<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>

def binary_search(target, nums_sort):
    left, right = 0, len(nums_sort) - 1
    while left <= right:
        middle = left + (right - left) // 2
        if nums_sort[middle] < target:
            left = middle + 1
        elif nums_sort[middle] > target:
            right = middle - 1
        else: 
            return middle
    return left

def fillPrefixSum(arr, n, prefixSum):
    # prefixSum[0] = arr[0]
    for j in range(1,n+1):
        prefixSum[j] = prefixSum[j-1] + arr[j-1]
class Solution:
    def minOperations(self, nums: List[int], queries: List[int]) -> List[int]:
        nums.sort()
        n = len(nums)
        prefixSum = [0 for ii in range(n+1)]
        fillPrefixSum(nums, n, prefixSum)
        ans = []
        for i in range(len(queries)):
            target = queries[i]
            target_idx = binary_search(target, nums)
            # print(target_idx, prefixSum)
            target_sum_pre = target * target_idx
            target_sum_post = target * (n - target_idx)
            ans_i_pre = target_sum_pre - prefixSum[target_idx] 
            ans_i_post = prefixSum[-1] - prefixSum[target_idx] - target_sum_post
            ans_i = ans_i_pre + ans_i_post
            ans.append(ans_i)
        return ans
            
981