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# Discover final component in Array fashioned from bitwise AND of array parts

Given an array A[] of measurement N, the duty is to seek out the final remaining component in a brand new array B containing all pairwise bitwise ANDs of parts from A i.e., B consists of N⋅(N − 1) / 2 parts, every of the shape Ai & Aj for some 1 ≤ i < j ≤ N. And we will carry out the next operation any variety of occasions on a brand new array until there is just one component remaining equivalent to:

• Let X and Y be the present most and minimal parts of array B respectively and take away X and Y from array B and insert X|Y into it.

Examples:

Enter: A[] = {2, 7, 1}
Output: 3
Rationalization: Array B might be [A1 & A2, A1 & A3, A2 & A3] = [2 & 7, 2 & 1, 7 & 1] = [2, 0, 1].
Then, we do the next operations on B:
Take away 2 and 0 from B and insert 2|0=2 into it. Now, B=[1, 2].
Take away 2 and 1 from B and insert 2|1=3 into it. Now, B=.
The final remaining component is thus 3.

Enter: A[] = {4, 6, 7, 2}
Output: 6

Method: The issue might be solved primarily based on the next commentary:

Observations:

• The property of bitwise or is that if we’re performing X | Y, bit i might be set if atleast one in all bit i in X or Y is ready.
• This leads us to essentially the most essential commentary of the issue, for each bit i, if bit i is ready in atleast one in all B1, B2, …BN⋅(N−1)/2, then bit i might be set within the last remaining component of B when all of the operations are carried out. We don’t want to fret about how the operations are carried out.
• For bit i to be set in atleast one component of B, we have to have atleast 2 parts in A say j and ok the place Aj and Aok each have bit i set. This units the bit i in Aj & Aok.
• Now we’ve the next resolution, iterate over each legitimate bit i, rely the variety of parts in array A which have bit i set. If this rely is bigger than 1, the ultimate reply may have bit i set else it will likely be unset.

Observe the steps talked about under to implement the thought:

• Create an array of measurement 32 to retailer the rely of bits set in ith place.
• Traverse the array and for every array component:
• Discover the positions through which the bit is ready.
• Increment the set bit rely for that place by 1.
• Traverse the array storing the set bit rely.
• If the rely is no less than 2, set that bit within the last reply.
• Return the quantity fashioned because the required reply.

Under is the implementation of the above strategy:

## Java

 ` `  `import` `java.io.*;` `import` `java.util.*;` ` `  `public` `class` `GFG {` ` `  `    ` `    ` `    ``public` `static` `int` `discover(``int` `a[], ``int` `n)` `    ``{` `        ``int` `rely = ``0``;` `        ``int` `b[] = ``new` `int``[``33``];` `        ``for` `(``int` `bit = ``30``; bit >= ``0``; bit--) {` `            ``for` `(``int` `i = ``0``; i < n; i++) {` `                ``if` `(((``1` `<< bit) & (a[i])) != ``0``) {` `                    ``b[bit]++;` `                ``}` `            ``}` `        ``}` `        ``for` `(``int` `bit = ``30``; bit >= ``0``; bit--) {` `            ``if` `(b[bit] > ``1``)` `                ``rely = rely + (``1` `<< bit);` `        ``}` `        ``return` `rely;` `    ``}` ` `  `    ` `    ``public` `static` `void` `important(String[] args)` `    ``{` `        ``int` `A[] = { ``2``, ``7``, ``1` `};` `        ``int` `N = A.size;` ` `  `        ` `        ``System.out.println(discover(A, N));` `    ``}` `}`

Time Complexity: O(N)
Auxiliary Area: O(1)

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