# 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
``````

## Analysis

Binary search variation.

## Solution

Binary Search - compare mid point determine search direction - O(logn) time - (10ms, 30.57%)

``````class Solution {
public int search(int[] nums, int target) {
int left = 0, right = nums.length - 1;
while (left <= right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) {
return mid;
}
if (nums[mid] > nums[right]) {
if (target < nums[mid] && target >= nums[left]) {
right = mid - 1;
} else {
left = mid + 1;
}
} else {
if (target > nums[mid] && target <= nums[right]) {
left = mid + 1;
} else {
right = mid - 1;
}
}
}
return -1;
}
}
``````

Binary Search - Find the pivot position first

``````class Solution {
public int search(int[] nums, int target) {
if(nums.length == 0){
return -1;
}

//find the pivot
int pivot = findPivot(nums, 0, nums.length - 1);
int end = nums.length - 1;
int pivotIndex = 0;

for(int i = 0; i < nums.length; i++){
if(nums[i] == pivot){
pivotIndex = i;
}
}

if(target == pivot){
return pivotIndex;
}
else if(target >= pivot && target <= nums[end]){
return binarySearch(nums, pivotIndex, end, target);
}
else {
return binarySearch(nums, 0, pivotIndex, target);
}
}

public int findPivot(int[] nums, int start, int end){
if(start == end){
return nums[start];
}

int mid = (start + end) / 2;

if(nums[mid] > nums[mid + 1]){
return nums[mid + 1];
}
else if(nums[start] < nums[mid]){
return findPivot(nums, mid + 1, end);
}
else{
return findPivot(nums, start, mid);
}
}

public int binarySearch(int[] nums, int start, int end, int target){
if(start == end){
if(nums[start] == target){
return start;
}
else{
return -1;
}
}

int mid = (start + end) / 2;

if(nums[mid] == target){
return mid;
}
else if(nums[mid] < target){
return binarySearch(nums, mid + 1, end, target);
}
else{
return binarySearch(nums, start, mid, target);
}

}
}
``````