Find in Mountain Array

Find in Mountain Array

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

Approach

Intuition

Algo

1. Binary find peak in the mountain. 852. Peak Index in a Mountain Array
2. Binary find the target in strict increasing array
3. Binary find the target in strict decreasing array
   
   

Tip

Personally,
If I want find the index, I always use while (left < right)
If I may return the index during the search, I'll use while (left <= right)

Complexity

Time O(logN) Space O(1)

Some Improvement

1. Cache the result of get, in case we make the same calls.
   In sacrifice of O(logN) space for the benefit of less calls.
2. Binary search of peak is unnecessary, just easy to write.
   
   

Java

    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;
    }