<p>Given an integer array <code>arr</code> and an integer <code>k</code>, modify the array by repeating it <code>k</code> times.</p>

<p>For example, if <code>arr = [1, 2]</code> and <code>k = 3 </code>then the modified array will be <code>[1, 2, 1, 2, 1, 2]</code>.</p>

<p>Return the maximum sub-array sum in the modified array. Note that the length of the sub-array can be <code>0</code> and its sum in that case is <code>0</code>.</p>

<p>As the answer can be very large, return the answer <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [1,2], k = 3
<strong>Output:</strong> 9
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,-2,1], k = 5
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [-1,-2], k = 7
<strong>Output:</strong> 0
</pre>

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

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

class Solution {
public:
    int kConcatenationMaxSum(vector<int>& arr, int k) 
    {
        long int best =INT_MIN , arrsum =0 , maxsum=0;

        for(int i =0 ; i<arr.size();i++)
        {
            arrsum += arr[i];
        }

        if (k ==1)
        {
            for (int i =0 ; i<arr.size();i++)
            {
                maxsum += arr[i];
                if (maxsum >best)
                    best = maxsum ;
                if (maxsum <0)
                    maxsum =0 ;
            }
            if (best>0)
                return best ;
            else 
                return 0 ;
        }    
        else if (arrsum <=0)
        {
            int B[2*arr.size()] ;
            int  j= 0;
            for (int i =0 ; i<2*arr.size();i++)
            {
                if (j== arr.size())
                    j=0;
                B[i] = arr[j];
                j++;
            }
            for (int i =0 ; i<2*arr.size();i++)
            {
                maxsum += B[i];
                if (maxsum >best)
                    best = maxsum ;
                if (maxsum <0)
                    maxsum =0 ;
            }
            if (best>0)
                return best ;
            else 
                return 0 ;
        }

        else
        {
            int B[2*arr.size()] ;
            int  j= 0;
            for (int i =0 ; i<2*arr.size();i++)
            {
                if (j==arr.size())
                    j=0;
                B[i] = arr[j];
                j++;
            }
            for (int i =0 ; i<2*arr.size();i++)
            {
                maxsum += B[i];
                if (maxsum > best)
                    best = maxsum ;
                if (maxsum <0)
                    maxsum =0 ;
            }
            return (best + (k-2)*arrsum)%(static_cast<int>(pow(10,9)+7));
        }
    }
};
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