1095. Find in Mountain Array

Problem

**(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

Solution

/**
 * // This is the MountainArray's API interface.
 * // You should not implement it, or speculate about its implementation
 * function MountainArray() {
 *     @param {number} index
 *     @return {number}
 *     this.get = function(index) {
 *         ...
 *     };
 *
 *     @return {number}
 *     this.length = function() {
 *         ...
 *     };
 * };
 */

/**
 * @param {number} target
 * @param {MountainArray} mountainArr
 * @return {number}
 */
var findInMountainArray = function(target, mountainArr) {
     var maxIndex = findMaxIndex(mountainArr);
     var leftIndex = findInLeft(target, mountainArr, maxIndex);
     if (leftIndex !== -1) return leftIndex;
     return findInRight(target, mountainArr, maxIndex);
};

var findMaxIndex = function(mountainArr) {
    var left = 0;
    var right = mountainArr.length() - 1;
    while (left < right) {
        var mid = left + Math.floor((right - left) / 2);
        if (mountainArr.get(mid) > mountainArr.get(mid + 1)) {
            right = mid;
        } else {
            left = mid + 1;
        }
    }
    return left;
};

var findInLeft = function(target, mountainArr, maxIndex) {
    var left = 0;
    var right = maxIndex;
    while (left <= right) {
        var mid = left + Math.floor((right - left) / 2);
        var midVal = mountainArr.get(mid);
        if (midVal === target) {
            return mid;
        } else if (midVal > target) {
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    return -1;
};

var findInRight = function(target, mountainArr, maxIndex) {
    var left = maxIndex;
    var right = mountainArr.length() - 1;
    while (left <= right) {
        var mid = left + Math.floor((right - left) / 2);
        var midVal = mountainArr.get(mid);
        if (midVal === target) {
            return mid;
        } else if (midVal < target) {
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    return -1;
};

Explain:

nope.

Complexity:

• Time complexity : O(log(n)).
• Space complexity : O(1).