Table of contents


Given problem

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.


Using Binary Search algorithm

  1. Using Template #1

     class Solution {
         public int findMin(int[] arr) {
             int len = arr.length;
             int left = 0;
             int right = len - 1;
    
             while (left <= right) {
                 if (arr[left] <= arr[right]) {  // Case 1: sorted array
                     return arr[left];
                 }
    
                 int mid = left + (right - left) / 2;
                 int next = (mid + 1) % len;
                 int prev = (mid + len - 1) % len;
                 if (arr[mid] <= arr[next] && arr[mid] <= arr[prev]) {   // Case 2: mid index points to the minimum element
                     return arr[mid];
                 } else if (arr[mid] <= arr[right]) {    // Case 4
                     right = mid - 1;
                 } else if (arr[mid] >= arr[left]) {     // Case 3
                     left = mid + 1;
                 }
             }
    
             return -1;
         }
     }
    
  2. Using Template #2

     class Solution {
         public int findMin(int[] nums) {
             if (nums == null) {
                 return -1;
             }
    
             int left = 0;
             int right = nums.length - 1;
    
             while (left < right) {
                 int mid = left + (right - left) / 2;
    
                 if (nums[mid] > nums[right]) {
                     left = mid + 1;
                 } else {
                     right = mid;
                 }
             }
    
             return nums[left];
         }
     }
    
  3. Using Template #3

     class Solution {
         public int findMin(int[] nums) {
             int low = 0;
             int high = nums.length - 1;
                
             while (low + 1 < high) {
                 int mid = low + (high - low) / 2;
                    
                 if (nums[mid] > nums[high]) {
                     low = mid;
                 } else {
                     high = mid;
                 }
             }
                
             return nums[low] < nums[high] ? nums[low] : nums[high];
         }
     }
    

The complexity of this solution:

  • Time complexity: O(logn)
  • Space complexity: O(1)


Wrapping up

  • Understanding how to use some templates of Binary Search algorithm.


Refer:

153. Find Minimum in Rotated Sorted Array