<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 solve2(vector<int>& arr) {
        vector<long long>v(arr.size(),0);
        stack<int>st;
        long long ans=0;
        int n=arr.size();
        v[n-1]=n;
        st.push(n-1);
        for(int i=n-2;i>=0;i--){
            while(!st.empty() && arr[st.top()]>arr[i])st.pop();
            if(st.empty())v[i]=n;
            else v[i]=st.top();
            st.push(i);
            
        }
        vector<long long>v2(arr.size(),0);
        while(!st.empty())st.pop();
        v2[0]=-1;
        st.push(0);
        for(int i=1;i<n;i++){
            while(!st.empty() && arr[st.top()]>=arr[i])st.pop();
            if(st.empty())v2[i]=-1;
            else v2[i]=st.top();st.push(i);
            
        }
        for(int i=0;i<n;i++){
            ans+=((long long)(i-v2[i])*(v[i]-i)*arr[i]);
        }
        return ans;



        
    }
    long long solve1(vector<int>& arr) {
        vector<long long>v(arr.size(),0);
        stack<int>st;
        long long ans=0;
        int n=arr.size();
        v[n-1]=n;
        st.push(n-1);
        for(int i=n-2;i>=0;i--){
            while(!st.empty() && arr[st.top()]<arr[i])st.pop();
            if(st.empty())v[i]=n;
            else v[i]=st.top();
            st.push(i);
            
        }
        vector<long long>v2(arr.size(),0);
        while(!st.empty())st.pop();
        v2[0]=-1;
        st.push(0);
        for(int i=1;i<n;i++){
            while(!st.empty() && arr[st.top()]<=arr[i])st.pop();
            if(st.empty())v2[i]=-1;
            else v2[i]=st.top();st.push(i);
            
        }
        for(int i=0;i<n;i++){
            ans+=((long long)(i-v2[i])*(v[i]-i)*arr[i]);
        }
        return ans;



        
    }
    long long subArrayRanges(vector<int>& nums) {
        return (long long)solve1(nums)-(long long)solve2(nums);
        
    }
};
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