<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>

#define ll int 
class Solution {
public:
    ll solve(ll l , ll r,  vector<vector<ll> >&jk){
         ll s= log2(r-l+1);
         return jk[l][s]&jk[r-pow(2,s)+1][s];
    }
    int closestToTarget(vector<int>& arr, int target) {
        vector<ll>v=arr;
        ll n = v.size();
        ll limit = log2(n+1)+1;
        vector<vector<ll> >jk(n+1,vector<ll>(limit+1));
        for(int i = 0;i<n;i++){
            jk[i][0]= v[i];
        }
        for(int i = 1;i<limit;i++){
            for(int j = 0;j+pow(2,i)<=n;j++){
                jk[j][i]= jk[j][i-1]&jk[j+pow(2,i-1)][i-1];
            }
        }
        ll ans = 1e9;
        for(int i = 0;i<n;i++){
            ll l = i;
            ll r = n-1;
            while(r>=l){
                ll mid = l+(r-l)/2;
                ll val = solve(i,mid,jk);
                ans = min(ans,abs(val-target));
                if(ans==0){
                      return 0;
                }
                else if(val>target){
                    l = mid+1;
                }
                else if(val<target){
                     r = mid-1;
                }
            }
        }
        return ans ;

    }
};
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