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

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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)

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