's a backTracking

@riddle-me-ruben 6 ай бұрын

as we go into the recursive calls and add things to the current path, when we return back to explore other paths, if we did not delete the stuff then our other paths we are trying to explore will contain those extra strings. Thus, we delete it so that the other paths have a fresh start. (And also if we do not delete, then at the end we will have one giant list).