<p>Given an array of integers arr, find the sum of <code>min(b)</code>, where <code>b</code> ranges over every (contiguous) subarray of <code>arr</code>. Since the answer may be 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 = [3,1,2,4]
<strong>Output:</strong> 17
<strong>Explanation:</strong> 
Subarrays are [3], [1], [2], [4], [3,1], [1,2], [2,4], [3,1,2], [1,2,4], [3,1,2,4]. 
Minimums are 3, 1, 2, 4, 1, 1, 2, 1, 1, 1.
Sum is 17.
</pre>

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

<pre>
<strong>Input:</strong> arr = [11,81,94,43,3]
<strong>Output:</strong> 444
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 3 * 10<sup>4</sup></code></li>
</ul>

class Solution
{
    vector<int> prev_smaller_idx(vector<int> &arr, int n)
    {
        vector<int> ans(n,-1);
        int j;
        for (int i = 1; i < n; i++)
        {
            for(j=i-1;j >=0;j--){
                if(arr[j] < arr[i]){
                    ans[i]=j;
                    break;
                }
            }
        }
        return ans;
    }
    vector<int> next_smaller_idx(vector<int> &arr, int n)
    {
        vector<int> ans(n,n);
        int j;
        for (int i = 0; i< n-1; i++)
        {
            for(j=i+1;j < n;j++){
                if(arr[j] <= arr[i]){
                    ans[i]=j;
                    break;
                }
            }
        }
        return ans;
    }

public:
    int sumSubarrayMins(vector<int> &arr)
    {
        ios::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
        int mod = 1e9 + 7;
        int n = arr.size();
        vector<int> prev(n);
        vector<int> next(n);
        prev = prev_smaller_idx(arr, n);
        next = next_smaller_idx(arr, n);
        long long sum = 0;
        for (int i = 0; i < n; i++)
        {
            long long left = (i - prev[i]);
            long long right = (next[i] - i);
            sum = (sum + (left * right * arr[i]) % mod) % mod;
        }
        return sum;
    }
};
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