<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>

#define ll long long
class Solution {
public:
    long long minCost(vector<int>& nums, vector<int>& cost) {
        vector<ll> pre(nums.size(), 0), suf(nums.size(), 0);
        pre[0] = 0; suf[nums.size() - 1] = 0;
        vector<vector<ll>> v;
        for(ll i = 0;i<nums.size();i++) v.push_back({nums[i], cost[i]});
        sort(v.begin(),v.end());
        ll sum = v[0][1];
        for(ll i = 1;i<nums.size();i++){
            pre[i] = pre[i-1] + (v[i][0] - v[i-1][0]) * sum;
            // cout << pre[i] << " ";
            sum += v[i][1];
        }
        // cout << endl;
        sum = v[nums.size() - 1][1];
        for(ll i = nums.size() -2;i>=0;i--){
            suf[i] = suf[i+1] + (v[i+1][0] - v[i][0]) * sum;
            // cout << suf[i] << " ";
            sum += v[i][1];
        }
        // cout << endl;
        ll ans = 1e18;
        for(ll i = 0;i<nums.size();i++){
            ans = min(ans, pre[i] + suf[i]);
        }
        return ans;
    }
};


// 1 2 3 5 
// 2 14 3 1 
// 0 2 18 56
// 24 6 2 0 
// => dp[3] = 8 
// => dp[2] => 1 * 1 + 3 * 4 => 12 + 1 => 13 
//     dp[2] => 1(1 + 4) + dp[3] => 5 + 8 => 13
    

// dp[i] = dp[i+1] + (a[i+1] - a[i]) (cost[i+1] + ...... + cost[n-1]) 
61200