<p>You are given an array <code>nums</code> consisting of positive integers.</p>

<p>You are also given an integer array <code>queries</code> of size <code>m</code>. For the <code>i<sup>th</sup></code> query, you want to make all of the elements of <code>nums</code> equal to<code> queries[i]</code>. You can perform the following operation on the array <strong>any</strong> number of times:</p>

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

<p>Return <em>an array </em><code>answer</code><em> of size </em><code>m</code><em> where </em><code>answer[i]</code><em> is the <strong>minimum</strong> number of operations to make all elements of </em><code>nums</code><em> equal to </em><code>queries[i]</code>.</p>

<p><strong>Note</strong> that after each query the array is reset to its original state.</p>

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

<pre>
<strong>Input:</strong> nums = [3,1,6,8], queries = [1,5]
<strong>Output:</strong> [14,10]
<strong>Explanation:</strong> For the first query we can do the following operations:
- Decrease nums[0] 2 times, so that nums = [1,1,6,8].
- Decrease nums[2] 5 times, so that nums = [1,1,1,8].
- Decrease nums[3] 7 times, so that nums = [1,1,1,1].
So the total number of operations for the first query is 2 + 5 + 7 = 14.
For the second query we can do the following operations:
- Increase nums[0] 2 times, so that nums = [5,1,6,8].
- Increase nums[1] 4 times, so that nums = [5,5,6,8].
- Decrease nums[2] 1 time, so that nums = [5,5,5,8].
- Decrease nums[3] 3 times, so that nums = [5,5,5,5].
So the total number of operations for the second query is 2 + 4 + 1 + 3 = 10.
</pre>

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

<pre>
<strong>Input:</strong> nums = [2,9,6,3], queries = [10]
<strong>Output:</strong> [20]
<strong>Explanation:</strong> We can increase each value in the array to 10. The total number of operations will be 8 + 1 + 4 + 7 = 20.
</pre>

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

<ul>
	<li><code>n == nums.length</code></li>
	<li><code>m == queries.length</code></li>
	<li><code>1 &lt;= n, m &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i], queries[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:  
        nums.sort()
        res = []
        prefix_sum = [0]
        n = len(nums)
        for num in nums:
            prefix_sum.append(prefix_sum[-1]+num)

        for query in queries:
            pos = self.binary_search(query, nums)
            # print(pos)
            left = query * (pos) - prefix_sum[pos]
            right = prefix_sum[n] - prefix_sum[pos] - query * (n-pos)
            # print(f"left is {left}, right is {right}")
            res.append(left+right)
        return res

    def binary_search(self, query, nums):
        l, r = 0, len(nums) - 1
       
        while r >= l:
            mid = (l + r) // 2
            if nums[mid] > query:
                r = mid - 1
            elif nums[mid] < query:
                l = mid + 1
            else:
                return mid + 1
    
        return l
# from bisect import bisect as t
# class Solution:
#     def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:
#         n = len(nums)
#         nums.sort()
#         s = list(accumulate(nums, initial=0))
#         print(s)
#         b=s[n]
#         a=[]
#         for q in queries:
#             j=t(nums, q)
#             print(j)
#             x=b-s[j]-s[j]-q*(n-j-j)
#             a.append(x)
#         return a
900