<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]:
        n = len(nums)
        ret = []
        def binary(l, r, target):
            while l < r:
                m = l + (r-l)//2
                if nums[m] == target:
                    return m
                if nums[m] < target:
                    l = m + 1
                else:
                    r = m
            return l
        
        nums.sort()
        print(nums)
        prefix_sum = [0]*len(nums)
        suffix_sum = [0]*len(nums)
        run_sum = 0
        for i,num in enumerate(nums):
            run_sum += num
            prefix_sum[i] = run_sum
        run_sum = 0
        for i in range(len(nums)-1, -1, -1):
            run_sum += nums[i]
            suffix_sum[i] = run_sum
        
        for q in queries:
            m = binary(0, n-1, q)
            #print(f"{q} {m} {nums[m]}")
            left = 0
            right = 0
            if m == q:
                right = suffix_sum[m+1] - (q * (n - (m + 1))) if m + 1 < n else 0
                left = (q * m) - prefix_sum[m-1] if m - 1 >= 0 else 0
            else:
                if q > nums[-1]:
                    left = (q * n) - prefix_sum[-1]
                elif q < nums[0]:
                    right = suffix_sum[0] - (q * n)
                else:
                    right = suffix_sum[m] - (q * (n  - m))
                    left = (q * m) - prefix_sum[m-1] if m - 1 >= 0 else 0
                #print(f"{left} {right}")
            #print(f"q:{q} m:{m} nums[m]:{nums[m]} l:{left} r:{right} s:{left + right}")
            ret.append(left + right)
        return ret
50000