README_EN.md

1064. Fixed Point

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Description

Given an array of distinct integers arr, where arr is sorted in ascending order, return the smallest index i that satisfies arr[i] == i. If there is no such index, return -1.

 

Example 1:

Input: arr = [-10,-5,0,3,7]
Output: 3
Explanation: For the given array, arr[0] = -10, arr[1] = -5, arr[2] = 0, arr[3] = 3, thus the output is 3.

Example 2:

Input: arr = [0,2,5,8,17]
Output: 0
Explanation: arr[0] = 0, thus the output is 0.

Example 3:

Input: arr = [-10,-5,3,4,7,9]
Output: -1
Explanation: There is no such i that arr[i] == i, thus the output is -1.

 

Constraints:

• 1 <= arr.length < 104
• -109 <= arr[i] <= 109

 

Follow up: The O(n) solution is very straightforward. Can we do better?

Solutions

Binary search.

Python3

class Solution:
    def fixedPoint(self, arr: List[int]) -> int:
        left, right = 0, len(arr) - 1
        while left < right:
            mid = (left + right) >> 1
            if arr[mid] >= mid:
                right = mid
            else:
                left = mid + 1
        return left if arr[left] == left else -1

Java

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

C++

class Solution {
public:
    int fixedPoint(vector<int>& arr) {
        int left = 0, right = arr.size() - 1;
        while (left < right) {
            int mid = left + right >> 1;
            if (arr[mid] >= mid) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return arr[left] == left ? left : -1;
    }
};

Go

func fixedPoint(arr []int) int {
	left, right := 0, len(arr)-1
	for left < right {
		mid := (left + right) >> 1
		if arr[mid] >= mid {
			right = mid
		} else {
			left = mid + 1
		}
	}
	if arr[left] == left {
		return left
	}
	return -1
}

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