Sum of Mutated Array Closest to Target
<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>
Array; Binary Search; Sorting
class Solution {
    public int findBestValue(int[] arr, int target) {
        Arrays.sort(arr);
        int n = arr.length;
        int[] s = new int[n + 1];
        int mx = 0;
        for (int i = 0; i < n; ++i) {
            s[i + 1] = s[i] + arr[i];
            mx = Math.max(mx, arr[i]);
        }
        int ans = 0, diff = 1 << 30;
        for (int value = 0; value <= mx; ++value) {
            int i = search(arr, value);
            int d = Math.abs(s[i] + (n - i) * value - target);
            if (diff > d) {
                diff = d;
                ans = value;
            }
        }
        return ans;
    }

    private int search(int[] arr, int x) {
        int left = 0, right = arr.length;
        while (left < right) {
            int mid = (left + right) >> 1;
            if (arr[mid] > x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}