Construct the Array using given bitwise AND, OR and XOR

Given Bitwise AND, OR, and XOR of N elements of an array denoted by a, b, c. The task is to find the elements of the array. If there exists no such array then print “-1”.
Examples:

Input: N = 3, a = 4, b = 6, c = 6. 
Output: {4, 4, 6} 
Explanation:
Bitwise AND of array = 4 & 4 & 6 = 4 
Bitwise OR of array = 4 | 4 | 6 = 6 
Bitwise XOR of array = 4 ^ 4 ^ 6 = 6
Input: N = 2, a = 4, b = 6, c = 6. 
Output: -1 

Approach:

1. For Bitwise AND, if i^th bit is set in a, then in the array every element must have i^th bit set because if even one element’s i^th bit is 0 then bitwise AND of the array will result in i^th bit to be 0.
2. Secondly, if i^th bit is not set in a, then OR and XOR values need to be handled simultaneously. If i^th bit is set in b, then at least one element must have i^th bit set. So, set i^th bit in the only first element of the array.
3. Now, if i^th bit was set in b then i^th bit must be checked in c. If that bit is set in c then there’s no problem as the first element’s i^th bit is already set so 1 ^ 0 = 1. If that bit is not set in c then set the i^th bit of the second element. Now, there will not be any effect in b and for c, 1 ^ 1 will be 0.
4. Then, just calculate Bitwise AND, OR, and XOR of the array to check if it’s equal or not. If results are not equal then the array is not possible else given array is the answer.

Below is the implementation of the approach: 

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the array
void findArray(int n, int a, int b, int c)
{
    int arr[n + 1] = {};

    // Loop through all bits in number
    for (int bit = 30; bit >= 0; bit--) {

        // If bit is set in AND
        // then set it in every element
        // of the array
        int set = a & (1 << bit);
        if (set) {
            for (int i = 0; i < n; i++)
                arr[i] |= set;
        }

        // If bit is not set in AND
        else {

            // But set in b(OR)
            if (b & (1 << bit)) {

                // Set bit position
                // in first element
                arr[0] |= (1 << bit);

                // If bit is not set in c
                // then set it in second
                // element to keep xor as
                // zero for bit position
                if (!(c & (1 << bit))) {
                    arr[1] |= (1 << bit);
                }
            }
        }
    }

    int aa = INT_MAX, bb = 0, cc = 0;

    // Calculate AND, OR
    // and XOR of array
    for (int i = 0; i < n; i++) {
        aa &= arr[i];
        bb |= arr[i];
        cc ^= arr[i];
    }

    // Check if values are equal or not
    if (a == aa && b == bb && c == cc) {
        for (int i = 0; i < n; i++)
            cout << arr[i] << " ";
    }

    // If not, then array
    // is not possible
    else
        cout << "-1";
}

// Driver Code
int main()
{
    // Given Bitwise AND, OR, and XOR
    int n = 3, a = 4, b = 6, c = 6;

    // Function Call
    findArray(n, a, b, c);

    return 0;
}

Java

// Java program for the above approach
import java.io.*;

class GFG{

// Function to find the array
static void findArray(int n, int a,
                      int b, int c)
{
    int arr[] = new int[n + 1];

    // Loop through all bits in number
    for(int bit = 30; bit >= 0; bit--)
    {

       // If bit is set in AND
       // then set it in every element
       // of the array
       int set = a & (1 << bit);
       if (set != 0)
       {
           for(int i = 0; i < n; i++)
              arr[i] |= set;
       }

       // If bit is not set in AND
       else
       {

           // But set in b(OR)
           if ((b & (1 << bit)) != 0)
           {

               // Set bit position
               // in first element
               arr[0] |= (1 << bit);

               // If bit is not set in c
               // then set it in second
               // element to keep xor as
               // zero for bit position
               if ((c & (1 << bit)) == 0)
               {
                   arr[1] |= (1 << bit);
               }
           }
       }
    }
    int aa = Integer.MAX_VALUE, bb = 0, cc = 0;

    // Calculate AND, OR
    // and XOR of array
    for(int i = 0; i < n; i++)
    {
       aa &= arr[i];
       bb |= arr[i];
       cc ^= arr[i];
    }

    // Check if values are equal or not
    if (a == aa && b == bb && c == cc)
    {
        for(int i = 0; i < n; i++)
           System.out.print(arr[i] + " ");
    }

    // If not, then array
    // is not possible
    else
        System.out.println("-1");
}

// Driver code
public static void main(String[] args)
{

    // Given Bitwise AND, OR, and XOR
    int n = 3, a = 4, b = 6, c = 6;

    // Function Call
    findArray(n, a, b, c);
}
}

// This code is contributed by Pratima Pandey

Python3

# Python3 program for
# the above approach
import sys

# Function to find the array
def findArray(n, a, b, c):

    arr = [0] * (n + 1)

    # Loop through all bits in number
    for bit in range (30, -1, -1):

        # If bit is set in AND
        # then set it in every element
        # of the array
        set = a & (1 << bit)
        if (set):
            for i in range (n):
                arr[i] |= set

        # If bit is not set in AND
        else :

            # But set in b(OR)
            if (b & (1 << bit)):

                # Set bit position
                # in first element
                arr[0] |= (1 << bit)

                # If bit is not set in c
                # then set it in second
                # element to keep xor as
                # zero for bit position
                if (not (c & (1 << bit))):
                    arr[1] |= (1 << bit)

    aa = sys.maxsize
    bb = 0
    cc = 0

    # Calculate AND, OR
    # and XOR of array
    for i in range (n):
        aa &= arr[i]
        bb |= arr[i]
        cc ^= arr[i]

    # Check if values are equal or not
    if (a == aa and b == bb and c == cc):
        for i in range (n):
            print (arr[i], end = " ")

    # If not, then array
    # is not possible
    else:
        print ("-1")

# Driver Code
if __name__ =="__main__":

    # Given Bitwise AND, OR, and XOR
    n = 3
    a = 4
    b = 6
    c = 6

    # Function Call
    findArray(n, a, b, c)

# This code is contributed by Chitranayal

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the array
static void findArray(int n, int a,
                      int b, int c)
{
    int []arr = new int[n + 1];

    // Loop through all bits in number
    for(int bit = 30; bit >= 0; bit--)
    {

       // If bit is set in AND
       // then set it in every element
       // of the array
       int set = a & (1 << bit);
       if (set != 0)
       {
           for(int i = 0; i < n; i++)
              arr[i] |= set;
       }

       // If bit is not set in AND
       else
       {

           // But set in b(OR)
           if ((b & (1 << bit)) != 0)
           {

               // Set bit position
               // in first element
               arr[0] |= (1 << bit);

               // If bit is not set in c
               // then set it in second
               // element to keep xor as
               // zero for bit position
               if ((c & (1 << bit)) == 0)
               {
                   arr[1] |= (1 << bit);
               }
           }
       }
    }
    int aa = int.MaxValue, bb = 0, cc = 0;

    // Calculate AND, OR
    // and XOR of array
    for(int i = 0; i < n; i++)
    {
       aa &= arr[i];
       bb |= arr[i];
       cc ^= arr[i];
    }

    // Check if values are equal or not
    if (a == aa && b == bb && c == cc)
    {
        for(int i = 0; i < n; i++)
           Console.Write(arr[i] + " ");
    }

    // If not, then array
    // is not possible
    else
        Console.WriteLine("-1");
}

// Driver code
public static void Main(String[] args)
{

    // Given Bitwise AND, OR, and XOR
    int n = 3, a = 4, b = 6, c = 6;

    // Function Call
    findArray(n, a, b, c);
}
}

// This code is contributed by gauravrajput1

Javascript

<script>
// Javascript program for the above approach

// Function to find the array
function findArray(n, a, b, c) {
    let arr = new Array(n + 1);

    // Loop through all bits in number
    for (let bit = 30; bit >= 0; bit--) {

        // If bit is set in AND
        // then set it in every element
        // of the array
        let set = a & (1 << bit);
        if (set) {
            for (let i = 0; i < n; i++)
                arr[i] |= set;
        }

        // If bit is not set in AND
        else {

            // But set in b(OR)
            if (b & (1 << bit)) {

                // Set bit position
                // in first element
                arr[0] |= (1 << bit);

                // If bit is not set in c
                // then set it in second
                // element to keep xor as
                // zero for bit position
                if (!(c & (1 << bit))) {
                    arr[1] |= (1 << bit);
                }
            }
        }
    }

    let aa = Number.MAX_SAFE_INTEGER, bb = 0, cc = 0;

    // Calculate AND, OR
    // and XOR of array
    for (let i = 0; i < n; i++) {
        aa &= arr[i];
        bb |= arr[i];
        cc ^= arr[i];
    }

    // Check if values are equal or not
    if (a == aa && b == bb && c == cc) {
        for (let i = 0; i < n; i++)
            document.write(arr[i] + " ");
    }

    // If not, then array
    // is not possible
    else
        document.write("-1");
}

// Driver Code

// Given Bitwise AND, OR, and XOR
let n = 3, a = 4, b = 6, c = 6;

// Function Call
findArray(n, a, b, c);

// This code is contributed by gfgking
</script>

Output:

6 4 4

Time Complexity: O(31*N)

Auxiliary Space: O(N)