<p>Given a <strong>sorted</strong> integer array <code>arr</code>, two integers <code>k</code> and <code>x</code>, return the <code>k</code> closest integers to <code>x</code> in the array. The result should also be sorted in ascending order.</p>

<p>An integer <code>a</code> is closer to <code>x</code> than an integer <code>b</code> if:</p>

<ul>
	<li><code>|a - x| &lt; |b - x|</code>, or</li>
	<li><code>|a - x| == |b - x|</code> and <code>a &lt; b</code></li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> arr = [1,2,3,4,5], k = 4, x = 3
<strong>Output:</strong> [1,2,3,4]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> arr = [1,2,3,4,5], k = 4, x = -1
<strong>Output:</strong> [1,2,3,4]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= arr.length</code></li>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>4</sup></code></li>
	<li><code>arr</code> is sorted in <strong>ascending</strong> order.</li>
	<li><code>-10<sup>4</sup> &lt;= arr[i], x &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution {
public:
    vector<int> findClosestElements(vector<int>& arr, int k, int x) 
    {
        vector<long long> diff;
        for(int i=0;i<arr.size();i++)
        {
            diff.push_back(abs(arr[i]-x));
        }
        //k so like i start at the end of my segment right
       // Start of the sliding window
    int i = 0;
    int j = k - 1;
    int left = i;
    int right = j;

    long long curr_sum = 0;
    for(int idx = i; idx <= j; idx++)
    {
        curr_sum += diff[idx];
    }

    long long min_sum = curr_sum;
    for(int idx = j + 1; idx < arr.size(); idx++)
    {
        curr_sum = curr_sum - diff[i] + diff[idx];
        i++;

        if(curr_sum < min_sum)
        {
            min_sum = curr_sum;
            left = i;
            right = idx;
        }
    }
    vector<int> ans;
    for(int idx = left; idx <= right; idx++)
    {
        ans.push_back(arr[idx]);
    }
        return ans;
    }
};
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