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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.
1interface MountainArray {
2 int get(int index);
3 int length();
4}
5
6public
The Java solution performs similarly to the C/C++ implementations. It identifies the peak using a binary search, which enables the array to be divided into two segments for additional binary searches to locate the target. Each binary search operation effectively reduces the time complexity to O(log n).
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 <vector>
using namespace std;
// Mockup of MountainArray, simulating interface.
class MountainArray {
public:
int get(int index) const;
int length() const;
};
int findInMountainArray(int target, const MountainArray &mountainArr) {
int len = mountainArr.length();
for (int i = 0; i < len; ++i) {
if (mountainArr.get(i) == target) {
return i;
}
}
return -1;
}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.