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154. Find Minimum in Rotated Sorted Array II

Description

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

• [4,5,6,7,0,1,4] if it was rotated 4 times.
• [0,1,4,4,5,6,7] 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 that may contain duplicates, return the minimum element of this array.

You must decrease the overall operation steps as much as possible.

 

Example 1:

Input: nums = [1,3,5]
Output: 1

Example 2:

Input: nums = [2,2,2,0,1]
Output: 0

 

Constraints:

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• nums is sorted and rotated between 1 and n times.

 

Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

 

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• Javascript

• class Solution {
      public int findMin(int[] nums) {
          int left = 0, right = nums.length - 1;
          while (left < right) {
              int mid = (left + right) >> 1;
              if (nums[mid] > nums[right]) {
                  left = mid + 1;
              } else if (nums[mid] < nums[right]) {
                  right = mid;
              } else {
                  --right;
              }
          }
          return nums[left];
      }
  }
  

• class Solution {
  public:
      int findMin(vector<int>& nums) {
          int left = 0, right = nums.size() - 1;
          while (left < right) {
              int mid = (left + right) >> 1;
              if (nums[mid] > nums[right])
                  left = mid + 1;
              else if (nums[mid] < nums[right])
                  right = mid;
              else
                  --right;
          }
          return nums[left];
      }
  };
  

• class Solution:
      def findMin(self, nums: List[int]) -> int:
          left, right = 0, len(nums) - 1
          while left < right:
              mid = (left + right) >> 1
              if nums[mid] > nums[right]:
                  left = mid + 1
              elif nums[mid] < nums[right]:
                  right = mid
              else:
                  right -= 1
          return nums[left]
  
  

• func findMin(nums []int) int {
  	left, right := 0, len(nums)-1
  	for left < right {
  		mid := (left + right) >> 1
  		if nums[mid] > nums[right] {
  			left = mid + 1
  		} else if nums[mid] < nums[right] {
  			right = mid
  		} else {
  			right--
  		}
  	}
  	return nums[left]
  }
  

• function findMin(nums: number[]): number {
      let left = 0,
          right = nums.length - 1;
      while (left < right) {
          const mid = (left + right) >> 1;
          if (nums[mid] > nums[right]) {
              left = mid + 1;
          } else if (nums[mid] < nums[right]) {
              right = mid;
          } else {
              --right;
          }
      }
      return nums[left];
  }
  
  

• /**
   * @param {number[]} nums
   * @return {number}
   */
  var findMin = function (nums) {
      let left = 0,
          right = nums.length - 1;
      while (left < right) {
          const mid = (left + right) >> 1;
          if (nums[mid] > nums[right]) {
              left = mid + 1;
          } else if (nums[mid] < nums[right]) {
              right = mid;
          } else {
              --right;
          }
      }
      return nums[left];
  };
  
  

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