给定n个元素的数组arr [] ，我们必须将索引i与另一个索引i + k交换任意次，并检查是否可以对给定的数组arr []进行排序。如果是，则打印“是”，否则打印“否”。
例子：
Input: K = 2, arr = [4, 3, 2, 6, 7]
Output: Yes
Explanation:
Choose index i = 0 and swap index i with i + k then the array becomes [2, 3, 4, 6, 7] which is sorted hence the output is “yes”.
Input : K = 2, arr = [4, 2, 3, 7, 6]
Output : No
Explanation:
It is not possible to obtain sorted array.
方法：
为了解决上述问题，我们必须从索引0开始采用元素，然后将K的倍数相加，即0、0 + k，0 +(2 * k)，依此类推。从0到K-1的所有索引交换这些位置，并检查最终数组是否已排序。如果是，则返回“是”，否则返回“否”。
下面是上述方法的实现：
C++
// CPP implementation to Check if it is possible to sort an
// array with conditional swapping of elements at distance K
#include
using namespace std;
// Function for finding if it possible
// to obtain sorted array or not
bool fun(int arr[], int n, int k)
{
vector v;
// Iterate over all elements until K
for (int i = 0; i < k; i++) {
// Store elements as multiples of K
for (int j = i; j < n; j += k) {
v.push_back(arr[j]);
}
// Sort the elements
sort(v.begin(), v.end());
int x = 0;
// Put elements in their required position
for (int j = i; j < n; j += k) {
arr[j] = v[x];
x++;
}
v.clear();
}
// Check if the array becomes sorted or not
for (int i = 0; i < n - 1; i++) {
if (arr[i] > arr[i + 1])
return false;
}
return true;
}
// Driver code
int main()
{
int arr[] = { 4, 2, 3, 7, 6 };
int K = 2;
int n = sizeof(arr) / sizeof(arr[0]);
if (fun(arr, n, K))
cout << "yes" << endl;
else
cout << "no" << endl;
return 0;
} Java
// Java implementation to check if it
// is possible to sort an array with
// conditional swapping of elements
// at distance K
import java.lang.*;
import java.io.*;
import java.util.*;
class GFG{
// Function for finding if it possible
// to obtain sorted array or not
public static boolean fun(int[] arr, int n,
int k)
{
Vector v = new Vector();
// Iterate over all elements until K
for(int i = 0; i < k; i++)
{
// Store elements as multiples of K
for(int j = i; j < n; j += k)
{
v.add(arr[j]);
}
// Sort the elements
Collections.sort(v);
int x = 0;
// Put elements in their
// required position
for(int j = i; j < n; j += k)
{
arr[j] = v.get(x);
x++;
}
v.clear();
}
// Check if the array becomes
// sorted or not
for(int i = 0; i < n - 1; i++)
{
if (arr[i] > arr[i + 1])
{
return false;
}
}
return true;
}
// Driver code
public static void main (String args[])
{
int[] arr = { 4, 2, 3, 7, 6 };
int K = 2;
int n = arr.length;
if (fun(arr, n, K))
{
System.out.println("yes");
}
else
{
System.out.println("no");
}
}
}
// This code is contributed by sayesha Python3
# Python3 implementation to Check if it is possible to sort an
# array with conditional swapping of elements at distance K
# Function for finding if it possible
# to obtain sorted array or not
def fun(arr, n, k):
v = []
# Iterate over all elements until K
for i in range(k):
# Store elements as multiples of K
for j in range(i, n, k):
v.append(arr[j]);
# Sort the elements
v.sort();
x = 0
# Put elements in their required position
for j in range(i, n, k):
arr[j] = v[x];
x += 1
v = []
# Check if the array becomes sorted or not
for i in range(n - 1):
if (arr[i] > arr[i + 1]):
return False
return True
# Driver code
arr= [ 4, 2, 3, 7, 6 ]
K = 2;
n = len(arr)
if (fun(arr, n, K)):
print("yes")
else:
print("no")
# This code is contributed by apurva rajC#
// C# implementation to check if it
// is possible to sort an array with
// conditional swapping of elements
// at distance K
using System;
using System.Collections.Generic;
class GFG{
// Function for finding if it possible
// to obtain sorted array or not
public static bool fun(int[] arr,
int n, int k)
{
List v = new List();
// Iterate over all elements until K
for(int i = 0; i < k; i++)
{
// Store elements as multiples of K
for(int j = i; j < n; j += k)
{
v.Add(arr[j]);
}
// Sort the elements
v.Sort();
int x = 0;
// Put elements in their
// required position
for(int j = i; j < n; j += k)
{
arr[j] = v[x];
x++;
}
v.Clear();
}
// Check if the array becomes
// sorted or not
for(int i = 0; i < n - 1; i++)
{
if (arr[i] > arr[i + 1])
{
return false;
}
}
return true;
}
// Driver code
public static void Main(String []args)
{
int[] arr = {4, 2, 3, 7, 6};
int K = 2;
int n = arr.Length;
if (fun(arr, n, K))
{
Console.WriteLine("yes");
}
else
{
Console.WriteLine("no");
}
}
}
// This code is contributed by shikhasingrajput 输出：
no