LeetCode 1064: Fixed Point Solution

Problem Explanation

Explanation

To solve this problem, we can iterate through the array and check if the value at any index is equal to the index itself. If we find such an index, it is a fixed point, and we can return that index. If we don't find any fixed point, we return -1.

The array is sorted in non-decreasing order, so we can utilize this property and optimize our solution by using binary search. By comparing the value at the mid index with the index itself, we can determine whether the fixed point lies to the left or right of the mid index.

Algorithm:

Initialize left to 0 and right to arr.length - 1.
While left <= right: a. Calculate mid as (left + right) / 2. b. If arr[mid] == mid, return mid. c. If arr[mid] < mid, set left = mid + 1. d. If arr[mid] > mid, set right = mid - 1.
If no fixed point is found, return -1.

Time Complexity

The time complexity of this solution is O(log n) as we are using binary search.

Space Complexity

The space complexity is O(1) as we are not using any extra space other than a few variables.

Solution Code

class Solution {
    public int fixedPoint(int[] arr) {
        int left = 0, right = arr.length - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (arr[mid] == mid) {
                return mid;
            } else if (arr[mid] < mid) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return -1;
    }
}

Frequently Asked Questions

How to solve LeetCode 1064 (Fixed Point)?

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What is the time complexity of LeetCode 1064 (Fixed Point)?

The time complexity for LeetCode 1064 (Fixed Point) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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1064. Fixed Point