The simplest way to solve the problem is to iterate over the array and find the first character that is greater than the target character. If such a character is not found by the end of the array, the function should return the first character of the array. This approach simply checks each character in order from left to right, comparing it with the target.
Time Complexity: O(n) where n is the size of the array since each element may be checked in the worst case.
Space Complexity: O(1) as only a constant amount of space is used.
1public class Solution {
2 public char NextGreatestLetter(char[] letters, char target) {
3 foreach (char letter in letters) {
4 if (letter > target) {
5 return letter;
6 }
7 }
8 return letters[0];
9 }
10}The method iterates through each letter sequentially. If one is found greater than target, it is returned immediately. Otherwise, the method returns the first letter.
Utilizing the sorted nature of the array, we can employ a binary search technique to pinpoint the smallest character that exceeds the target. By continually narrowing the search range, we can efficiently determine the desired character. If the binary search concludes without finding a suitable character, the array's initial character is returned.
Time Complexity: O(log n), hinging on halving the array.
Space Complexity: O(1), processing is constant-space.
1#include <vector>
2using namespace std;
3
4char nextGreatestLetter(vector<char>& letters, char target) {
5 int low = 0, high = letters.size() - 1;
6 while (low <= high) {
7 int mid = low + (high - low) / 2;
8 if (letters[mid] <= target) {
9 low = mid + 1;
10 } else {
11 high = mid - 1;
12 }
13 }
14 return letters[low % letters.size()];
15}Standard binary search is deployed here, working through the ascending letters. It calculates the midpoint, adjusts the range based on comparison, and concludes with the smallest suitable letter based on the circular nature and calculates position with a modulus operation.