Comment on page

# search-in-rotated-sorted-array

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

This problem asked to implement in

`O(logn)`

. Binary search should come into your mind when you see `O(logn)`

, sorted and search keywords.- 1.Intuitive solution is simple, iterate through array, find target in nums return pos, not, return -1,
`O(n)`

- 2.Binary search, given array is rotated sorted array, meaning there are 2 sorted array. when do binary search, need to carefully comparetarget and nums[mid] and nums[lo] to chose whether go right (higher) or left (lower).
- define
`lo, hi, and mid = (lo + hi)/2`

. - if
`target == nums[mid]`

found, return mid and exit. - if
`nums[mid] >= nums[lo]`

meaning mid is within sorted parts, now compare target,- if
`target >= nums[lo] && target <= nums[md]`

, target is located in`[lo, mid]`

sorted part, search left,`hi = mid - 1`

- else target located in
`[mid, hi]`

, search right,`lo = mid + 1`

- else check compare target with hi and mid value
- if
`target >= nums[mid] && target <= nums[hi]`

, target is located in`[mid, hi]`

part, search right,`lo = mid + 1`

else target located in`[lo, mid]`

part, search left,`hi = mid - 1`

For example:

Search in Rotated Sorted Array

**Time Complexity:**

`O(log(N))`

**Space Complexity:**

`O(1)`

- N - the length of array nums

class Solution {

public int search(int[] nums, int target) {

if (nums == null || nums.length == 0) {

return -1;

}

int lo = 0;

int hi = nums.length - 1;

while (lo < hi) {

int mid = lo + (hi - lo) / 2;

// found, return mid, exit

if (nums[mid] == target) {

return mid;

}

// mid is within sorted parts

if (nums[lo] <= nums[mid]) {

// target located at [lo, mid] part, search left

if (target >= nums[lo] && target <= nums[mid]) {

hi = mid - 1;

} else {

lo = mid + 1;

}

} else {

// target located at [mid, hi] part, search right

if (target >= nums[mid] && target <= nums[hi]) {

lo = mid + 1;

} else {

hi = mid - 1;

}

}

}

return nums[lo] == target ? lo : -1;

}

}

**Naive solution - O(n)**

class Solution {

public int search(int[] nums, int target) {

if (nums == null || nums.length == 0) return -1;

for (int i = 0; i < nums.length; i++) {

if (nums[i] == target) return i;

}

return -1;

}

}

Last modified 3yr ago