Introduction Link to heading
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]).
You are given a target value to search. If found in the array return its index, otherwise return -1.
You may assume no duplicate exists in the array.
Your algorithm’s runtime complexity must be in the order of O(log n).
Example 1:
Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
Example 2:
Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
Solution Link to heading
Golang:
func search(nums []int, target int) int {
lo, hi := 0, len(nums)
for lo < hi {
mi := lo + (hi - lo) / 2
if nums[mi] == target {
return mi
} else if nums[lo] < nums[mi] {
if nums[lo] <= target && target < nums[mi] {
return binarySearch(nums[lo:mi], lo, target)
} else {
lo = mi + 1
}
} else {
if nums[mi] <= target && target <= nums[hi-1] {
return binarySearch(nums[mi+1:hi], mi+1, target)
} else {
hi = mi
}
}
}
return -1
}
func binarySearch(nums []int, offset int, target int) int {
lo, hi := 0, len(nums)
for lo < hi {
mi := lo + (hi - lo) / 2
if nums[mi] == target {
return mi + offset
} else if nums[mi] < target {
lo = mi + 1
} else {
hi = mi
}
}
return -1
}
Explanation Link to heading
This problem could be transformed to binary-search sub-problem or reduced to the same search problem, total complexity is O(log(n)).
In each iteration we calculate mi middle value between lo low and hi high in interval [lo, hi). By splitting array in to two sub-arrays we end-up with two situations: one sub-array is sorted, another one is the half-problem of existing problem. We need to check what part of sub-array is sorted by comparing ’nums[lo] and nums[mi]’, for sorted left part they will complain with condition ’nums[lo] < nums[mi]’. If target value is between nums[lo] and nums[mi] then this left sub-array is a binary search sub-problem, overwise we select another sub-array as sub-problem of original search, and so on.