# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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; ` `    ``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; ` `    ``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 `

Output:

```3 5 3 1
```

Time Complexity: O(N)

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