<p><img alt="" src="https://assets.leetcode.com/uploads/2020/07/09/change.png" style="width: 635px; height: 312px;" /></p>

<p>Winston was given the above mysterious function <code>func</code>. He has an integer array <code>arr</code> and an integer <code>target</code> and he wants to find the values <code>l</code> and <code>r</code> that make the value <code>|func(arr, l, r) - target|</code> minimum possible.</p>

<p>Return <em>the minimum possible value</em> of <code>|func(arr, l, r) - target|</code>.</p>

<p>Notice that <code>func</code> should be called with the values <code>l</code> and <code>r</code> where <code>0 &lt;= l, r &lt; arr.length</code>.</p>

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

<pre>
<strong>Input:</strong> arr = [9,12,3,7,15], target = 5
<strong>Output:</strong> 2
<strong>Explanation:</strong> Calling func with all the pairs of [l,r] = [[0,0],[1,1],[2,2],[3,3],[4,4],[0,1],[1,2],[2,3],[3,4],[0,2],[1,3],[2,4],[0,3],[1,4],[0,4]], Winston got the following results [9,12,3,7,15,8,0,3,7,0,0,3,0,0,0]. The value closest to 5 is 7 and 3, thus the minimum difference is 2.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1000000,1000000,1000000], target = 1
<strong>Output:</strong> 999999
<strong>Explanation:</strong> Winston called the func with all possible values of [l,r] and he always got 1000000, thus the min difference is 999999.
</pre>

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

<pre>
<strong>Input:</strong> arr = [1,2,4,8,16], target = 0
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= arr[i] &lt;= 10<sup>6</sup></code></li>
	<li><code>0 &lt;= target &lt;= 10<sup>7</sup></code></li>
</ul>

class BitCounter:
    def __init__(self, bitLen: int = 32):
        self.vals = [0] * bitLen
        self.length = 0
    
    def add_num(self, val: int) -> int:
        i = 0
        self.length += 1
        while 1 << i <= val:
            if val & 1 << i:
                self.vals[i] += 1
                if self.vals[i] < self.length:
                    val ^= 1 << i
            i += 1
        return val

    def remove_num(self, val: int) -> int:
        i = 0
        self.length -= 1
        assert self.length > 0
        ret = 0
        for i in range(len(self.vals)):
            if val & 1 << i:
                self.vals[i] -= 1
            if self.vals[i] == self.length:
                ret |= 1 << i
        return ret


class Solution:
    def closestToTarget(self, arr: List[int], target: int) -> int:
        counter = BitCounter()
        p, best = 0, 10000000000
        for q in range(len(arr)):
            v = counter.add_num(arr[q])
            best = min(best, abs(v - target))
            while p < q and v < target:
                v = counter.remove_num(arr[p])
                best = min(best, abs(target - v))
                p += 1
        return best
27300