<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>

// https://leetcode.com/problems/k-concatenation-maximum-sum/
class Solution {
public:
    long long kadane(vector<int> a)
    {
        long long sum=INT_MIN;
        long long max_sum=0;
        long long mod = 1e9+7;
        for(int i=0;i<a.size();i++)
        {
            if(sum<0)
                sum=a[i];
            else
                sum=sum+a[i];
            
            if(sum>max_sum)
            {
                max_sum=sum;
            }
        }

        return max_sum;
    }

    int kConcatenationMaxSum(vector<int>& arr, int k) {
        if(arr.size()==0)  return 0;
        long long mod=1e9+7;
        long long single_max = kadane(arr);
        long long total_sum = 0;
        long long prefix_sum=0,suffix_sum=0,max_prefix_sum=0,max_suffix_sum=0;
        for (int num : arr) {
            total_sum += num;
            prefix_sum += num;
            max_prefix_sum = max(max_prefix_sum, prefix_sum);
        }

        for (int i = arr.size() - 1; i >= 0; --i) {
            suffix_sum += arr[i];
            max_suffix_sum = max(max_suffix_sum, suffix_sum);
        }

        if (k == 1) {
            return single_max % mod;
        }

        // Case when k > 1
        long long result = max(single_max, max_suffix_sum + max_prefix_sum);
        if (total_sum > 0) {
            result = max(result, max_suffix_sum + max_prefix_sum + (k - 2) * total_sum);
        }

        return result % mod;
    }
};
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