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 <iostream>
2#include <vector>
3using namespace std;
4
5// Mockup of MountainArray, simulating interface.
6class MountainArray {
7public:
8 int get(int index) const;
9 int length() const;
10};
11
12int findInMountainArray(int target, const MountainArray &mountainArr) {
13 int len = mountainArr.length();
14 for (int i = 0; i < len; ++i) {
15 if (mountainArr.get(i) == target) {
16 return i;
17 }
18 }
19 return -1;
20}The C++ brute-force solution iterates with a loop, accessing each value one-by-one, and checking for equality with the target value. It returns the index if a match is found and -1 if the entire array is traversed without finding the target. This is straightforward but is less optimized since it calls the get function repeatedly and traverses each element.