<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 fun(vector<vector<int>>& sparse, int l, int r) 
    {
        int ans = (1<<30)-1;
        int idx = l; 
        for(int i=30; i>=0; i--) {
            if((r-l+1)&(1<<i)){
                ans &= sparse[idx][i]; 
                idx += (1<<i); 
            }   
        }
        return ans; 
    }

    int closestToTarget(vector<int>& arr, int target) {
        int n = arr.size(); 
        int ans = 1e9;

        vector<vector<int>> sparse(n, vector<int>(32)); 
        for(int i=0; i<n; i++) {
            sparse[i][0] = arr[i]; 
        }

        for(int i=n-1; i>=0; i--) {
            for(int j=1; j<=30; j++) {
                if(i + (1<<(j-1)) <= n-1) {
                    sparse[i][j] = sparse[i][j-1] & sparse[i + (1<<(j-1))][j-1]; 
                }
            }
        }

        // for(int i=0; i<n; i++) {
        //     for(int j=0; j<3; j++) cout << sparse[i][j] << " "; 
        //     cout << endl; 
        // }
        // cout << endl; 

        for(int ll = 0; ll < n; ll++) {
            int l = ll, r = n-1;
            ans = min(ans, abs(fun(sparse, l, r) - target)); 
            while(l <= r) {
                int mid = (l+r)/2; 
                int v1 = abs(fun(sparse, l, mid) - target); 
                int v2 = abs(fun(sparse, l, mid+1) - target);

                if(v1 <= v2) {
                    ans = min(ans, min(v1, v2)); 
                    r = mid-1; 
                }
                else 
                {
                    if(mid+1 <= n-1)
                        ans = min(ans, min(v1, v2)); 
                    ans = min(ans, v1); 
                    l = mid+1; 
                }
            }
        }

        return ans; 
    }
};
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