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.
1interface MountainArray {
2 int get(int index);
3 int length();
4}
5
6public class Solution {
7 public int findInMountainArray(int target, MountainArray mountainArr) {
8 int len = mountainArr.length();
9 for (int i = 0; i < len; i++) {
10 if (mountainArr.get(i) == target) {
11 return i;
12 }
13 }
14 return -1;
15 }
16}The Java implementation performs a linear search across the mountain array elements, comparing each element against the target. As soon as the target is found, its index is returned, otherwise, the loop runs through all elements. Despite being simple, the efficiency is reduced compared to optimized techniques due to increased function calls and potential worst-case traversal across all elements.