package LeetCode_154 /** * 154. Find Minimum in Rotated Sorted Array II * https://leetcode.com/problems/find-minimum-in-rotated-sorted-array-ii/ * * Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand. (i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]). Find the minimum element. The array may contain duplicates. Example 1: Input: [1,3,5] Output: 1 Example 2: Input: [2,2,2,0,1] Output: 0 Note: 1. This is a follow up problem to Find Minimum in Rotated Sorted Array. 2. Would allow duplicates affect the run-time complexity? How and why? * */ class Solution { /* * solution: Divide and conquer, check left side and right side of array, * Time complexity: O(logn), because at least half of array is sorted, * Space complexity: O(logn), there are logn levels recursion, * */ fun findMin(nums: IntArray): Int { return findMin(nums, 0, nums.size - 1) } private fun findMin(nums: IntArray, left: Int, right: Int): Int { //if just one or two elements if (left == right) { return nums[left] } //is sorted, because array contains duplicate, so cannot check if equals if (nums[left] < nums[right]) { return nums[left] } //compare left and right side val mid = left + (right - left) / 2 val leftSide = findMin(nums, left, mid) val rightSide = findMin(nums, mid + 1, right) return Math.min(leftSide, rightSide) } }