<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:
vector<int>findnse(vector<int>&arr){
    int n=arr.size();
    vector<int>nse(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);
    }
    return nse;
}
vector<int>findpse(vector<int>&arr){
    int n=arr.size();
    vector<int>pse(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);
    }
    return pse;
}
long long sum_min(vector<int>&arr){
    long long total=0;
    int n=arr.size();
    vector<int>pse(n);
    vector<int>nse(n);
    nse=findnse(arr);
    pse=findpse(arr);
    for(int i=0;i<n;i++){
    
    long long  left=i-pse[i];
    long long right=nse[i]-i;
    total=(total+(right*left*arr[i]));
    }
    return total;
}
vector<int>findnge(vector<int>&arr){
    int n=arr.size();
    vector<int>nge(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);
    }
    return nge;
}
vector<int>findpge(vector<int>&arr){
    int n=arr.size();
    vector<int>pge(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);
    }
    return pge;
}
long long sum_max(vector<int>&arr){
    long long total=0;
    int n=arr.size();
    vector<int>pge(n);
    vector<int>nge(n);
    nge=findnge(arr);
    pge=findpge(arr);
    for(int i=0;i<n;i++){
    
    long long  left=i-pge[i];
    long long right=nge[i]-i;
    total=(total+(right*left*arr[i]));
    }
    return total;
}
    long long subArrayRanges(vector<int>& nums) {
        return sum_max(nums)-sum_min(nums);
    }
};
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