<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 subArrayRanges(vector<int>& arr) {
        //BRUTE FORCE  TC=O(N^2)    SC=O(1)
        //Going through all SubArrays
        /*
        int mini,maxi;
        int n=arr.size();
        long long ans=0;
        for(int i=0;i<n;i++)
        {
            mini = arr[i];
            maxi = arr[i];
            for(int j=i+1;j<n;j++)
            {
                mini = min(mini,arr[j]);
                maxi = max(maxi,arr[j]);
                ans = ans + maxi - mini;
            }
        }
        return ans;
        */


        //Using (NGE, PGEE) for summation of upperbounds and (NSE, PSEE) for summation of lower bounds
        vector<int> nse = find_NSE(arr);
        vector<int> pse = find_PSE(arr);
        vector<int> nge = find_NGE(arr);
        vector<int> pge = find_PGE(arr);
        long long maxi_sum=0,mini_sum=0;
        int n=arr.size();
        for(int i=0;i<n;i++)
        {
            maxi_sum += (long long)(nge[i]-i)*(i-pge[i])*arr[i];
            mini_sum += (long long)(nse[i]-i)*(i-pse[i])*arr[i];
        }
        return maxi_sum-mini_sum;
    }
    

    //Return Indices not values
    vector<int> find_NSE(vector<int>&arr)
    {
        int n=arr.size();
        vector<int> ans(n);
        stack<int> st;
        for(int i=n-1;i>=0;i--)
        {
            while(!st.empty() && arr[i]<=arr[st.top()]) st.pop();
            ans[i] = st.empty()?n:st.top();
            st.push(i);
        }
        return ans;
    }

    //Return Indices not values
    vector<int> find_PSE(vector<int>&arr)
    {
        int n=arr.size();
        vector<int> ans(n);
        stack<int> st;
        for(int i=0;i<n;i++)
        {
            while(!st.empty() && arr[i]<arr[st.top()]) st.pop();
            ans[i] = st.empty()?-1:st.top();
            st.push(i);
        }
        return ans;
    }

    //Return Indices not values
    vector<int> find_NGE(vector<int>&arr)
    {
        int n=arr.size();
        vector<int> ans(n);
        stack<int> st;
        for(int i=n-1;i>=0;i--)
        {
            while(!st.empty() && arr[i]>=arr[st.top()]) st.pop();
            ans[i] = st.empty()?n:st.top();
            st.push(i);
        }
        return ans;
    }

    //Return Indices not values
    vector<int> find_PGE(vector<int>&arr)
    {
        int n=arr.size();
        vector<int> ans(n);
        stack<int> st;
        for(int i=0;i<n;i++)
        {
            while(!st.empty() && arr[i]>arr[st.top()]) st.pop();
            ans[i] = st.empty()?-1:st.top();
            st.push(i);
        }
        return ans;
    }
};
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