<p>Given an integer array <code>arr</code> and a target value <code>target</code>, return the integer <code>value</code> such that when we change all the integers larger than <code>value</code> in the given array to be equal to <code>value</code>, the sum of the array gets as close as possible (in absolute difference) to <code>target</code>.</p>

<p>In case of a tie, return the minimum such integer.</p>

<p>Notice that the answer is not neccesarilly a number from <code>arr</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [4,9,3], target = 10
<strong>Output:</strong> 3
<strong>Explanation:</strong> When using 3 arr converts to [3, 3, 3] which sums 9 and that&#39;s the optimal answer.
</pre>

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

<pre>
<strong>Input:</strong> arr = [2,3,5], target = 10
<strong>Output:</strong> 5
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [60864,25176,27249,21296,20204], target = 56803
<strong>Output:</strong> 11361
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= arr[i], target &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def findBestValue(self, arr: List[int], target: int) -> int:
        
        arr.sort(reverse=True)
        cur = sum(arr)
        if cur<=target:
            return arr[0]
        if len(arr)*arr[-1]>target:
            k,r = divmod(target,len(arr))
            return k if 2*r<=len(arr) else k+1
        
        pre = arr[0]
        i = 0
        while cur>target:
            val = arr[i]
            cur -= i*(pre-val)
            pre = val
            i += 1
        k,r = divmod(target-cur,i-1)
        return k+pre if 2*r<=i-1 else k+pre+1
17600