<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:
    def minCost(self, nums: List[int], cost: List[int]) -> int:
        # zipped = zip(nums,cost)
        # ordered = sorted(zipped)
        # forward_cost =  list(accumulate([c for _,c in ordered]))
        # backward_cost = list(accumulate(reversed([c for _,c in ordered])))
        # backward_cost.reverse()
        # prefix = [0]
        # total =  0
        # for i,n in enumerate(ordered[1:]):
        #     i+=1
        #     total += (n[0]-ordered[i-1][0])*forward_cost[i-1]
        #     prefix.append(total)
        # total = 0
        # for i in range(len(nums)-2,-1,-1):
        #     total += (ordered[i+1][0]- ordered[i][0])*backward_cost[i+1]
        #     prefix[i]+=total
        # return min(prefix)


        num_and_cost = sorted([num, c] for num, c in zip(nums, cost))
        n = len(cost)
        
        prefix_cost = [0] * n
        prefix_cost[0] = num_and_cost[0][1]
        for i in range(1, n): 
            prefix_cost[i] = num_and_cost[i][1] + prefix_cost[i - 1]

        total_cost = 0
        for i in range(1, n): 
            total_cost += num_and_cost[i][1] * (num_and_cost[i][0] - num_and_cost[0][0])
        answer = total_cost

        for i in range(1, n):
            gap = num_and_cost[i][0] - num_and_cost[i - 1][0]
            total_cost += prefix_cost[i - 1] * gap
            total_cost -= gap * (prefix_cost[n - 1] - prefix_cost[i - 1])
            answer = min(answer, total_cost)
            
        return answer

        arr = sorted(zip(nums, cost))
        total, cnt = sum(cost), 0
        for num, c in arr:
            cnt += c
            if cnt > total // 2:
                target = num
                break
        print(target)
        return sum(c * abs(num - target) for num, c in arr)
  


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