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 get(index) {
3 // Returns the element of the array at index 'index'.
4 }
5
6 length() {
7 // Returns the number of elements in the mountain array.
8 }
9}
10
11class Solution {
12 findPeakIndex(mountainArr) {
13 let left = 0;
14 let right = mountainArr.length() - 1;
15 while (left < right) {
16 const mid = Math.floor(left + (right - left) / 2);
17 if (mountainArr.get(mid) < mountainArr.get(mid + 1)) {
18 left = mid + 1;
19 } else {
20 right = mid;
21 }
22 }
23 return left;
24 }
25
26 binarySearch(mountainArr, start, end, target, isAscending) {
27 while (start <= end) {
28 const mid = Math.floor(start + (end - start) / 2);
29 const value = mountainArr.get(mid);
30 if (value === target) {
31 return mid;
32 }
33 if (isAscending) {
34 if (value < target) {
35 start = mid + 1;
36 } else {
37 end = mid - 1;
38 }
39 } else {
40 if (value > target) {
41 start = mid + 1;
42 } else {
43 end = mid - 1;
44 }
45 }
46 }
47 return -1;
48 }
49
50 findInMountainArray(target, mountainArr) {
51 const peak = this.findPeakIndex(mountainArr);
52 let index = this.binarySearch(mountainArr, 0, peak, target, true);
53 if (index !== -1) {
54 return index;
55 }
56 return this.binarySearch(mountainArr, peak + 1, mountainArr.length() - 1, target, false);
57 }
58}The JavaScript solution employs a binary search strategy first to find the peak of the mountain array, then to conduct two additional binary searches on both sections. This ensures that the target is found efficiently with a minimal index. The approach is efficient due to the logarithmic complexity of the binary search.
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.