<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]:

        # Naive Approach

        # ans = []

        # for i in queries:
        #     temp = nums            
        #     count = 0
        #     for j in range(len(temp)):
        #         count += abs(i - temp[j])
            
        #     ans.append(count)
        
        # return ans

        # Optimized Approach

        nums.sort()

        prefix_sum = [0]*len(nums)
        prefix_sum[0] = nums[0]

        for i in range(1, len(prefix_sum)):
            prefix_sum[i] += prefix_sum[i-1] + nums[i]

        def binary_search(arr, target):
            low = 0
            high = len(arr) - 1
            res = -1
            
            while low <= high:
                mid = (low + high) // 2

                if arr[mid] > target:
                    res = mid
                    high = mid - 1

                else:
                    low = mid + 1

            return res
        
        ans = []

        for i in queries:
            count = 0
            ind_elem = binary_search(nums, i)

            if ind_elem == -1:
                ind_elem = len(nums)
                less_than_sum = prefix_sum[ind_elem - 1]
                count += i * (ind_elem) - less_than_sum
                ans.append(count)

            elif ind_elem == 0:
                more_than_sum = prefix_sum[-1]
                count += more_than_sum - i*(len(nums)-ind_elem)
                ans.append(count)


            else:
                less_than_sum = prefix_sum[ind_elem-1]
                count += i * (ind_elem) - less_than_sum
                more_than_sum = prefix_sum[-1] - prefix_sum[ind_elem-1]
                count += more_than_sum - i*(len(nums)-ind_elem)
                ans.append(count)


        return ans
42100