How do you solve Find Minimum in Rotated Sorted Array on LeetCode?

Find Minimum in Rotated Sorted Array (LeetCode #153) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log n) time complexity and O(1) space complexity. Key insight: The array is sorted but rotated, which means the minimum element is the point where the order breaks.

What is the time complexity of Find Minimum in Rotated Sorted Array?

The optimal solution for Find Minimum in Rotated Sorted Array runs in O(log n) time and uses O(1) space. This complexity is O(log n) because we are halving the search space with each iteration, similar to binary search.

What is the brute force solution for Find Minimum in Rotated Sorted Array?

The brute force approach for Find Minimum in Rotated Sorted Array is Brute Force, with O(n) time complexity and O(1) space complexity. The simplest way to find the minimum is to scan through the entire array and keep track of the smallest element. This approach is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(log n).

What algorithm or data structure is used to solve Find Minimum in Rotated Sorted Array?

Find Minimum in Rotated Sorted Array uses the following concepts: Array, Binary Search. The recommended patterns to study are: Binary Search, Array.

What are the interview tips for Find Minimum in Rotated Sorted Array?

Always clarify the constraints and properties of the input data before jumping into coding. Think about how you can reduce the search space with each decision you make. Practice explaining your thought process clearly, as communication is key during interviews.

What are common mistakes when solving Find Minimum in Rotated Sorted Array?

Not recognizing the properties of the rotated array and attempting a linear search instead of binary search. Confusing the conditions for moving left and right pointers in the binary search.

#153

Find Minimum in Rotated Sorted Array

Medium

ArrayBinary SearchBinary SearchArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(log n)
Space	O(1)	O(1)

💡

Intuition

Time O(log n)Space O(1)

By leveraging the properties of a rotated sorted array, we can use a binary search approach to find the minimum element efficiently in O(log n) time. We can determine which half of the array is sorted and decide which half to search next.

⚙️

Algorithm

5 steps

1Step 1: Initialize two pointers, left at 0 and right at the last index of the array.
2Step 2: While left is less than right, calculate the mid index.
3Step 3: If nums[mid] > nums[right], the minimum must be in the right half. Move left to mid + 1.
4Step 4: If nums[mid] <= nums[right], the minimum is in the left half or could be mid. Move right to mid.
5Step 5: When left equals right, return nums[left] as the minimum.

solution.py14 lines

1# Full working Python code
2
3def findMin(nums):
4    left, right = 0, len(nums) - 1
5    while left < right:
6        mid = (left + right) // 2
7        if nums[mid] > nums[right]:
8            left = mid + 1
9        else:
10            right = mid
11    return nums[left]
12
13# Example usage
14print(findMin([3,4,5,1,2]))  # Output: 1

ℹ

Complexity note: This complexity is O(log n) because we are halving the search space with each iteration, similar to binary search.

1The array is sorted but rotated, which means the minimum element is the point where the order breaks.
2Using binary search allows us to efficiently narrow down the search space.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.