<p>You are given an integer array <code>nums</code>. The <strong>range</strong> of a subarray of <code>nums</code> is the difference between the largest and smallest element in the subarray.</p>

<p>Return <em>the <strong>sum of all</strong> subarray ranges of </em><code>nums</code><em>.</em></p>

<p>A subarray is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>

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

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0 
[2], range = 2 - 2 = 0
[3], range = 3 - 3 = 0
[1,2], range = 2 - 1 = 1
[2,3], range = 3 - 2 = 1
[1,2,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 1 + 1 + 2 = 4.</pre>

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

<pre>
<strong>Input:</strong> nums = [1,3,3]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The 6 subarrays of nums are the following:
[1], range = largest - smallest = 1 - 1 = 0
[3], range = 3 - 3 = 0
[3], range = 3 - 3 = 0
[1,3], range = 3 - 1 = 2
[3,3], range = 3 - 3 = 0
[1,3,3], range = 3 - 1 = 2
So the sum of all ranges is 0 + 0 + 0 + 2 + 0 + 2 = 4.
</pre>

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

<pre>
<strong>Input:</strong> nums = [4,-2,-3,4,1]
<strong>Output:</strong> 59
<strong>Explanation:</strong> The sum of all subarray ranges of nums is 59.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 1000</code></li>
	<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow-up:</strong> Could you find a solution with <code>O(n)</code> time complexity?</p>

class Solution {
public:
    void findNSE(vector<int>& arr, vector<int>& nse, int n) {
        stack<int>st;
        for(int i = n - 1;i >= 0;i--) {
            while(!st.empty() && arr[st.top()] >= arr[i]) {
                st.pop();
            }
            nse[i] = st.empty() ? n : st.top();
            st.push(i);
        }
    }
    void findPSE(vector<int>& arr, vector<int>& pse, int n) {
        stack<int>st;
        for(int i = 0;i < n;i++) {
            while(!st.empty() && arr[st.top()] > arr[i]) {
                st.pop();
            }
            pse[i] = st.empty() ? -1 : st.top();
            st.push(i);
        }
    }
    void findNGE(vector<int>& arr, vector<int>& nge, int n) {
        stack<int>st;
        for(int i = n - 1;i >= 0;i--) {
            while(!st.empty() && arr[st.top()] <= arr[i]) {
                st.pop();
            }
            nge[i] = st.empty() ? n : st.top();
            st.push(i);
        }
    }
    void findPGE(vector<int>& arr, vector<int>& pge, int n) {
        stack<int>st;
        for(int i = 0;i < n;i++) {
            while(!st.empty() && arr[st.top()] < arr[i]) {
                st.pop();
            }
            pge[i] = st.empty() ? -1 : st.top();
            st.push(i);
        }
    }
    long long subArrayRanges(vector<int>& arr) {
        long long sum = 0;
        int n = arr.size();
        vector<int> nse(n);
        vector<int> pse(n);
        vector<int> nge(n);
        vector<int> pge(n);
        findNSE(arr, nse,n);
        findPSE(arr, pse, n);
        findNGE(arr, nge,n);
        findPGE(arr, pge, n);
        long long minSum = 0;
        long long maxSum = 0;
        for(int i = 0; i < n;i++) {
            //We need to find how many times each element is contributed to subarray as a minimum
            int leftSmaller = i - pse[i];
            int rightSmaller = nse[i] - i;
            int leftlarger = i - pge[i];
            int rightLarger = nge[i] - i;
            minSum = (minSum + (leftSmaller * rightSmaller * 1ll * arr[i]));
            maxSum = (maxSum + (leftlarger * rightLarger * 1ll * arr[i]));
        }
        sum = abs (maxSum - minSum);
        return sum;

    }
};
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