# Question

Formatted question description: https://leetcode.ca/all/1064.html

 1064. Fixed Point

Given an array A of distinct integers sorted in ascending order,
return the smallest index i that satisfies A[i] == i.  Return -1 if no such i exists.

Example 1:

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

Example 2:

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

Example 3:

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

Note:

1 <= A.length < 10^4
-10^9 <= A[i] <= 10^9


# Algorithm

The basic idea of ​​binary search is to divide n elements into two roughly equal parts, and compare a[n/2] with x.

• If x=a[n/2], then find x and the algorithm stops;
• if x<a[n/2], as long as you continue to search for x in the left half of array a,
• if x>a[n/2], then as long as you search for x in the right half of array a.

# Code

Java

• 
public class Fixed_Point {

// binary search, logN
class Solution {
public int fixedPoint(int[] A) {
//Run binary search
int lo = 0;
int hi = A.length - 1;

while (lo <= hi){
int mi = lo + (hi - lo) / 2;
if (A[mi] == mi) return mi;
if (mi < A[mi]) hi = mi - 1;
else lo = mi+1;
}

return -1;
}
}

class Solution_bruteForce {
public int fixedPoint(int[] A) {
for (int i = 0; i < A.length; i++) {
if (A[i] == i) {
return i;
}
}
return -1;
}
}

}


• // OJ: https://leetcode.com/problems/fixed-point/
// 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;
}
};

• print("Todo!")