Sponsored
Sponsored
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.
1var rotateString = function(s, goal) {
2 if (s.length !== goal.length) return false;
3 return (s + s).includes(goal);
4};The JavaScript solution checks if the lengths of s and goal are the same. If not, it returns false. Otherwise, it checks if goal is a substring of s + s using the includes method.
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
The C solution iteratively rotates the string by slicing it and recomposing it character by character to check if it matches goal.