重新排列数组元素，使得前 N – 1 个元素的按位与等于最后一个元素

给定一个由N 个正整数组成的数组arr[] ，任务是找到一个排列，使得前N – 1 个元素的按位与等于最后一个元素。如果没有这样的安排是可能的，那么输出将是-1 。
例子：

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. 
 

方法：

• 让p = x & y然后p ≤ min(x, y)这意味着按位与是一个非递增函数。如果对某些元素执行按位 AND，则该值将减少或保持不变。
• 因此，很明显将最小元素放在最后一个索引处，然后检查最后一个元素是否等于前N – 1 个元素的按位与。如果是，则打印所需的排列，否则打印-1 。

下面是上述方法的实现：

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[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

 
// 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.

输出：

3 5 3 1

时间复杂度： O(N)

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