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In this approach, we use the property that a rotated version of a string s can be found as a substring of s + s. This is because rotating doesn't change the length of the string and by concatenating the string to itself, every possible rotation is covered.
Time Complexity: O(n^2), where n is the length of the string, due to using strstr.
Space Complexity: O(n), for the doubled string.
1#include <stdbool.h>
2#include <string.h>
3
4bool rotateString(char *s, char *goal) {
5 int lenS = strlen(s);
6 int lenGoal = strlen(goal);
7 if (lenS != lenGoal) return false;
8 char *doubled = (char *)malloc(2 * lenS + 1);
9 strcpy(doubled, s);
10 strcat(doubled, s);
11 bool result = strstr(doubled, goal) != NULL;
12 free(doubled);
13 return result;
14}This C solution first checks if the lengths of s and goal are the same. If not, it immediately returns false. Then it creates a new doubled string by concatenating s with itself. It checks if goal is a substring of the concatenated string. If it is, true is returned; otherwise, false is returned.
This approach simulates rotating the string s multiple times (equal to its length) to check if it equals goal at any point. This is a straightforward but less efficient method compared to the concatenation method.
Time Complexity: O(n^2), because we check for each possible rotation of s for a match with goal.
Space Complexity: O(1), as no additional storage is used beyond some counters.
1
class Solution {
public:
bool rotateString(std::string s, std::string goal) {
if (s.length() != goal.length()) return false;
size_t len = s.length();
for (size_t i = 0; i < len; ++i) {
bool match = true;
for (size_t j = 0; j < len; ++j) {
if (s[(i + j) % len] != goal[j]) {
match = false;
break;
}
}
if (match) return true;
}
return false;
}
};In the C++ solution, we iterate over all possible rotations by using modular arithmetic to simulate a left rotation at each step in the inner loop till a match is found with goal.