In this approach, the task is broken into three main steps:
arr[i] > arr[i+1] and arr[i] > arr[i-1].Time Complexity: O(log n) for peak finding and 2 * O(log n) for the two binary searches.
Total: O(log n).
Space Complexity: O(1) because we use a constant amount of space.
1class MountainArray:
2 def get(self, index: int) -> int:
3 # Return the element at index in the mountain array.
4 pass
5
6 def length(self) -> int:
7 # Return the length of the mountain array.
8 pass
9
10class Solution:
11 def findPeakIndex(self, mountainArr: 'MountainArray') -> int:
12 left, right = 0, mountainArr.length() - 1
13 while left < right:
14 mid = left + (right - left) // 2
15 if mountainArr.get(mid) < mountainArr.get(mid + 1):
16 left = mid + 1
17 else:
18 right = mid
19 return left
20
21 def binarySearch(self, mountainArr: 'MountainArray', start: int, end: int, target: int, is_ascending: bool) -> int:
22 while start <= end:
23 mid = start + (end - start) // 2
24 value = mountainArr.get(mid)
25 if value == target:
26 return mid
27 if is_ascending:
28 if value < target:
29 start = mid + 1
30 else:
31 end = mid - 1
32 else:
33 if value > target:
34 start = mid + 1
35 else:
36 end = mid - 1
37 return -1
38
39 def findInMountainArray(self, target: int, mountainArr: 'MountainArray') -> int:
40 peak = self.findPeakIndex(mountainArr)
41 index = self.binarySearch(mountainArr, 0, peak, target, True)
42 if index != -1:
43 return index
44 return self.binarySearch(mountainArr, peak + 1, mountainArr.length() - 1, target, False)The Python implementation follows the same logic as the previously described solutions: a binary search to find the peak, followed by two additional binary searches on the segregated parts of the array to locate the target efficiently.
This approach involves directly accessing each element linearly until the condition is satisfied (even though this is not allowed by the problem constraints). It is less optimal and efficient compared to the above implementations, requiring traversal of the entire array.
Time Complexity: O(n) as each element could potentially be checked once.
Space Complexity: O(1) as no extra space is used except for variables.
1#include <stdio.h>
2
3// Forward declaration of the MountainArray API.
4// You don't need to implement this interface, it is provided by the problem constraints.
5struct MountainArray {
6 int (*get)(struct MountainArray*, int index);
7 int (*length)(struct MountainArray*);
8};
9
10int findInMountainArray(int target, struct MountainArray* mountainArr) {
11 int len = mountainArr->length(mountainArr);
12 int index = -1;
13 for (int i = 0; i < len; ++i) {
14 if (mountainArr->get(mountainArr, i) == target) {
15 return i;
16 }
17 }
18 return index;
19}The brute-force approach directly scans through the mountain array from the start to the finish, checking each element against the target. If an element equals the target, it returns the index. Otherwise, it returns -1. This basic method is straightforward but inefficient due to its time complexity.