214.Shortest Palindrome || Leetcode || Solved&Explained || C++ 

you tried well Understanding a problem and effectively explaining it to others are two entirely different skills.

thanks ma'am for making this concept so easy

Best explanation so far

My search for it's solution finally came to an end . Thanks for such a great explanation .

she just taught KMP, good explanation

Thank You !!!

Thank you ma'am for such a great explanation, I had been struggling with KMP too, but now concepts are crystal clear. Thanks a lot!!!

thanks for make it. your explanation is very well 😊😊 i hope you will continue ..& grow

thanks for the algorithm,since code part is way too easy and self implementable...

Finally After 2 days got h pretty good solution!! Understood❤

Great explanation! Only tip is we can reuse the "t" variable instead of creating a new shorter string. return t.substr(0, s.size()-i)+s;

Is it possible to solve Using longest common substring between s and rev(s) and after that return the total length -s

Very Nice Explanation Mam, Thank you !

very very good explanation, thank you so much you are the great❤

Memory Limit Exceeded!!😭😭

10:06 Here we should not increment "j". There is a mistake in explanation. Check the code we are not incrementing "j" there in else statement. we are only changing the pointer "i".

class Solution: def shortestPalindrome(self, s: str) -> str: if s == s[::-1]: return s n = len(s) # Loop to find the largest palindrome prefix for i in range(n, 0, -1): if s[:i] == s[:i][::-1]: break # Add the reverse of the suffix to the start of the string suffix = s[i:] return suffix[::-1] + s

Understood💯❤

todays daily problem

thanks sister explained very well🙌🙌

loved the approach and awesome explanation

Thanks a lot ma'am, cleared my kmp concepts well.

But it doesn't pass all test cases.. for string "aabba" the ans is "abbaabba" how is it possible And also, "abb" -> ans: "bbabb"

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you are soo smart

keep going🤘

Nice approach Helpful✋

Clear explaination❤