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)