Search in Rotated Sorted Array 2 - Check if number exists - Duplicates allowed

Problem

There is an integer array nums sorted in non-decreasing order (not necessarily with distinct values).

Before being passed to your function, nums is rotated at an unknown pivot index k (0 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,4,4,5,6,6,7] might be rotated at pivot index 5 and become [4,5,6,6,7,0,1,2,4,4].

Given the array nums after the rotation and an integer target, return true if target is in nums, or false if it is not in nums.

You must decrease the overall operation steps as much as possible.

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

Solution

Method 1 - Iterative Binary Search

This problem is very similar to [Search in Rotated Sorted Array](search-in-rotated-sorted-array).

Code

Java

class Solution {
    public boolean search(int[] nums, int target) {
        int left = 0;
        int right = nums.length - 1;

        while (left <= right) {
            int mid = left + (right - left) / 2;
            if(nums[mid] == target || nums[left] == target || nums[right] == target) {
                return true;
            } else if(nums[left] < nums[mid]) { // left array sorted
                if (nums[left] < target && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else if (nums[mid] < nums[right]) { // right array sorted
                if(nums[mid] < target && target < nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else {
	            right--;
            }
        }

        return false;
    }
}

Complexity

• Time: O(log n)
• Space: O(1)