<p>You are given an integer array <code>nums</code> consisting of <code>n</code> elements, and an integer <code>k</code>.</p>

<p>Find a contiguous subarray whose <strong>length is equal to</strong> <code>k</code> that has the maximum average value and return <em>this value</em>. Any answer with a calculation error less than <code>10<sup>-5</sup></code> will be accepted.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,12,-5,-6,50,3], k = 4
<strong>Output:</strong> 12.75000
<strong>Explanation:</strong> Maximum average is (12 - 5 - 6 + 50) / 4 = 51 / 4 = 12.75
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5], k = 1
<strong>Output:</strong> 5.00000
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>1 &lt;= k &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>-10<sup>4</sup> &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    double findMaxAverage(vector<int>& nums, int k) {
        double sum =0;
        for (int i=0;i<k;i++){
            sum+=nums[i];
        }
        double res = sum;
        for (int i =k;i<nums.size();i++){
            sum+=nums[i] - nums[i-k];
            res = max(res,sum);
        }
        return res/k;
    }
};
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