33. Search in Rotated Sorted Array

Question

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= 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, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

Example 1

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Solution 1: Hash

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        arrayHash = {}
        for i, v in enumerate(nums):
            arrayHash[v] = i
        
        return arrayHash[target] if target in arrayHash else -1

Time Complexity: O(N)
Space Complexity: O(N)

Solution 2: Binary Search (Middle & Right)

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        lft = 0
        rht = len(nums) - 1
        while lft <= rht:
            mid = (lft + rht) // 2
            if nums[mid] == target:
                return mid
            
            ### The right side is ordered
            # 5, 6, 7, 0, 1, 2, 4
            # 6, 7, 0, 1, 2, 4, 5
            # 7, 0, 1, 2, 4, 5, 6
            # 0, 1, 2, 4, 5, 6, 7
            if nums[mid] < nums[rht]:
                if target > nums[mid] and target <= nums[rht]:
                    lft = mid + 1
                else:
                    rht = mid - 1
            ### The left side is ordered
            # 4, 5, 6, 7, 0, 1, 2
            # 1, 2, 4, 5, 6, 7, 0
            # 2, 4, 5, 6, 7, 0, 1
            else:
                if target < nums[mid] and nums[lft] <= target:
                    rht = mid - 1
                else:
                    lft = mid + 1
        return -1

Time Complexity: O(logN)
Space Complexity: O(1)

Solution 3: Binary Search (Middle & Left)

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        lft = 0
        rht = len(nums) - 1
        while lft <= rht:
            mid = (lft + rht) // 2
            if nums[mid] == target:
                return mid
            
            ### The left side is ordered
            # 4, 5, 6, 7, 0, 1, 2
            # 1, 2, 4, 5, 6, 7, 0
            # 2, 4, 5, 6, 7, 0, 1
            if nums[mid] >= nums[lft]: # NOTE: for the edge case [3, 1], we have to use ">="
                if target < nums[mid] and nums[lft] <= target:
                    rht = mid - 1
                else:
                    lft = mid + 1
           
            ### The right side is ordered
            # 5, 6, 7, 0, 1, 2, 4
            # 6, 7, 0, 1, 2, 4, 5
            # 7, 0, 1, 2, 4, 5, 6
            # 0, 1, 2, 4, 5, 6, 7
            else:
                if nums[mid] <= target and target <= nums[rht]:
                    lft = mid + 1
                else:
                    rht = mid - 1
        return -1

Time Complexity: O(logN)
Space Complexity: O(1)

Reference

https://www.cnblogs.com/grandyang/p/4325648.html

Follow-up: Pour Water Between Buckets to Make Water Levels Equal

TODO

Time Complexity: O()
Space Complexity: O()