<p>You are given two <strong>0-indexed</strong> arrays <code>nums</code> and <code>cost</code> consisting each of <code>n</code> <strong>positive</strong> integers.</p>

<p>You can do the following operation <strong>any</strong> number of times:</p>

<ul>
	<li>Increase or decrease <strong>any</strong> element of the array <code>nums</code> by <code>1</code>.</li>
</ul>

<p>The cost of doing one operation on the <code>i<sup>th</sup></code> element is <code>cost[i]</code>.</p>

<p>Return <em>the <strong>minimum</strong> total cost such that all the elements of the array </em><code>nums</code><em> become <strong>equal</strong></em>.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,5,2], cost = [2,3,1,14]
<strong>Output:</strong> 8
<strong>Explanation:</strong> We can make all the elements equal to 2 in the following way:
- Increase the 0<sup>th</sup> element one time. The cost is 2.
- Decrease the 1<sup><span style="font-size: 10.8333px;">st</span></sup> element one time. The cost is 3.
- Decrease the 2<sup>nd</sup> element three times. The cost is 1 + 1 + 1 = 3.
The total cost is 2 + 3 + 3 = 8.
It can be shown that we cannot make the array equal with a smaller cost.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,2,2,2,2], cost = [4,2,8,1,3]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All the elements are already equal, so no operations are needed.
</pre>

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

<ul>
	<li><code>n == nums.length == cost.length</code></li>
	<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i], cost[i] &lt;= 10<sup>6</sup></code></li>
	<li>Test cases are generated in a way that the output doesn&#39;t exceed&nbsp;2<sup>53</sup>-1</li>
</ul>

class Solution {
public:
    long long minCost(vector<int>& nums, vector<int>& cost) {
        long long ans=LLONG_MAX;
        vector<long long> pref;
        vector<pair<long, long>> p;
        
        for(int i=0; i<cost.size(); i++){
            p.push_back({nums[i], cost[i]});
        }
        
        sort(p.begin(), p.end());
        
        pref.push_back(0);
        for(int i=0; i<p.size(); i++){
            pref.push_back(pref[i]+p[i].second);
        }
        
        long long left=0, right=0;
        for(int i=0; i<nums.size(); i++){
           
            if(i==0){
                left=0;
                for(int j=1; j<cost.size(); j++){
                    right+=(long long)(p[j].first-p[0].first)*(long long)p[j].second;
                }
                if(ans>right+left){
                    ans=left+right;
                }
                continue;
            }
            long long elavation=p[i].first-p[i-1].first;
    
            right-=elavation*(long long)(p[i].second);
            right-=elavation*(pref[pref.size()-1]-pref[i+1]);
            
        
            left+=elavation*(pref[i]-pref[0]);
            
          
            if(ans>right+left){
                    ans=left+right;
            }
        }
        
        return ans;
    }
};
52000