LeetCode in Kotlin

153. Find Minimum in Rotated Sorted Array

Medium

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.

Solution

class Solution {
    private fun findMinUtil(nums: IntArray, l: Int, r: Int): Int {
        if (l == r) {
            return nums[l]
        }
        val mid = (l + r) / 2
        if (mid == l && nums[mid] < nums[r]) {
            return nums[l]
        }
        if (mid - 1 >= 0 && nums[mid - 1] > nums[mid]) {
            return nums[mid]
        }
        if (nums[mid] < nums[l]) {
            return findMinUtil(nums, l, mid - 1)
        } else if (nums[mid] > nums[r]) {
            return findMinUtil(nums, mid + 1, r)
        }
        return findMinUtil(nums, l, mid - 1)
    }

    fun findMin(nums: IntArray): Int {
        val l = 0
        val r = nums.size - 1
        return findMinUtil(nums, l, r)
    }
}

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