<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) {
    int len = (int)arr.size();
    vector<vector<int> > tab(30);
    for(int i = 0; i < len; i++)
        for(int j = 0; j < 30; j++) if((arr[i] >> j) & 1)
            tab[j].push_back(i);
    
    vector<int> dp(len);
    int ans = abs(arr[0] - target);
    for(int i = len - 1; i >= 0; i--) //In round i, dp[i], dp[i+1]...dp[N-1] is our new sequence of arr[i], arr[i]&arr[i+1],...,arr[i]&...&arr[N-1]
    {
        for(int j = 0; j < 30; j++) if(!((arr[i] >> j) & 1))
            while(!tab[j].empty() && tab[j].back() > i) //Find elements whose j'th bit need to be set to zero
            {
                dp[tab[j].back()] -= 1 << j;
                tab[j].pop_back(); //After set this element's j'th bit to zero, we need not consider this bit again
            }
        dp[i] = arr[i];
		
		//Binary search
        int l = i, r = len;
        while(l < r)
        {
            int m = l + (r - l) / 2;
            if(dp[m] > target) l = m + 1;
            else r = m;
        }
        if(l == len) ans = min(ans, abs(target - dp[l - 1]));
        else if(l == i) ans = min(ans, abs(target - dp[l]));
        else ans = min(ans, min(abs(target - dp[l]), abs(target - dp[l - 1])));
    }
    return ans;
    }
};
130600