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.
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 findInMountainArray(target, mountainArr) {
13 let len = mountainArr.length();
14 for (let i = 0; i < len; i++) {
15 if (mountainArr.get(i) === target) {
16 return i;
17 }
18 }
19 return -1;
20 }
21}Using JavaScript's brute-force search, the solution deploys a for loop to walk through each element of the mountain array. It checks for a match with the target element, returning the index if found, or -1 after scanning the full array, making it less favorable for large inputs due to its lengthy evaluation process.