81. Search in Rotated Sorted Array II

Description

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,0,1,2,2,5,6] might become [2,5,6,0,0,1,2]).

You are given a target value to search. If found in the array return true, otherwise return false.

Constraints

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: nums = [2, 5, 6, 0, 0, 1, 2], target = 0

Output: true

Example 2

Input: nums = [2, 5, 6, 0, 0, 1, 2], target = 3

Output: false

Solutions

Solution 1

/**
 * Time complexity : O(n)
 * Space complexity : O(1)
 */

class Solution {
    public boolean search(int[] nums, int target) {
        for(int i = 0; i < nums.length; i++) {
            if(nums[i] == target) return true;
        }
        return false;
    }
}

Solution 2

/**
 * Time complexity : O(logN)
 * Space complexity : O(1)
 */

class Solution {
    public boolean search(int[] nums, int target) {
        int left = 0, right = nums.length-1;
        while(left <= right) {
            int mid = left + (right-left)/2;
            if(target == nums[mid]) return true;
            if(nums[left] < nums[mid]) {
                if(target >= nums[left] && target <= nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else if(nums[left] > nums[mid]) {
                if(target >= nums[mid] && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else {
                left++;
            }
        }
        return false;
    }
}

Follow up

• This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.

• Would this affect the run-time complexity? How and why?