Algorithm
Problem Name: 1095. Find in Mountain Array
(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
iwith0 < i < arr.length - 1such that:arr[0] < arr[1] < ... < 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 -1.
You cannot access the mountain array directly. You may only access the array using a MountainArray interface:
MountainArray.get(k)returns the element of the array at indexk(0-indexed).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.
Example 1:
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.
Example 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.
Constraints:
3 <= mountain_arr.length() <= 1040 <= target <= 1090 <= mountain_arr.get(index) <= 109
Code Examples
#1 Code Example with Javascript Programming
Code -
Javascript Programming
const findInMountainArray = function(target, mountainArr) {
const p = findPeak(mountainArr)
if (mountainArr.get(p) === target) {
return p
}
const left = binarySeach(mountainArr, 0, p, target, 'asc')
if (left > -1) {
return left
}
return binarySeach(mountainArr, p + 1, mountainArr.length(), target, 'dsc')
}
function findPeak(arr) {
let left = 0
let right = arr.length()
while (left < right) {
const mid = Math.floor((left + right) / 2)
if (arr.get(mid) < arr.get(mid + 1)) {
left = mid + 1
} else {
right = mid
}
}
return left
}
function binarySeach(mountainArr, start, end, target, order) {
let left = start
let right = end
while (left < right) {
const mid = Math.floor((left + right) / 2)
if (target === mountainArr.get(mid)) {
return mid
} else if (
(target > mountainArr.get(mid) && order === 'asc') ||
(target < mountainArr.get(mid) && order === 'dsc')
) {
left = mid + 1
} else {
right = mid
}
}
return -1
}
Input
Output
#2 Code Example with Python Programming
Code -
Python Programming
# """
# 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:
l, r = 0, mountain_arr.length() - 1
while l <= r:
mid = (l + r) // 2
md = mountain_arr.get(mid)
if mid and mountain_arr.get(mid - 1) < md > mountain_arr.get(mid + 1):
pivot = mid
break
elif md < mountain_arr.get(mid + 1):
l = mid + 1
else:
r = mid - 1
l, r = 0, pivot
while l <= r:
mid = (l + r) // 2
md = mountain_arr.get(mid)
if md == target:
return mid
elif md < target:
l = mid + 1
else:
r = mid - 1
l, r = pivot, mountain_arr.length() - 1
while l <= r:
mid = (l + r) // 2
md = mountain_arr.get(mid)
if md == target:
return mid
elif md > target:
l = mid + 1
else:
r = mid - 1
return -1
Input
Output