Fixed Point

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

The solution provided implements a method to find a "fixed point" in an array using the binary search algorithm. A fixed point in an array is an index i such that data[i] is equal to i.

• Initialize start to 0 and end to the last index of data.
• Initialize result to -1 to represent the absence of a fixed point initially.
• Use a while loop to conduct the binary search as long as start is less than or equal to end:
  • Calculate the middle index middle.
  • Check if the element at this middle index is the same as its index. If true, store this index in result and narrow the search to the left half to find the smallest fixed point (if any).
  • If the middle element is less than its index, move the start index to middle + 1, shifting the search range to the right.
  • If the middle element is greater than its index, update the end index to middle - 1, focusing the search on the left part.
• At the end of the loop, return result. If a fixed point exists, result will contain the smallest index that satisfies the fixed point condition. Otherwise, it will return -1, indicating no fixed point was found.

This implementation efficiently finds and confirms the presence of a fixed point using binary search, making it optimal for sorted arrays, ensuring a time complexity of O(log n), where n is the number of elements in the array.

class Solution {
    public int searchFixedPoint(int[] data) {
        int lo = 0, hi = data.length - 1;
        int result = -1;
        
        while (lo <= hi) {
            int mid = (lo + hi) / 2;
            
            if (data[mid] == mid) {
                result = mid;
                hi = mid - 1; // Shift search area to left
            } else if (data[mid] < mid) {
                lo = mid + 1; // Search in right half
            } else {
                hi = mid - 1; // Search in left half
            }
        }
        
        return result;
    }
}

The provided Java code implements a method to find a fixed point in an array. A fixed point is an index where the element at that index is equal to the index itself, e.g., data[i] == i.

The method named searchFixedPoint employs a binary search algorithm to efficiently locate the fixed point. Here’s how it functions:

1. Initialize two pointers, lo and hi, representing the start and end of the array segment being searched.
2. Initiate a loop that continues as long as lo is less than or equal to hi.
3. Calculate the midpoint of the current segment using (lo + hi) / 2. This position is continuously updated based on the results of each comparison.
4. If the element at this midpoint position equals the midpoint (i.e., data[mid] == mid), then:
   • Assign this position as the result.
   • Adjust hi to search the left side of the array for any earlier fixed point occurrence.
5. If the element at the midpoint is less than the midpoint (data[mid] < mid), move the lo pointer to mid + 1 to search the right half of the array segments.
6. If the element at the midpoint is greater than the midpoint, adjust the hi pointer to mid - 1 to search the left half.

The method returns the first occurrence of the fixed point if it exists, returning -1 if no fixed point is found. This return value effectively indicates that no index in the array satisfies the condition data[index] == index.

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

class Solution:
    def findFixedPoint(self, values: List[int]) -> int:
        start, end = 0, len(values) - 1
        result = -1

        while start <= end:
            middle = (start + end) // 2

            if values[middle] == middle:
                result = middle
                end = middle - 1
            elif values[middle] < middle:
                start = middle + 1
            else:
                end = middle - 1

        return result

Discover the method to identify a fixed point in an array where the element's value equals its index, using Python. The provided code implements a binary search technique to efficiently locate such a position within a sorted list of integers.

• The function findFixedPoint initializes two pointers, start and end, to represent the range currently being searched.
• A variable result is initialized to -1, to store the result if a fixed point is found.
• A while loop continues until start exceeds end. Inside this loop:
  • Calculate the midpoint of the current range.
  • If the element at the midpoint index is equal to the index itself, update result with this index and narrow the search to the left half to possibly find a lower fixed point.
  • If the element is less than the midpoint index, adjust the start pointer to focus on the right half; otherwise, adjust the end pointer to focus on the left half.
• The process concludes by returning result, which will either be -1 (indicating no fixed point was found) or the index of a fixed point, with preference given to the lowest fixed point if multiple are present.

Implement this solution to determine the fixed point in your data, ensuring efficient performance even with sizable input arrays.