前N个自然数的排列增加

给定前N个自然数的排列{P ₁ ，P ₂ ，P ₃ ，….. P _N ) 。任务是检查是否有可能通过交换任意两个数字来增加排列。如果已经按升序排列，则什么也不做。
例子：

Input: a[] = {5, 2, 3, 4, 1} 
Output: Yes 
Swap 1 and 5
Input: a[] = {1, 2, 3, 4, 5} 
Output: Yes 
Already in increasing order
Input: a[] = {5, 2, 1, 4, 3} 
Output: No 
 

方法：设K为P ₁ ≠i (基于1的索引)的位置i的数量。如果K = 0 ，则答案为“是”，因为排列可以保持原样。如果K = 2 ，则答案也为“是” ：交换两个放错位置的元素。 (注意，如果将任何元素放置在错误的位置，永远不可能使K = 1 ，原本应该放置在该位置的元素也必须放错位置。)如果K> 2，则答案为否：一次交换只能影响两个元素，因此最多只能校正两个错位。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function that returns true if it is
// possible to make the permutation
// increasing by swapping any two numbers
bool isPossible(int a[], int n)
{
    // To count misplaced elements
    int k = 0;
 
    // Count all misplaced elements
    for (int i = 0; i < n; i++) {
        if (a[i] != i + 1)
            k++;
    }
 
    // If possible
    if (k <= 2)
        return true;
 
    return false;
}
 
// Driver code
int main()
{
    int a[] = { 5, 2, 3, 4, 1 };
    int n = sizeof(a) / sizeof(a[0]);
 
    if (isPossible(a, n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// Function that returns true if it is
// possible to make the permutation
// increasing by swapping any two numbers
static boolean isPossible(int a[], int n)
{
    // To count misplaced elements
    int k = 0;
 
    // Count all misplaced elements
    for (int i = 0; i < n; i++)
    {
        if (a[i] != i + 1)
            k++;
    }
 
    // If possible
    if (k <= 2)
        return true;
 
    return false;
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 5, 2, 3, 4, 1 };
    int n = a.length;
 
    if (isPossible(a, n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by Code_Mech

Python3

# Python3 implementation of the approach
 
# Function that returns true if it is
# possible to make the permutation
# increasing by swapping any two numbers
def isPossible(a, n) :
 
    # To count misplaced elements
    k = 0;
 
    # Count all misplaced elements
    for i in range(n) :
        if (a[i] != i + 1) :
            k += 1;
 
    # If possible
    if (k <= 2) :
        return True;
 
    return False;
 
# Driver code
if __name__ == "__main__" :
 
    a = [5, 2, 3, 4, 1 ];
    n = len(a);
 
    if (isPossible(a, n)) :
        print("Yes");
    else :
        print("No");
 
# This code is contributed by AnkitRai01

C#

// C# implementation of the approach
using System;
     
class GFG
{
     
// Function that returns true if it is
// possible to make the permutation
// increasing by swapping any two numbers
static Boolean isPossible(int []a, int n)
{
    // To count misplaced elements
    int k = 0;
 
    // Count all misplaced elements
    for (int i = 0; i < n; i++)
    {
        if (a[i] != i + 1)
            k++;
    }
 
    // If possible
    if (k <= 2)
        return true;
 
    return false;
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = { 5, 2, 3, 4, 1 };
    int n = a.Length;
 
    if (isPossible(a, n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by 29AjayKumar

Javascript

 
// Javascript implementation of the approach
 
// Function that returns true if it is
// possible to make the permutation
// increasing by swapping any two numbers
function isPossible(a, n)
{
 
    // To count misplaced elements
    let k = 0;
 
    // Count all misplaced elements
    for (let i = 0; i < n; i++) {
        if (a[i] != i + 1)
            k++;
    }
 
    // If possible
    if (k <= 2)
        return true;
 
    return false;
}
 
// Driver code
    let a = [ 5, 2, 3, 4, 1 ];
    let n = a.length;
 
    if (isPossible(a, n))
        document.write("Yes");
    else
        document.write("No");
 
// This code is contributed by Manoj

输出：

Yes