Find Smallest Letter Greater Than Target

class Solution {
public:
    char findNextChar(vector<char>& letters, char target) {
        int start = 0, end = letters.size() - 1, middle;
        while (start <= end) {
            middle = (start + end) / 2;
            if (letters[middle] <= target) {
                start = middle + 1;
            } else {
                end = middle - 1;
            }
        }
    
        return start == letters.size() ? letters[0] : letters[start];
    }
};

The provided C++ solution addresses the problem of finding the smallest letter greater than a given target from a sorted list of letters. The approach effectively utilizes a binary search method to optimize the search process. Here's an outline of how the solution operates:

• Define a class Solution with a public member function findNextChar, which takes a vector of characters letters and a character target as parameters.
• Initialize two integer variables start and end to represent the beginning and the end of the search range within the list.
• Commence a while loop that continues as long as start is less than or equal to end. Each iteration adjusts the start and end based on comparisons of the middle character in the current search range with the target.
  • Calculate the middle position of the current range.
  • If the character at the middle position is less than or equal to the target, move the start point up (to middle + 1), discarding the first half of the current search range as all these characters are less than or equal to the target.
  • Otherwise, adjust the end point down (to middle - 1), keeping possible candidates in the range.
• After exiting the loop, check if the start index has exceeded the size of the list, which means the target character is greater than any character in the list. In this situation, return the first character from the list, effectively implementing a wrap-around as the list is sorted.
• If not, return the character found at the start index in the list as it will be the smallest character greater than the target.

This solution ensures a time complexity of O(log n), due to the binary search approach, making it efficient for large datasets.

class Solution {
    public char findNextChar(char[] sequence, char reference) {
        int start = 0, end = sequence.length - 1, pivot;
        while (start <= end) {
            pivot = (start + end) / 2;
            if (sequence[pivot] <= reference) {
                start = pivot + 1;
            } else {
                end = pivot - 1;
            }
        }
    
        return start == sequence.length ? sequence[0] : sequence[start];
    }
}

The Java solution presented efficiently finds the smallest character in a sorted array greater than a given target character utilizing binary search. Follow these steps to understand the underlying logic and execution flow of the method findNextChar:

1. Initialize Pointers: Two pointers, start and end, are initialized to the beginning and end of the array respectively.

2. Binary Search Process: Use a binary search loop wherein:

   • Calculate the middle index pivot.
   • If the element at pivot is less than or equal to the reference character, move the start index to pivot + 1.
   • If the element is greater, adjust the end index to pivot - 1.

3. Determine Result: After exiting the loop, check if start has moved beyond the last index of the array. If so, the smallest letter is wrapped around to the first element of the array (sequence[0]). Otherwise, return the element at the start index.

This method ensures an optimal O(log n) time complexity, making it efficient even for large datasets. It handles scenarios where no direct match is found, and even when all characters in the array are less than or equal to the target character.

class Solution:
    def next_larger_character(self, chars: List[str], ref_char: str) -> str:
        low = 0
        high = len(chars) - 1
    
        while low <= high:
            middle = (low + high) // 2
            if chars[middle] <= ref_char:
                low = middle + 1
            else:
                high = middle - 1
    
        if low == len(chars):
            return chars[0]
        else:
            return chars[low]

This solution uses a binary search algorithm to efficiently solve the problem of finding the smallest letter greater than a given target within a sorted list of characters. The function next_larger_character takes two parameters: a list of characters chars and a reference character ref_char.

Follow these steps to understand the implementation within a Python environment:

1. Initialize two pointers, low and high, which represent the starting and ending bounds of your search area within the list of characters. low starts at 0 and high starts at the last valid index, len(chars) - 1.

2. Implement a binary search loop:

   • Calculate the middle index by averaging low and high.
   • Compare the middle character chars[middle] with the reference character ref_char:
     • If chars[middle] is less than or equal to ref_char, it means the desired letter must be further in the list. Therefore, set low to middle + 1 to search the upper half.
     • Otherwise, the smallest larger character might be at middle or to its left, so set high to middle - 1 to explore the lower half.

3. After the loop, check if low equals the length of the list:

   • If true, this indicates that all characters in the list are less than or equal to the ref_char. Therefore, return the first character in the list as it is circularly greater.
   • Otherwise, return the character at the low index, which is the smallest character greater than ref_char.

This approach efficiently pinpoints the desired character by reducing the search space in half with each iteration, allowing it to run in logarithmic time complexity relative to the size of the input list. Moreover, by leveraging the sorted nature of the list, this solution guarantees that the first element found greater than ref_char after all iterations is the smallest such element.