<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 Solution {
public:
    int closestToTarget(vector<int>& arr, int target) {
        // goal : find the closet value to target.
        //        value = func(arr, l, r) = bitwise and of arr[l+1] to arr[r]
        //        => find the subarray s.t. bitwise and value is closet to target
        
        // 1. iterate all possible func(l, r) in O(n)
        //    Ex. arr = [9, 12, 3, 7, 15] with target = 5
        //    iterate array with index i
        //    i = 0, [[9]]
        //    [9 - 5] = 4
        //    i = 1, [[12], [12 & 9]]
        //    [12 - 9] = 3, ...
        //    i = 2, [[3], [3 & 12], [3 & 12 & 9]]
        //    ...
        //    i = 3, [[7], [7 & 3], [7 & 3 & 12], [7 & 3 & 12 & 9]]
        //    ...
        //    i = 4, [[15], [15 & 7], [15 & 7 & 3 & 12], [15 & 7 & 3 & 12 & 9]]
        //    ...
        //    => find the minimum value as answer.
        //    => due to bitwise and value will be decreasing if number of bitwise value 
        //       increase. So there will be at most 32 entry for 32-bit value

        unordered_set<int> cur, next;
        int ret = numeric_limits<int>::max();

        for (int& a : arr) {
            cur.insert(a);
            for (auto& val : cur) {
                int andValue = val & a;
                next.insert(andValue);
                ret = min(ret, abs(andValue - target));
            }
            cur = next;
            next.clear();
        }

        return ret;
    }
};
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