LeetCode in Kotlin

154. Find Minimum in Rotated Sorted Array II

Hard

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?

Solution

class Solution {
    fun findMin(nums: IntArray): Int {
        return if (nums.isEmpty()) {
            0
        } else find(0, nums.size - 1, nums)
    }

    private fun find(left: Int, right: Int, nums: IntArray): Int {
        if (left + 1 >= right) {
            return Math.min(nums[left], nums[right])
        }
        val mid = left + (right - left) / 2
        if (nums[left] == nums[right] && nums[left] == nums[mid]) {
            return Math.min(find(left, mid, nums), find(mid, right, nums))
        }
        return if (nums[left] >= nums[right]) {
            if (nums[mid] >= nums[left]) {
                find(mid, right, nums)
            } else {
                find(left, mid, nums)
            }
        } else {
            find(left, mid, nums)
        }
    }
}

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