Description
https://leetcode.com/problems/fixed-point/
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 suchi
thatarr[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?
Explanation
Just check if the element’s value equals to its index.
Python Solution
class Solution:
def fixedPoint(self, arr: List[int]) -> int:
for i, num in enumerate(arr):
if num == i:
return i
return -1
- Time Complexity: O(N).
- Space Complexity: O(1).