Solution Explanation

This problem asks to find the smallest index i in a sorted array arr such that arr[i] == i. A simple linear scan would work, but the problem hints at a more efficient solution using binary search. This is possible because the array is sorted.

Approach:

We can leverage the sorted nature of the array. If arr[mid] >= mid, it means that the fixed point (if it exists) must be in the left half or at mid itself because the array is sorted in ascending order. If arr[mid] < mid, then the fixed point (if it exists) must be in the right half. This allows us to perform a binary search.

Algorithm:

1. Initialization: Set left = 0 and right = arr.length - 1.
2. Binary Search: While left < right:
   • Calculate mid = (left + right) / 2 (or (left + right) >> 1 for bitwise efficiency).
   • If arr[mid] >= mid: The fixed point (if any) lies in the left half (including mid), so update right = mid.
   • Otherwise (arr[mid] < mid): The fixed point (if any) lies in the right half, so update left = mid + 1.
3. Check Result: After the loop, left will point to the potential fixed point. Check if arr[left] == left. If true, return left; otherwise, return -1.

Time Complexity Analysis:

The solution uses binary search, which has a time complexity of O(log n), where n is the length of the array. This is significantly better than the O(n) time complexity of a linear scan.

Space Complexity Analysis:

The algorithm uses only a constant amount of extra space for variables like left, right, and mid. Therefore, the space complexity is O(1).

Code Examples (with explanations inline):

The provided code examples in Python, Java, C++, Go, and TypeScript all implement the same algorithm. Let's examine the Python code in detail:

class Solution:
    def fixedPoint(self, arr: List[int]) -> int:
        left, right = 0, len(arr) - 1 # Initialize left and right pointers
        while left < right:           # Binary search loop
            mid = (left + right) >> 1 # Calculate middle index efficiently (bitwise right shift)
            if arr[mid] >= mid:      # If arr[mid] is greater than or equal to mid
                right = mid          # Search in the left half
            else:                    # Otherwise
                left = mid + 1       # Search in the right half
        return left if arr[left] == left else -1 # Return index if found, otherwise -1
 

The other languages' implementations follow the same logic with minor syntactic differences. They all achieve the same O(log n) time complexity and O(1) space complexity.