Binary Search Solution for smallest letter greater than the target

This solution uses binary search for an efficient O(logn) time complexity. The key steps are:

1. Initialize start and end pointers to the beginning and end of the array.
2. Use binary search to check the mid-point:
   • If target < letters[mid], move the end pointer left to look for smaller valid letters.
   • Otherwise, move the start pointer right to skip smaller or equal letters.
3. After the search, start contains the index of the smallest letter greater than the target. To handle wrap-around cases, use the modulo operator: start % letters.length.

The code implementation is as follows:

class Solution {
    public char nextGreatestLetter(char[] letters, char target) {
        int start = 0;
        int end = letters.length - 1;

        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (target < letters[mid]) {
                end = mid - 1;
            } else {
                start = mid + 1;
            }
        }

        return letters[start % letters.length];
    }
}

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