LeetCode 81. Search in Rotated Sorted Array II

 81. Search in Rotated Sorted Array II

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.

 

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

 

Constraints:

• 1 <= nums.length <= 5000
• -104 <= nums[i] <= 104
• nums is guaranteed to be rotated at some pivot.
• -104 <= target <= 104

 

Follow up: This problem is similar to Search in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

	class Solution {
	public:
	bool search(vector<int>& nums, int target) {
	int l = 0;
	int r = nums.size() - 1;
	while(l <= r)
	{
	int mid = l + (r-l) / 2;
	if (nums[mid] == target)
	return true;
	// with duplicates we can have this contdition, just update left & right
	if((nums[l] == nums[mid]) && (nums[r] == nums[mid]))
	{
	l++;
	r--;
	}
	// first half
	// first half is in order
	else if(nums[l] <= nums[mid])
	{
	// target is in first half
	if((nums[l] <= target) && (nums[mid] > target))
	r = mid - 1;
	else
	l = mid + 1;
	}
	// second half
	// second half is order
	// target is in second half
	else
	{
	if((nums[mid] < target) && (nums[r]>= target))
	l = mid + 1;
	else
	r = mid - 1;
	}
	}
	return false;
	}
	};

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