检查数组是否可以通过交换对与设置位计数等于最小数组元素的 GCD 的对进行排序

给定一个由N 个整数组成的数组arr[] ，任务是检查是否可以使用以下交换操作对数组进行排序：

Swapping of two numbers is valid only if the Greatest Common Divisor of count of set bits of the two numbers is equal to the number of set bits in the smallest element of the array. 
 

如果可以通过仅执行上述交换对数组进行排序，则打印“Yes” 。否则，打印“否” 。

例子：

Input: arr[] = {2, 3, 5, 7, 6} 
Output: Yes 
Explanation: 
7 and 6 are needed to be swapped to make the array sorted. 
7 has 3 set bits and 6 has 2 set bits. 
Since GCD(2, 3) = 1, which is equal to the number of set bits in the smallest integer from the array i.e., 2 (1 set bit). 
Therefore, the array can be sorted by swapping (7, 6).

Input: arr[] = {3, 3, 15, 7} 
Output: No 
Explanation: 
15 and 7 are needed to be swapped to make the array sorted. 
15 has 4 set bits and 7 has 3 set bits. GCD(4, 3) = 1, which is not equal to the number of set bits in the smallest integer from the array i.e., 3(2 set bit). 
Therefore, the array cannot be sorted. 

处理方法：按照以下步骤解决问题：

1. 对给定数组进行排序并将其存储在辅助数组中(比如dup[] )。
2. 遍历数组，对于每个元素，检查它是否在arr[]和dup[] 中的索引相同。如果发现为真，则继续下一个索引。
3. 否则，如果需要在arr[] 中交换^第i^个和^第j^个位置元素，则计算arr[i]的设置位与arr[j]的设置位的GCD并检查它是否等于设置位的计数是否在数组的最小元素中。
4. 如果不允许任何此类交换，则打印“No” 。否则，打印“是” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to count number of set
// bits in an integer
int calculateSetBit(int n)
{
    int count = 0;
 
    // Traverse every bits
    for (int i = 0; i < 32; i++) {
        if (n & 1)
            count++;
 
        // Right shift by 1
        n = n >> 1;
    }
 
    return count;
}
 
// Function to check if sorting the
// given array is possible or not
void sortPossible(int arr[], int n)
{
    // Duplicate array
    int dup[n];
 
    for (int i = 0; i < n; i++)
        dup[i] = arr[i];
 
    // Sorted array to check if the
    // original array can be sorted
    sort(dup, dup + n);
 
    // Flag variable to check
    // if possible to sort
    bool flag = 1;
 
    // Calculate bits of smallest
    // array element
    int bit = calculateSetBit(dup[0]);
 
    // Check every wrong placed
    // integer in the array
    for (int i = 0; i < n; i++) {
        if (arr[i] != dup[i]) {
 
            // Swapping only if GCD of set
            // bits is equal to set bits in
            // smallest integer
            if (__gcd(
                    calculateSetBit(arr[i]),
                    bit)
                != bit) {
                flag = 0;
                break;
            }
        }
    }
 
    // Print the result
    if (flag) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
 
    return;
}
 
// Driver Code
int main()
{
    int arr[] = { 3, 9, 12, 6 };
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    sortPossible(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG{
  
// Recursive function to return
// gcd of a and b
static int gcd(int a, int b)
{
     
    // Everything divides 0 
    if (a == 0)
        return b;
    if (b == 0)
         return a;
        
    // Base case
    if (a == b)
        return a;
        
    // a is greater
    if (a > b)
        return gcd(a - b, b);
    return gcd(a, b - a);
}
     
// Function to count number of set
// bits in an integer
static int calculateSetBit(int n)
{
    int count = 0;
  
    // Traverse every bits
    for(int i = 0; i < 32; i++)
    {
        if ((n & 1) != 0)
            count++;
  
        // Right shift by 1
        n = n >> 1;
    }
    return count;
}
  
// Function to check if sorting the
// given array is possible or not
static void sortPossible(int arr[], int n)
{
     
    // Duplicate array
    int dup[] = new int[n];
  
    for(int i = 0; i < n; i++)
        dup[i] = arr[i];
  
    // Sorted array to check if the
    // original array can be sorted
    Arrays.sort(dup);
  
    // Flag variable to check
    // if possible to sort
    int flag = 1;
  
    // Calculate bits of smallest
    // array element
    int bit = calculateSetBit(dup[0]);
  
    // Check every wrong placed
    // integer in the array
    for(int i = 0; i < n; i++)
    {
        if (arr[i] != dup[i])
        {
             
            // Swapping only if GCD of set
            // bits is equal to set bits in
            // smallest integer
            if (gcd(calculateSetBit(
                arr[i]), bit) != bit)
            {
                flag = 0;
                break;
            }
        }
    }
  
    // Print the result
    if (flag != 0)
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
    return;
}
  
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 3, 9, 12, 6 };
  
    int N = arr.length;
  
    // Function call
    sortPossible(arr, N);
}
}
 
// This code is contributed by sanjoy_62

Python3

# Python3 program for the above approach
from math import gcd
 
# Function to count number of set
# bits in an eger
def calculateSetBit(n):
 
    count = 0
 
    # Traverse every bits
    for i in range(32):
        if (n & 1):
            count += 1
 
        # Right shift by 1
        n = n >> 1
         
    return count
 
# Function to check if sorting the
# given array is possible or not
def sortPossible(arr, n):
 
    # Duplicate array
    dup = [0] * n
 
    for i in range(n):
        dup[i] = arr[i]
 
    # Sorted array to check if the
    # original array can be sorted
    dup = sorted(dup)
 
    # Flag variable to check
    # if possible to sort
    flag = 1
 
    # Calculate bits of smallest
    # array element
    bit = calculateSetBit(dup[0])
 
    # Check every wrong placed
    # eger in the array
    for i in range(n):
        if (arr[i] != dup[i]):
 
            # Swapping only if GCD of set
            # bits is equal to set bits in
            # smallest eger
            if (gcd(calculateSetBit(arr[i]), bit) != bit):
                flag = 0
                break
 
    # Print the result
    if (flag):
        print("Yes")
    else:
        print("No")
 
    return
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 3, 9, 12, 6 ]
 
    N = len(arr)
 
    # Function call
    sortPossible(arr, N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach 
using System;
 
class GFG{
  
// Recursive function to return
// gcd of a and b
static int gcd(int a, int b)
{
     
    // Everything divides 0 
    if (a == 0)
        return b;
    if (b == 0)
         return a;
        
    // Base case
    if (a == b)
        return a;
        
    // a is greater
    if (a > b)
        return gcd(a - b, b);
    return gcd(a, b - a);
}
     
// Function to count number of set
// bits in an integer
static int calculateSetBit(int n)
{
    int count = 0;
  
    // Traverse every bits
    for(int i = 0; i < 32; i++)
    {
        if ((n & 1) != 0)
            count++;
  
        // Right shift by 1
        n = n >> 1;
    }
    return count;
}
  
// Function to check if sorting the
// given array is possible or not
static void sortPossible(int[] arr, int n)
{
     
    // Duplicate array
    int[] dup = new int[n];
  
    for(int i = 0; i < n; i++)
        dup[i] = arr[i];
  
    // Sorted array to check if the
    // original array can be sorted
    Array.Sort(dup);
  
    // Flag variable to check
    // if possible to sort
    int flag = 1;
  
    // Calculate bits of smallest
    // array element
    int bit = calculateSetBit(dup[0]);
  
    // Check every wrong placed
    // integer in the array
    for(int i = 0; i < n; i++)
    {
        if (arr[i] != dup[i])
        {
             
            // Swapping only if GCD of set
            // bits is equal to set bits in
            // smallest integer
            if (gcd(calculateSetBit(
                arr[i]), bit) != bit)
            {
                flag = 0;
                break;
            }
        }
    }
  
    // Print the result
    if (flag != 0)
    {
        Console.WriteLine("Yes");
    }
    else
    {
        Console.WriteLine("No");
    }
  
    return;
}
  
// Driver Code
public static void Main()
{
    int[] arr = { 3, 9, 12, 6 };
  
    int N = arr.Length;
  
    // Function call
    sortPossible(arr, N);
}
}
 
// This code is contributed by sanjoy_62

Javascript

 
// JavaScript program to implement
// the above approach
 
// Recursive function to return
// gcd of a and b
function gcd(a, b)
{
      
    // Everything divides 0
    if (a == 0)
        return b;
    if (b == 0)
         return a;
         
    // Base case
    if (a == b)
        return a;
         
    // a is greater
    if (a > b)
        return gcd(a - b, b);
    return gcd(a, b - a);
}
      
// Function to count number of set
// bits in an leteger
function calculateSetBit(n)
{
    let count = 0;
   
    // Traverse every bits
    for(let i = 0; i < 32; i++)
    {
        if ((n & 1) != 0)
            count++;
   
        // Right shift by 1
        n = n >> 1;
    }
    return count;
}
   
// Function to check if sorting the
// given array is possible or not
function sortPossible(arr, n)
{
      
    // Duplicate array
    let dup = [];
   
    for(let i = 0; i < n; i++)
        dup[i] = arr[i];
   
    // Sorted array to check if the
    // original array can be sorted
    dup.sort();
   
    // Flag variable to check
    // if possible to sort
    let flag = 1;
   
    // Calculate bits of smallest
    // array element
    let bit = calculateSetBit(dup[0]);
   
    // Check every wrong placed
    // leteger in the array
    for(let i = 0; i < n; i++)
    {
        if (arr[i] != dup[i])
        {
              
            // Swapping only if GCD of set
            // bits is equal to set bits in
            // smallest leteger
            if (gcd(calculateSetBit(
                arr[i]), bit) != bit)
            {
                flag = 0;
                break;
            }
        }
    }
   
    // Print the result
    if (flag != 0)
    {
        document.write("Yes");
    }
    else
    {
        document.write("No");
    }
    return;
}
 
// Driver code
    let arr = [ 3, 9, 12, 6 ];
    let N = arr.length;
   
    // Function call
    sortPossible(arr, N);
      
     // This code is contributed by target_2.

输出：

Yes

时间复杂度： O(N)
辅助空间： O(N)

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