给定一个由N个正整数组成的数组arr [] ,任务是找到一种安排,使前N个元素至第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,那么输出将为-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:
- 令p = x&y,则p≤min(x,y) ,这意味着按位与是非递增函数。如果对某些元素执行按位与运算,则该值将减小或保持不变。
- 因此,显而易见的是,将最小的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,则打印所需的布置,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
输出:
3 5 3 1
时间复杂度: O(N)