1095. Find in Mountain Array

(This problem is an interactive problem.)

You may recall that an array A is a mountain array if and only if:

• A.length >= 3
• There exists some i with 0 < i < A.length - 1 such that:
  • A[0] < A[1] < ... A[i-1] < A[i]
  • A[i] > A[i+1] > ... > A[A.length - 1]

Given a mountain array mountainArr, return the minimum index such that mountainArr.get(index) == target. If such an index doesn't exist, return -1.

You can 't 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 index k (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() <= 10000
• 0 <= target <= 10^9
• 0 <= mountain_arr.get(index) <= 10^9

# @lc code=start
using LeetCode

struct MountainArray
    vec::Vector{Int}
end

Base.length(ma::MountainArray) = length(ma.vec)
Base.getindex(ma::MountainArray, i) = ma.vec[i]

function find_in_mountain_array(ma::MountainArray, target::Int)
    len = length(ma)
    i, j = 1, len
    while i != j
        mid = (i + j) >> 1
        ma[mid] > ma[mid + 1] ? (j = mid) : (i = mid + 1)
    end
    idx1 = searchsortedfirst(1:i, i; by=i -> ma[i] >= target)
    ma[idx1] == target && return idx1
    idx2 = searchsortedfirst((i + 1):len, i; by=i -> ma[i] <= target) + i
    ma[idx2] == target && return idx2
    return -1
end
# @lc code=end

find_in_mountain_array (generic function with 1 method)

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