LeetCode-in-Swift

153. Find Minimum in Rotated Sorted Array

Medium

Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

• [4,5,6,7,0,1,2] if it was rotated 4 times.
• [0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

Example 1:

Input: nums = [3,4,5,1,2]

Output: 1

Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

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

Output: 0

Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]

Output: 11

Explanation: The original array was [11,13,15,17] and it was rotated 4 times.

Constraints:

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• All the integers of nums are unique.
• nums is sorted and rotated between 1 and n times.

To solve the problem of finding the minimum element in a rotated sorted array of unique elements, we can leverage the binary search algorithm due to its efficiency of ( O(log n) ) time complexity.

Steps to Solve the Problem

1. Initialize Pointers:
   • Use two pointers, left and right, initialized to the start and end indices of the array respectively.
2. Binary Search:
   • Perform a binary search to find the minimum element:
     • Calculate the middle index mid as (left + right) / 2.
     • Compare the element at mid with the elements at left and right.
     • Adjust the search range based on the comparisons:
       • If nums[mid] < nums[right], the minimum element must be in the left half including mid, so move right to mid.
       • Otherwise, the minimum element must be in the right half excluding mid, so move left to mid + 1.
3. Terminate and Return:
   • When left equals right, it will point to the minimum element.
   • Return nums[left] or nums[right], as they will both point to the minimum element.

Swift Implementation

Here’s the Swift implementation of the Solution class using the described approach:

class Solution {
    func findMin(_ nums: [Int]) -> Int {
        var left = 0
        var right = nums.count - 1
        
        while left < right {
            let mid = left + (right - left) / 2
            
            // Check if the middle element is less than the rightmost element
            if nums[mid] < nums[right] {
                right = mid // Minimum must be in the left half including mid
            } else {
                left = mid + 1 // Minimum must be in the right half excluding mid
            }
        }
        
        // When left == right, it points to the minimum element
        return nums[left]
    }
}

Explanation of the Swift Code

• Initialization: Start with left at the beginning of the array (0) and right at the end of the array (nums.count - 1).

• Binary Search Loop:
  • Compute mid as the midpoint between left and right.
  • Compare nums[mid] with nums[right]:
    • If nums[mid] < nums[right], this means the minimum element is in the left half (including mid). So, update right to mid.
    • Otherwise, the minimum element is in the right half (excluding mid). So, update left to mid + 1.
• Termination:
  • When left equals right, it indicates that left (or right) points to the minimum element in the array.
• Return:
  • Return nums[left] or nums[right], which represents the minimum element found after the binary search completes.

This implementation efficiently finds the minimum element in a rotated sorted array of unique elements in ( O(log n) ) time complexity, meeting the problem’s requirements.

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