/**

 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


 @tag-array

 */
public class Find_in_Mountain_Array {
    /**
     * // This is MountainArray's API interface.
     * // You should not implement it, or speculate about its implementation
     * interface MountainArray {
     *     public int get(int index) {}
     *     public int length() {}
     * }
     */

    class Solution {
        int findInMountainArray(int target, MountainArray A) {
            int n = A.length(), l, r, m, peak = 0;
            // find index of peak
            l  = 0;
            r = n - 1;
            while (l < r) {
                m = (l + r) / 2;
                if (A.get(m) < A.get(m + 1))
                    l = peak = m + 1;
                else
                    r = m;
            }
            // find target in the left of peak
            l = 0;
            r = peak;
            while (l <= r) {
                m = (l + r) / 2;
                if (A.get(m) < target)
                    l = m + 1;
                else if (A.get(m) > target)
                    r = m - 1;
                else
                    return m;
            }
            // find target in the right of peak
            l = peak;
            r = n - 1;
            while (l <= r) {
                m = (l + r) / 2;
                if (A.get(m) > target)
                    l = m + 1;
                else if (A.get(m) < target)
                    r = m - 1;
                else
                    return m;
            }
            return -1;
        }

    }

    class MountainArray {
        public int get(int x) { return 0; }
        public int length() { return 0; }
    }
}

// OJ: https://leetcode.com/problems/find-in-mountain-array/
// Time: O(logN)
// Space: O(1)
class Solution {
    int binarySearch(int target, MountainArray &A, int L, int R, int dir) {
        while (L <= R) {
            int M = (L + R) / 2, val = A.get(M);
            if (val == target) return M;
            bool left = (dir == 1 && val < target) || (dir == -1 && val > target);
            if (left) L = M + 1;
            else R = M - 1;
        }
        return -1;
    }
    int findTop(MountainArray &A, int N) {
        int L = 0, R = N - 1;
        while (L <= R) {
            int M = (L + R) / 2, left = M - 1 >= 0 ? A.get(M - 1) : INT_MIN, val = A.get(M), right = M + 1 < N ? A.get(M + 1) : INT_MIN;
            if (val > left && val > right) return M;
            else if (val < left) R = M - 1;
            else L = M + 1;
        }
        return -1;
    }
public:
    int findInMountainArray(int target, MountainArray &A) {
        int N = A.length(), top = findTop(A, N);
        int a = binarySearch(target, A, 0, top, 1);
        int b = binarySearch(target, A, top + 1, N - 1, -1);
        return a != -1 ? a : b;
    }
};

# """
# 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, nums: 'nums') -> int:
        N = nums.length()
        peek = self.findPeek(target, nums)
        left_index = self.findInAscOrder(target, nums, 0, peek)
        right_index = self.findInDecOrder(target, nums, peek, N - 1)
        if left_index != -1:
            return left_index
        else:
            return right_index

    def findPeek(self, target, nums):
        N = nums.length()
        left, right = 1, N - 2
        while left <= right:
            mid = left + (right - left) // 2
            if nums.get(mid - 1) < nums.get(mid) > nums.get(mid + 1):
                return mid
            elif nums.get(mid - 1) < nums.get(mid) < nums.get(mid + 1):
                left = mid + 1
            else:
                right = mid - 1
        return left
    
    def findInAscOrder(self, target, nums, begin, end):
        left, right = begin, end
        while left <= right:
            mid = left + (right - left) // 2
            print(left, right, mid)
            cur = nums.get(mid)
            if cur == target:
                return mid
            elif cur < target:
                left = mid + 1
            else:
                right = mid - 1
        return -1
    
    def findInDecOrder(self, target, nums, begin, end):
        left, right = begin, end
        while left <= right:
            mid = left + (right - left) // 2
            cur = nums.get(mid)
            if cur == target:
                return mid
            elif cur < target:
                right = mid - 1
            else:
                left = mid + 1
        return -1

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