Leetcode#81

Problem Statement

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?

Logic

• if we find left , mid and right value are equal,
  • Just shrink the array by 1
  • left++ & right — ; continue
• Please read normal Binary search code for finding target and its comparison with rotated array.
• This will help to build foundation.
  • Binary Search vs Rotated Binary Search

Golang Solution

func search(nums []int, target int) bool {

	left, right := 0, len(nums)-1

	for left <= right {
		middle := left + (right-left)/2

		// check for target comparison
		if nums[middle] == target {
			return true
		}

		// Special check for duplicates
		// shrink the array by 1 place from both directions
		if nums[left] == nums[middle] && nums[middle] == nums[right] {
			left++
			right--
			continue
		}

		// left half is sorted
		// compare left and middle
		if nums[left] <= nums[middle] {
			// compare left, target & middle
			if nums[left] <= target && target < nums[middle] {
				right = middle - 1
			} else {
				left = middle + 1
			}
		} else {
			// right half is sorted
			if nums[middle] < target && target <= nums[right] {
				left = middle + 1
			} else {
				right = middle - 1
			}
		}
	}

	return false
}

Output

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

Output
true

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