Search in Rotated Sorted Array

• Difficulty: Medium
• Primary pattern: Arrays & Strings
• Tags: Array, Binary Search
• Time taken: 18:00
• LeetCode Link

Key Idea

• Use Pointers, Left and Right
• From there, think about how to move the pointers to close the gap, where the “mid” is your target index

Solution

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        # Understanding
            # nums in ascending order, with DISTINCT values
            # might be left rotated resulting array not in ascending value
            # You need to find the index of target if it is in nums, else -1
        
        left = 0
        right = len(nums) -1
        
        while left <= right:
            mid = (left + right) // 2
            if nums[mid] == target:
                return mid
            if nums[left] <= nums[mid]:
                if nums[left] <= target < nums[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            else:
                # Right half is sorted
                if nums[mid] < target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1
        return -1

Complexity

• Time: O(log n)
• Space: O(1)