Rotate String - Given two strings s and goal, return true if and only if s can become goal after some number of shifts on s. A shift on s consists of moving the leftmost character of s to the rightmost position.

In-depth solution and explanation for LeetCode 796. Rotate String in Python, Java, C++ and more. Intuitions, example walk through, and complexity analysis. Better than official and forum solutions.

public class Solution { /** * @param str: An array of char * @param offset: An integer * @return: nothing */ public void rotateString(char[] str, int offset) { if(str.length <=0) return; offset = offset % str.length; char[] str1 = new char[str.length + offset]; for(int i = str1.length-1;i>= offset; i--){ str1[i] = str[i - offset]; } for(int i = 0; i< offset; i++){ str1[i] = str1[i+str.length]; } for(int i = 0; i< str.length; i++){ str[i] = str1[i]; } } }

class Solution { public: bool rotateString(string s, string goal) { return s.size() == goal.size() && strstr((s + s).data(), goal.data()); } };

A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can become B after some number of shifts ...

LeetCode Solutions in C++23, Java, Python, MySQL, and TypeScript.

May 13, 2023 - Therefore, string was welcomed again in this right solution which beats 98% solutions in leetcode: class Solution: def rotateString(self, s: str, goal: str) -> bool: #convert a string into an array #for every value change the first index into last index #return True if change is equal to goal for i,char in enumerate(s): s+=s[0] #print(s) s=s[1:] #print(s) if s==goal: return True return False

A_shift onA_consists of taking stringAand moving the leftmost character to the rightmost position. For example, ifA = 'abcde', then it will be'bcdea'after one shift onA. ReturnTrueif and only ifAcan becomeBafter some number of shifts onA. class Solution { public boolean rotateString(String A, String B) { if(A == null && B == null) return true; if(A == null || B == null) return false; if(A.length() != B.length()) return false; if(A.equals(B)) return true; int j = 0; while(j < B.length()) { while(j < B.length() && B.charAt(j) != A.charAt(0)) j++; int i = 0; for(; i < A.length(); i++) { if(j + i < B.length()) { if(A.charAt(i) != B.charAt(i + j)) break; } else { if(A.charAt(i) != B.charAt((i + j) % (B.length()))) break; } } if(i == A.length()) return true; j++; } return false; } }

public class GroupRotatedStrings { public static void main(String[] args) { List<String> input = Arrays.asList("abc", "bca", "cab", "xyz", "yzx", "cba", "aaaa"); List<List<String>> groups = groupRotatedStrings(input); System.out.println(groups); } public static List<List<String>> groupRotatedStrings(List<String> strings) { return v1(strings); // return v2(strings); } private static List<List<String>> v1(List<String> strings) { for (int i = 0; i < strings.size(); i++) { String s1 = strings.get(i); for (int j = i+1; j < strings.size(); j++) { String s2 = strings.get(j); System.out.println(s1 + " " + s2 + " " + isRotatedVersion(s1, s2)); } } return new ArrayList<>(); } private static List<List<String>> v2(List<String> strings) { Map<String, List<String>> group = new HashMap<>(); for(String s : strings) group.compute(genHash(s), (k, v) -> v == null ?

A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can become B after some number of shifts on A. Example 1: Input: A = 'abcde', B = 'cdeab' ...

/** * @param {string} A * @param ... */ var rotateString = function(A, B) { if(!A && !B) { return true } let current = "" for(let i = 0; i < A.length; i++) { current += A[i] if(A.slice(i+1) + current === B) { return true } } return false };...

string B2 = "abced"; print_bool(Solution().rotateString(A2, B2)); return 0; } /// Source : https://leetcode.com/problems/rotate-string/description/ /// Author : liuyubobobo · /// Time : 2018-03-11 · #include <iostream> #include <vector> #include <string> using namespace std; /// Brute Force ·

June 3, 2020 - def rotateString(self, A: str, B: str) -> bool: return len(A) == len(B) and B in A + A · Iterate the element of the string once, so total time complexity is O(n)

class Solution { public: bool isFlipedString(string s1, string s2) { return s1.size() == s2.size() && (s1 + s1).find(s2) != string::npos; } };