# 0033-search-in-rotated-sorted-array


Try it on <a href='https://leetcode.com/problems/0033-search-in-rotated-sorted-array'>leetcode</a>

## Description
<div class="description">
<p>There is an integer array <code>nums</code> sorted in ascending order (with <strong>distinct</strong> values).</p>

<p>Prior to being passed to your function, <code>nums</code> is <strong>possibly left rotated</strong> at an unknown index <code>k</code> (<code>1 &lt;= k &lt; nums.length</code>) such that the resulting array is <code>[nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]]</code> (<strong>0-indexed</strong>). For example, <code>[0,1,2,4,5,6,7]</code> might be left rotated by&nbsp;<code>3</code>&nbsp;indices and become <code>[4,5,6,7,0,1,2]</code>.</p>

<p>Given the array <code>nums</code> <strong>after</strong> the possible rotation and an integer <code>target</code>, return <em>the index of </em><code>target</code><em> if it is in </em><code>nums</code><em>, or </em><code>-1</code><em> if it is not in </em><code>nums</code>.</p>

<p>You must write an algorithm with <code>O(log n)</code> runtime complexity.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> nums = [4,5,6,7,0,1,2], target = 0
<strong>Output:</strong> 4
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [4,5,6,7,0,1,2], target = 3
<strong>Output:</strong> -1
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [1], target = 0
<strong>Output:</strong> -1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 5000</code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li>All values of <code>nums</code> are <strong>unique</strong>.</li>
	<li><code>nums</code> is an ascending array that is possibly rotated.</li>
	<li><code>-10<sup>4</sup> &lt;= target &lt;= 10<sup>4</sup></code></li>
</ul>

</div>

## Solution(Python)
```Python
class Solution:
    def search(self, nums: List[int], target: int) -> int:
        # 4 5 6 7 0 1 2
        #      7 > 0
        #  l r 4 2
        #  mid 7 mid + 1 0 7 > 0 return mid + 1
        #  l = mid + 1
        # 456012
        def findpivot(l,r):
            while l < r:
                mid = (l + r) // 2
                if nums[r] < nums[mid]:
                    l = mid + 1
                else:
                    r = mid
            return r
        def binarysearch(l ,r):
            while l <= r:
                mid = (l + r) // 2
                if nums[mid] == target:
                    return mid
                elif nums[mid] > target:
                    r = mid -1
                else:
                    l = mid + 1
            return -1

        n = len(nums)
        if n == 1:
            return 0 if nums[0] == target else -1
        k = findpivot(0, n - 1)
        if k == -1:
            k = n -1
        if nums[k] <= target <= nums[n-1]:
            return binarysearch(k, n-1)
        else:
            return binarysearch(0, k-1)
```