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This approach leverages the characteristics of a mountain array to achieve an O(log n) time complexity using binary search. The concept is to look for the part of the array where the increase turns into a decrease. We will perform a binary search for the peak index by comparing middle elements and deciding the side that can have the peak element.
Time complexity: O(log n) since we are performing a binary search.
Space complexity: O(1) as we are using a constant amount of extra space.
1class Solution {
2 public int peakIndexInMountainArray(int[] arr) {
3 int low = 0, high = arr.length - 1;
4 while (low < high) {
5 int mid = low + (high - low) / 2;
6 if (arr[mid] < arr[mid + 1]) {
7 low = mid + 1;
8 } else {
9 high = mid;
10 }
11 }
12 return low;
13 }
14
15 public static void main(String[] args) {
16 Solution sol = new Solution();
17 int[] arr = {0, 2, 1, 0};
18 System.out.println("Peak Index: " + sol.peakIndexInMountainArray(arr));
19 }
20}This Java solution replicates the binary search strategy to find the peak in a mountain array. We initialize low as 0 and high as the last index of the array, iterating until they meet. The peak index is found by leveraging comparisons within the loop.
This approach involves a simple linear scan to find the peak element. Although not achieving the necessary O(log n) time complexity, it serves as a straightforward solution to verify correctness.
Time complexity: O(n) as each element is visited once.
Space complexity: O(1).
1#
The C solution performs a linear scan from index 1 to arrSize - 2 (exclusive). For each element, it checks if it is larger than its neighbors. If true, the index is returned as the peak index.