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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 rotated4times.[0,1,2,4,5,6,7]if it was rotated7times.
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.length1 <= n <= 5000-5000 <= nums[i] <= 5000- All the integers of
numsare unique. numsis sorted and rotated between1andntimes.
Solution
object Solution {
def findMin(nums: Array[Int]): Int = {
def findMinUtil(l: Int, r: Int): Int = {
if (l == r) {
nums(l)
} else {
val mid = (l + r) / 2
if (mid == l && nums(mid) < nums(r)) {
nums(l)
} else if (mid - 1 >= 0 && nums(mid - 1) > nums(mid)) {
nums(mid)
} else if (nums(mid) < nums(l)) {
findMinUtil(l, mid - 1)
} else if (nums(mid) > nums(r)) {
findMinUtil(mid + 1, r)
} else {
findMinUtil(l, mid - 1)
}
}
}
val l = 0
val r = nums.length - 1
findMinUtil(l, r)
}
}