Swap Kth Elements
Given an array ‘ARR’ of size ‘N,’ swap the Kth element from beginning with the Kth element from the end.

For example:

If ‘N’ = 5 and K = 2 [1, 2, 3, 4, 5] Then the output will be [1, 4, 3, 2, 5].

Input format :

The first line of input contains an integer T denoting the number of test cases. The first line of each test case contains two space-separated integers N, and K, where N is the number of elements of the array and K is the index. The second line of each test case contains ‘N’ space-separated integers, denoting the array elements.

Output format :

For each test case, print the array after swapping the Kth element from the start and the Kth element from the end. The output of each test case will be printed in a separate line.

Note:

You do not need to print anything; it has already been taken care of. Just implement the given function.

Constraints:

1

Question

Swap Kth Elements

Given an array ‘ARR’ of size ‘N,’ swap the Kth element from beginning with the Kth element from the end.

For example:

If ‘N’ = 5 and K = 2
[1, 2, 3, 4, 5]

Then the output will be [1, 4, 3, 2, 5].

Input format :

The first line of input contains an integer T denoting the number of test cases.

The first line of each test case contains two space-separated integers N, and K, where N is the number of elements of the array and K is the index.

The second line of each test case contains ‘N’ space-separated integers, denoting the array elements.

Output format :

For each test case, print the array after swapping the  Kth element from the start and the Kth element from the end.

The output of each test case will be printed in a separate line.

Note:

You do not need to print anything; it has already been taken care of. Just implement the given function.

Constraints:

1 <= T <= 5
1 <= K < N <= 5000

Time Limit : 1 sec

Accepted Answer

Step 1: Find the kth node from the start and the kth node from the end (n-k+1th node from start)
Step 2: Swap both the nodes

Accepted Answer

Traverse till Kth index.

Initialize two pointers, one from the beginning and on the other end of the array.
Move the beginning pointer K times and simultaneously move the end pointer backward K times.
Swap the elements that are being pointed by the begging pointer and the end pointer.

Space Complexity: O(1)Explanation:

O(1)

Since we are not using any extra space.

Time Complexity: O(n)Explanation:

O(N), where N is the size of the given array.

At worst, K can be equal to N, and then we have to traverse the array completely. Therefore, O(K + K) = O(N + N) = O(N).

Python (3.5)

''' 

    Time Complexity : O(N)

	Space Complexity : O(1)



	Where N is the size of the array.

'''



def swapKthElement(arr: list, k: int):



    n = len(arr)

    

    start = -1

    end = n

    ctr = 0

    

    while ctr < k:

        start+=1

        end-=1

        ctr+=1

        

    arr[start], arr[end] = arr[end], arr[start]

    

    return arr

Java (SE 1.8)

/*

	Time Complexity : O(N)

	Space Complexity : O(1)



	Where N is the size of the array.

*/



import java.util.Collections;

import java.util.ArrayList;



public class Solution 

{

    public static ArrayList swapKthElement(ArrayList arr,Integer k)

    {

        int n = arr.size() ;



        int start = -1, end = n ;

        int ctr = 0 ;



        while (ctr < k) {

            start++;

            end-- ;

            ctr++ ;

        }



        Collections.swap(arr, start, end);



        return arr;

    }

}

C++ (g++ 5.4)

/*

	Time Complexity : O(N)

	Space Complexity : O(1)



	Where N is the size of the array.

*/





vector swapKthElement(vector&arr, int k) {

	int n = arr.size();



	int start = -1, end = n ;

	int ctr = 0 ;



	while (ctr < k) {

		start++, end-- ;

		ctr++ ;

	}



	swap(arr[start], arr[end]);



	return arr;

}

Accepted Answer

Direct Swap

We can directly swap the elements at the Kth position and (N - K)th position.

Space Complexity: O(1)Explanation:

O(1)

Since we are not using any extra space.

Time Complexity: O(1)Explanation:

O(1)

Because swapping is a constant time work.

Python (3.5)

''' 

    Time Complexity: O(1)

    Space Complexity: O(1)

'''



def swapKthElement(arr: list, k: int):



    n = len(arr)

    

    arr[k - 1], arr[n - k] = arr[n - k], arr[k - 1]

    

    return arr

C++ (g++ 5.4)

/*

	Time Complexity : O(1)

	Space Complexity : O(1)

*/





vector swapKthElement(vector arr, int k) 

{

	int n = arr.size() ;



	swap(arr[k - 1], arr[n - k]) ;



	return arr;

}

Java (SE 1.8)

/*

	Time Complexity : O(1)

	Space Complexity : O(1)	

*/



import java.util.Collections;

import java.util.ArrayList;



public class Solution 

{

    public static ArrayList swapKthElement(ArrayList arr,Integer k)

    {

        int n = arr.size() ;



        Collections.swap(arr, k - 1, n - k) ;



        return arr;

    }

}

Given an array ‘ARR’ of size ‘N,’ swap the Kth element from beginning with the Kth element from the end.

For example:

Input format :

Output format :

Note:

Constraints:

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