Find Minimum in Rotated Sorted Array II LeetCode Solution

Minimum Cost to Merge Stones
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Find Minimum in Rotated Sorted Array II 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.

Find Minimum in Rotated Sorted Array II Solutions

Time: O(logn)→O(n)
Space: O(1)

C++

class Solution {
 public:
  int findMin(vector<int>& nums) {
    int l = 0;
    int r = nums.size() - 1;

    while (l < r) {
      const int m = l + (r - l) / 2;
      if (nums[m] == nums[r])
        --r;
      else if (nums[m] < nums[r])
        r = m;
      else
        l = m + 1;
    }

    return nums[l];
  }
};

Java

 class Solution {
  public int findMin(int[] nums) {
    int l = 0;
    int r = nums.length - 1;

    while (l < r) {
      final int m = l + (r - l) / 2;
      if (nums[m] == nums[r])
        --r;
      else if (nums[m] < nums[r])
        r = m;
      else
        l = m + 1;
    }

    return nums[l];
  }
}

Python


class Solution:
  def findMin(self, nums: List[int]) -> int:
    l = 0
    r = len(nums) - 1

    while l < r:
      m = (l + r) // 2
      if nums[m] == nums[r]:
        r -= 1
      elif nums[m] < nums[r]:
        r = m
      else:
        l = m + 1

    return nums[l]

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