1095. Find in Mountain Array

Problem Statement

(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 i with 0 < i < arr.length - 1 such 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 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() <= 104
0 <= target <= 109
0 <= mountain_arr.get(index) <= 109

Intuition

Approach:

Find the peak and do Binary search on both sides

Only one edge case
While finding peak,
3 5 4 3 2

let
3 5 4 3 2
l h
m 

3 < 3 < 5 is false, so we will not be able to judge the increasing and decreasing 
side so 
it goes in else, high = mid-1
returns -1

hence finding at index -1, gives error in == target statement


so, here there is confirm peak element 
so we search peak in the range 1 to len-2 to avoid this

Links

https://leetcode.com/problems/find-in-mountain-array/description/?envType=daily-question&envId=2023-10-12

Video Links

Approach 1:

C++

class Solution {
public:
    int find_peak(int low, int high, MountainArray &arr){
        while(low<=high){
            int mid = low+(high-low)/2;
            int num = arr.get(mid);
            if(arr.get(mid-1) < num and num > arr.get(mid+1) )
                return mid;
            else if(arr.get(mid-1) < num and num < arr.get(mid+1))
                low = mid+1;
            else
                high = mid-1;
        }

        return -1;
    }

    int binary_search(int low, int high, MountainArray &arr, int target, bool check){
        while(low<=high){
            int mid = low + (high-low)/2;
            int num = arr.get(mid);

            if(num == target)
                return mid;
            else if(num > target){
                if(check)
                    high = mid-1;
                else
                    low = mid+1;
            }
            else{
                if(check)
                    low = mid+1;    
                else
                    high = mid-1;
            }
        }

        return -1;
    }

    
    int findInMountainArray(int target, MountainArray &arr) {
        int len = arr.length();
        int peak = find_peak(1, len-2, arr);

        if(arr.get(peak) == target)
            return peak;

        int left = binary_search(0, peak-1, arr, target, true);
        int right = binary_search(peak+1, len-1, arr, target, false);

        if(left == -1 and right == -1)
            return -1;
        else if(left != -1 and right != -1)
            return min(left, right);

        return max(left, right);
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++