1064

1064. Fixed Point (Easy)

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?

Companies:
Uber

Related Topics:
Array, Binary Search

Solution 1. Binary Search (L <= R)

// OJ: https://leetcode.com/problems/fixed-point/
// Author: github.com/lzl124631x
// Time: O(logN)
// Space: O(1)
class Solution {
public:
    int fixedPoint(vector<int>& A) {
        int N = A.size(), L = 0, R = N - 1;
        while (L <= R) {
            int M = (L + R) / 2;
            if (A[M] < M) L = M + 1;
            else R = M - 1;
        }
        return L < N && A[L] == L ? L : -1;
    }
};

Solution 2. Binary Search (L < R)

// OJ: https://leetcode.com/problems/fixed-point/
// Author: github.com/lzl124631x
// Time: O(logN)
// Space: O(1)
class Solution {
public:
    int fixedPoint(vector<int>& A) {
        int N = A.size(), L = 0, R = N - 1;
        while (L < R) {
            int M = (L + R) / 2;
            if (A[M] < M) L = M + 1;
            else R = M;
        }
        return L < N && A[L] == L ? L : -1;
    }
};