<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:
        def get_sum(val):
            return sum([min(v, val) for v in arr])

        if sum(arr) <= target:
            return max(arr)

        left = 0
        right = max(arr)

        while left + 1 < right:
            mid = (left + right) // 2
            val = get_sum(mid)
            if val >= target:
                right = mid 
            else:
                left = mid 
        lval = abs(get_sum(left) - target)
        rval = abs(get_sum(right) - target)
        print(left, right)
        print(lval, rval)
        if lval <= rval:
            return left
        else:
            return right        
17800