<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 {
public:
   vector<int> nse(vector<int>& arr) {
    vector<int> ans(arr.size(), arr.size());  // Initialize to arr.size() if no smaller element on the right
    stack<int> st;
    
    for (int i = 0; i < arr.size(); i++) {
        // Look for strictly smaller elements on the right
        while (!st.empty() && arr[i] < arr[st.top()]) {
            ans[st.top()] = i;
            st.pop();
        }
        st.push(i);
    }
    
    return ans;
}

vector<int> pse(vector<int>& arr) {
    vector<int> ans(arr.size(), -1);  // Initialize to -1 if no smaller element on the left
    stack<int> st;
    
    for (int i = arr.size() - 1; i >= 0; i--) {
        // Look for strictly smaller elements on the left
        while (!st.empty() && arr[i] <= arr[st.top()]) {
            ans[st.top()] = i;
            st.pop();
        }
        st.push(i);
    }
    
    return ans;
}

int sumSubarrayMins(vector<int>& arr) {
    vector<int> prevSmaller = pse(arr);
    vector<int> nextSmaller = nse(arr);
    
    long long total = 0;
    int MOD = 1e9 + 7;

    // Calculate contribution for each element
    for (int i = 0; i < arr.size(); i++) {
        int lc = i - prevSmaller[i];   // How far left it can extend
        int rc = nextSmaller[i] - i;   // How far right it can extend

        total = (total + (long long)arr[i] * lc * rc) % MOD;  // Add contribution of arr[i]
    }
    
    return total;
}

};
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