<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:
        # medium: https://leetcode.com/problems/minimum-moves-to-equal-array-elements-ii/
        # Binary search - convex function
        # convex
        # N^2?? then binary
        # O(NlogK) k: the difference between max and min in nums, O(1)

        # Get the cost of making every element equals base.
        def get_cost(base):
            return sum(abs(base - num) * c for num, c in zip(nums, cost))
        
        # Initialize the left and the right boundary of the binary search.
        left, right = min(nums), max(nums) # not idx, value############ as we can choose 4, that is not in the array
        #res = get_cost(nums[0])
        res = 0
        
        # As shown in the previous picture, if F(mid) > F(mid + 1), then the minimum
        # is to the right of mid, otherwise, the minimum is to the left of mid.
        while left <= right:
            mid = (left + right) // 2
            cost_1 = get_cost(mid)
            cost_2 = get_cost(mid + 1)
            #res = min(cost_1, cost_2)
            
            if cost_1 <= cost_2:
                right = mid - 1
            else:
                left = mid + 1
        #return res
        return get_cost(left)
31700