Rotate String | Something to learn | Leetcode 796 | codestorywithMIK
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This is the 57th Video of our Playlist "Leetcode Easy : Popular Interview Problems" by codestorywithMIK
In this video we will try to solve an easy string based Problem : Rotate String | Something to learn | Leetcode 796 | codestorywithMIK
NOTE - You will learn a few good things in this video which will help you in future qns.
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Rotate String | Something to learn | Leetcode 796 | codestorywithMIK
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My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/strings/Easy Tagged/Rotate String.cpp
Leetcode Link : https://leetcode.com/problems/rotate-string
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Summary :
Brute Force Approach (Iterative Rotations):
Logic: Rotate the string s one character at a time and check if it matches goal.
Steps:
Use a loop to rotate s up to m times (where m is the length of s).
For each rotation, shift s left by one character.
If s matches goal at any step, return true.
Time Complexity: O(m2) (where m is the length of s), due to rotating the string and comparing it m times.
Space Complexity: O(1) if modifying s in place.
Optimized Approach (Double String Method):
Logic: Concatenate s with itself (i.e., s + s) and check if goal is a substring of this concatenated string.
Steps:
Check if the lengths of s and goal are equal.
Create s + s and use .contains(goal) in Java or .find(goal) != string::npos in C++ to see if goal is a substring.
If goal is found, return true; otherwise, return false.
Time Complexity: O(m), as substring search in s + s is linear.
Space Complexity: O(m) due to the space needed for s + s.
✨ Timelines✨
00:00 - Introduction
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