Skip to content

Latest commit

 

History

History
138 lines (106 loc) · 3.3 KB

20210716.md

File metadata and controls

138 lines (106 loc) · 3.3 KB

Algorithm

1095. Find in Mountain Array

Description

(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

Solution

/**
 * // 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 {
    public int findInMountainArray(int target, MountainArray mountainArr) {
        int l = 0;
        int length = mountainArr.length() - 1;
        int r = length;
        int mid = 0;

        //find mountain peak
        //if peak is target return peak
        //because peak is unique
        while(r>l){
            mid = (r + l) / 2;
            int mider = mountainArr.get(mid);
            int higher = mountainArr.get(mid + 1);
            int lower = mountainArr.get(mid - 1);
            if(mider > higher && mider > lower) {
                if(mider == target) {
                    return mid;
                }
                break;
            }
            if(mider<lower){
                r = mid;
            }
            if(mider<higher){
                l = mid + 1;
            }
        }

        //check left side
        l = 0;
        int r1 = mid;
        while(r1 > l){
            int mid1 = (r1 + l) / 2;
            int mider = mountainArr.get(mid1);

            if(mider == target) {
                return mid1;
            }

            if(mider > target) {
                r1 = mid1;                
            }

            if(mider < target) {
                l = mid1 + 1;                
            }
        }

        //check right side
        int l2 = mid + 1;
        int r2 = length;
        while (l2 <= r2) {
            int mid2 = (l2 + r2)  / 2;
            int mider = mountainArr.get(mid2);

            if (mider == target){
                return mid2;
            }

            if (mider < target){
                r2 = mid2 - 1;    
            }else{
                l2 = mid2 + 1;
            }
        }

        return -1;
    }
}

Discuss

Review

Tip

Share