# 7. 153. Find Minimum in Rotated Sorted Array

153\. Find Minimum in Rotated Sorted ArrayMedium11K496Companies

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 rotated `4` times.
* `[0,1,2,4,5,6,7]` if it was rotated `7` times.

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:**

<pre><code><strong>Input: nums = [3,4,5,1,2]
</strong><strong>Output: 1
</strong><strong>Explanation: The original array was [1,2,3,4,5] rotated 3 times.
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: nums = [4,5,6,7,0,1,2]
</strong><strong>Output: 0
</strong><strong>Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: nums = [11,13,15,17]
</strong><strong>Output: 11
</strong><strong>Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 
</strong></code></pre>

**Constraints:**

* `n == nums.length`
* `1 <= n <= 5000`
* `-5000 <= nums[i] <= 5000`
* All the integers of `nums` are **unique**.
* `nums` is sorted and rotated between `1` and `n` times.

{% embed url="<https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/>" %}

> Time: O(n)

> Space: O(1)

```javascript
/**
 * @param {number[]} nums
 * @return {number}
 */
var findMin = function(nums) {
    let s = 0;
    let e = nums.length - 1;

    while (s < e) {
        const m = s + Math.floor((e - s) / 2);

        if (nums[m] <= nums[e]) {
            e = m;
        } else {
            s = m + 1;
        }

    }

    return nums[e];
};
```


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