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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

  • 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];
};


  • impl Solution {
    pub fn find_min(nums: Vec<i32>) -> i32 {
        let (mut l, mut r) = (0, nums.len() - 1);
        while l < r {
            let mid = (l + r) >> 1;
            if nums[mid] > nums[r] {
                l = mid + 1;
            } else if nums[mid] == nums[r] {
                r -= 1;
            } else {
                r = mid;
            }
        }
        nums[l]
    }
}

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