<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:
  long long subarraymaxi(vector<int>& arr)
    {
        int n=arr.size();
        long long pge[n];
        long long nge[n];
        stack<long long> st;
        for(int i=0;i<n;i++)
        {
            while(!st.empty()&&arr[i]>arr[st.top()])
            {
                st.pop();
            }

            if(st.empty())
            pge[i]=-1;
            else
             pge[i]=st.top();

            st.push(i);
        }

        stack<long long>st1;
        for(int i=n-1;i>=0;i--)
        {
            while(!st1.empty()&&arr[i]>=arr[st1.top()])
            {
                st1.pop();
            }

             if(st1.empty())
              nge[i]=n;
            else
             nge[i]=st1.top();

            st1.push(i);
        }

        long long ans=0;
        for(int i=0;i<n;i++)
        {
            long long l=i-pge[i];
            long long r=nge[i]-i;

           ans+=(l*r*arr[i]);
        }

        return ans;
    }

    long long subarraymini(vector<int>& arr)
    {
        int n=arr.size();
        long long pse[n];
        long long nse[n];
        stack<long long> st;
        for(int i=0;i<n;i++)
        {
            while(!st.empty()&&arr[i]<arr[st.top()])
            {
                st.pop();
            }

            if(st.empty())
            pse[i]=-1;
            else
             pse[i]=st.top();

            st.push(i);
        }

        stack<long long>st1;
        for(int i=n-1;i>=0;i--)
        {
            while(!st1.empty()&&arr[i]<=arr[st1.top()])
            {
                st1.pop();
            }

             if(st1.empty())
              nse[i]=n;
            else
             nse[i]=st1.top();

            st1.push(i);
        }

        long long ans=0;
        for(int i=0;i<n;i++)
        {
            long long l=i-pse[i];
            long long r=nse[i]-i;

           ans+=(l*r*arr[i]);
        }

        return ans;
    }

    long long subArrayRanges(vector<int>& nums) {
        
        return (subarraymaxi(nums)-subarraymini(nums));

    }
};
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