(This problem is an interactive problem.)
You may recall that an array
arr is a mountain array if and only if:
arr.length >= 3
- There exists some
0 < i < arr.length - 1such that:
arr < arr < ... < arr[i - 1] < arr[i]
arr[i] > arr[i + 1] > ... > arr[arr.length - 1]
Given a mountain array
mountainArr, return the minimum
index such that
mountainArr.get(index) == target. If such an
index does not exist, return
You cannot access the mountain array directly. You may only access the array using a
MountainArray.get(k)returns the element of the array at index
MountainArray.length()returns the length of the array.
Submissions making more than
100 calls to
MountainArray.get will be judged Wrong Answer. Also, any solutions that attempt to circumvent the judge will result in disqualification.
Input: array = [1,2,3,4,5,3,1], target = 3 Output: 2 Explanation: 3 exists in the array, at index=2 and index=5. Return the minimum index, which is 2.
Input: array = [0,1,2,4,2,1], target = 3 Output: -1 Explanation: 3 does not exist in
the array,so we return -1.
3 <= mountain_arr.length() <= 104
0 <= target <= 109
0 <= mountain_arr.get(index) <= 109
Find the peak of the mountain first. If the target is the peak of the mountain, return the peak index. If the target is greater than the peak, return -1. Otherwise, use binary search on the uphill of the mountain first, if the target is still not found, search on the downhill of the mountain.
# """ # This is MountainArray's API interface. # You should not implement it, or speculate about its implementation # """ #class MountainArray: # def get(self, index: int) -> int: # def length(self) -> int: class Solution: def findInMountainArray(self, target: int, mountain_arr: 'MountainArray') -> int: if mountain_arr.get(0) == target: return 0 peak_idx = self.find_peak(mountain_arr) if mountain_arr.get(peak_idx) < target: return -1 left = self.binary_search(target, mountain_arr, 0, peak_idx - 1, True) if left != -1: return left if mountain_arr.get(peak_idx) == target: return peak_idx right = self.binary_search(target, mountain_arr, peak_idx, mountain_arr.length() - 1, False) if right != -1: return right return -1 def find_peak(self, mountain_arr): peak = -1 start = 0 end = mountain_arr.length() - 1 while start + 1 < end: mid = (start + end) // 2 if mountain_arr.get(mid - 1) < mountain_arr.get(mid) and mountain_arr.get(mid) > mountain_arr.get(mid + 1): return mid elif mountain_arr.get(mid) < mountain_arr.get(mid - 1): end = mid elif mountain_arr.get(mid) < mountain_arr.get(mid + 1): start = mid if mountain_arr.get(start) > mountain_arr.get(end): peak = start else: peak = end return peak def binary_search(self, target, mountain_arr, start, end, is_increasing=True): while start + 1 < end: mid = (start + end) // 2 if mountain_arr.get(mid) == target: end = mid elif mountain_arr.get(mid) < target: if is_increasing: start = mid else: end = mid else: if is_increasing: end = mid else: start = mid if mountain_arr.get(start) == target: return start if mountain_arr.get(end) == target: return end return -1
- Time Complexity: O(log(N)).
- Space Complexity: O(1).