# 1064. Fixed Point

## 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.

• 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;
}
}

• 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;
}
};

• 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


• 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
}

• function fixedPoint(arr: number[]): number {
let left = 0;
let right = arr.length - 1;
while (left < right) {
const mid = (left + right) >> 1;
if (arr[mid] >= mid) {
right = mid;
} else {
left = mid + 1;
}
}
return arr[left] === left ? left : -1;
}