Rearrange array elements such that Bitwise AND of first N – 1 elements is equal to last element

Given an array arr[] of N positive integers, the task is to find an arrangement such that Bitwise AND of the first N – 1 elements is equal to the last element. If no such arrangement is possible then output will be -1.
Examples:

Input: arr[] = {1, 5, 3, 3} 
Output: 3 5 3 1 
(3 & 5 & 3) = 1 which is equal to the last element.
Input: arr[] = {2, 3, 7} 
Output: -1 
No such arrangement is possible. 

Approach:

• Let p = x & y then p ≤ min(x, y) which means Bitwise AND is a non-increasing function. If bitwise AND is performed on some elements then the value will be decreasing or remain the same.
• So, it is obvious to put the smallest element at the last index and then check if the last element is equal to the bitwise AND of the first N – 1 elements or not. If yes, then print the required arrangement otherwise print -1.

Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Utility function to print
// the elements of an array
void printArr(int arr[], int n)
{
    for (int i = 0; i < n; i++)
        cout << arr[i] << " ";
}

// Function to find the required arrangement
void findArrangement(int arr[], int n)
{

    // There has to be atleast 2 elements
    if (n < 2) {
        cout << "-1";
        return;
    }

    // Minimum element from the array
    int minVal = *min_element(arr, arr + n);

    // Swap any occurrence of the minimum
    // element with the last element
    for (int i = 0; i < n; i++) {
        if (arr[i] == minVal) {
            swap(arr[i], arr[n - 1]);
            break;
        }
    }

    // Find the bitwise AND of the
    // first (n - 1) elements
    int andVal = arr[0];
    for (int i = 1; i < n - 1; i++) {
        andVal &= arr[i];
    }

    // If the bitwise AND is equal
    // to the last element then
    // print the arrangement
    if (andVal == arr[n - 1])
        printArr(arr, n);
    else
        cout << "-1";
}

// Driver code
int main()
{
    int arr[] = { 1, 5, 3, 3 };
    int n = sizeof(arr) / sizeof(int);

    findArrangement(arr, n);

    return 0;
}

Java

// Java implementation of the approach
import java.util.*;

class GFG
{

// Utility function to print
// the elements of an array
static void printArr(int []arr, int n)
{
    for (int i = 0; i < n; i++)
        System.out.print(arr[i] + " ");
}

// Function to find the required arrangement
static void findArrangement(int arr[], int n)
{

    // There has to be atleast 2 elements
    if (n < 2)
    {
        System.out.print("-1");
        return;
    }

    // Minimum element from the array
    int minVal = Arrays.stream(arr).min().getAsInt();

    // Swap any occurrence of the minimum
    // element with the last element
    for (int i = 0; i < n; i++)
    {
        if (arr[i] == minVal)
        {
            swap(arr, i, n - 1);
            break;
        }
    }

    // Find the bitwise AND of the
    // first (n - 1) elements
    int andVal = arr[0];
    for (int i = 1; i < n - 1; i++)
    {
        andVal &= arr[i];
    }

    // If the bitwise AND is equal
    // to the last element then
    // print the arrangement
    if (andVal == arr[n - 1])
        printArr(arr, n);
    else
        System.out.print("-1");
}

static int[] swap(int []arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}

// Driver code
public static void main(String []args)
{
    int arr[] = { 1, 5, 3, 3 };
    int n = arr.length;

    findArrangement(arr, n);
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 implementation of the approach

# Utility function to print
# the elements of an array
def printArr(arr, n) :

    for i in range(n) :
        print(arr[i], end = " ");

# Function to find the required arrangement
def findArrangement(arr, n) :

    # There has to be atleast 2 elements
    if (n < 2) :
        print("-1", end = "");
        return;

    # Minimum element from the array
    minVal = min(arr);

    # Swap any occurrence of the minimum
    # element with the last element
    for i in range(n) :
        if (arr[i] == minVal) :
            arr[i], arr[n - 1] = arr[n - 1], arr[i];
            break;

    # Find the bitwise AND of the
    # first (n - 1) elements
    andVal = arr[0];
    for i in range(1, n - 1) :
        andVal &= arr[i];

    # If the bitwise AND is equal
    # to the last element then
    # print the arrangement
    if (andVal == arr[n - 1]) :
        printArr(arr, n);
    else :
        print("-1");

# Driver code
if __name__ == "__main__" :

    arr = [ 1, 5, 3, 3 ];
    n = len(arr);

    findArrangement(arr, n);

# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
using System.Linq;

class GFG
{

// Utility function to print
// the elements of an array
static void printArr(int []arr, int n)
{
    for (int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
}

// Function to find the required arrangement
static void findArrangement(int []arr, int n)
{

    // There has to be atleast 2 elements
    if (n < 2)
    {
        Console.Write("-1");
        return;
    }

    // Minimum element from the array
    int minVal = arr.Min();

    // Swap any occurrence of the minimum
    // element with the last element
    for (int i = 0; i < n; i++)
    {
        if (arr[i] == minVal)
        {
            swap(arr, i, n - 1);
            break;
        }
    }

    // Find the bitwise AND of the
    // first (n - 1) elements
    int andVal = arr[0];
    for (int i = 1; i < n - 1; i++)
    {
        andVal &= arr[i];
    }

    // If the bitwise AND is equal
    // to the last element then
    // print the arrangement
    if (andVal == arr[n - 1])
        printArr(arr, n);
    else
        Console.Write("-1");
}

static int[] swap(int []arr, int i, int j)
{
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
    return arr;
}

// Driver code
public static void Main(String []args)
{
    int []arr = { 1, 5, 3, 3 };
    int n = arr.Length;

    findArrangement(arr, n);
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// Javascript implementation of the approach

// Utility function to print
// the elements of an array
function printArr(arr, n)
{
    for (var i = 0; i < n; i++)
        document.write( arr[i] + " ");
}

// Function to find the required arrangement
function findArrangement(arr, n)
{

    // There has to be atleast 2 elements
    if (n < 2) {
        document.write( "-1");
        return;
    }

    // Minimum element from the array
    var minVal =  arr.reduce((a,b)=> Math.min(a,b));

    // Swap any occurrence of the minimum
    // element with the last element
    for (var i = 0; i < n; i++) {
        if (arr[i] == minVal) {
            [arr[i], arr[n-1]] = [arr[n-1], arr[i]];
            break;
        }
    }

    // Find the bitwise AND of the
    // first (n - 1) elements
    var andVal = arr[0];
    for (var i = 1; i < n - 1; i++) {
        andVal &= arr[i];
    }

    // If the bitwise AND is equal
    // to the last element then
    // print the arrangement
    if (andVal == arr[n - 1])
        printArr(arr, n);
    else
        document.write( "-1");
}

// Driver code
var arr = [1, 5, 3, 3];
var n = arr.length;
findArrangement(arr, n);

// This code is contributed by rrrtnx.
</script>

Output:

3 5 3 1

Time Complexity: O(N)

Auxiliary Space: O(1)