# Find a permutation of 2N numbers such that the result of given expression is exactly 2K

Given two integers N and K, the task is to find a permutation of first 2*N natural numbers such that the following equation is satisfied.

Note: The value of K will always be less than or equal to N.

Examples:

Input : N = 1,  K = 0
Output : 1 2
The result of the above expression will be:
|1-2|-|1-2| =0

Input : N = 2,  K = 1
Output : 2 1 3 4
The result of the above expression will be:
(|2-1|+|3-4|)-(|2-1+3-4|) = 2


## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

Consider the sorted permutation:

1, 2, 3, 4, 5, 6....


The result of the expression will come out to be exactly 0. If we swap any 2 indices 2i-1 and 2i, the result will increase by exactly 2. So we need to make K such swaps.

Below is the implementation of the above approach:

## C++

 // C++ program to find the required permutation  // of first 2*N natural numbers     #include  using namespace std;     // Function to find the required permutation  // of first 2*N natural numbers  void printPermutation(int n, int k)  {      // Iterate in blocks of 2      for (int i = 1; i <= n; i++) {          int x = 2 * i - 1;          int y = 2 * i;             // We need more increments, so print in reverse order          if (i <= k)              cout << y << " " << x << " ";             // We have enough increments, so print in same order          else             cout << x << " " << y << " ";      }  }     // Driver Code  int main()  {      int n = 2, k = 1;         printPermutation(n, k);         return 0;  }

## Python3

 # Python3 program to find the required   # permutation of first 2*N natural numbers      # Function to find the required permutation   # of first 2*N natural numbers   def printPermutation(n, k) :             # Iterate in blocks of 2       for i in range(1, n + 1) :          x = 2 * i - 1;           y = 2 * i;              # We need more increments,           # so print in reverse order           if (i <= k) :              print(y, x, end = " ");              # We have enough increments,           # so print in same order           else :              print(x, y, end = " ");      # Driver Code   if __name__ == "__main__" :       n = 2; k = 1;          printPermutation(n, k);          # This code is contributed by Ryuga

Output:

2 1 3 4


Time Complexity: O(N)
Auxiliary Space: O(1)

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