Swap Alternate Boundary Pairs
Given an array arr[] of N integers, the task is to swap the first and the last element then the third and the third last element then fifth and fifth last and so on. Print the final array after all the valid operations.
Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: 6 2 4 3 5 1
Operation 1: Swap 1 and 6
Operation 2: Swap 3 and 4
Input: arr[] = {5, 54, 12, 63, 45}
Output: 45 54 12 63 5
Approach: Initialize pointer i = 0 and j = N – 1 then swap the elements at these pointers and update i = i + 2 and j = j – 2. Repeat these steps while i < j. Finally print the updated array.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void printArr( int arr[], int n)
{
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
}
void UpdateArr( int arr[], int n)
{
int i = 0, j = n - 1;
while (i < j) {
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i += 2;
j -= 2;
}
printArr(arr, n);
}
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6 };
int n = sizeof (arr) / sizeof (arr[0]);
UpdateArr(arr, n);
return 0;
}
|
Java
class GFG
{
static void printArr( int arr[], int n)
{
for ( int i = 0 ; i < n; i++)
System.out.print(arr[i] + " " );
}
static void UpdateArr( int arr[], int n)
{
int i = 0 , j = n - 1 ;
while (i < j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i += 2 ;
j -= 2 ;
}
printArr(arr, n);
}
public static void main (String[] args)
{
int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 };
int n = arr.length;
UpdateArr(arr, n);
}
}
|
Python3
def printArr(arr, n):
for i in range (n):
print (arr[i], end = " " );
def UpdateArr(arr, n):
i = 0 ;
j = n - 1 ;
while (i < j):
temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i + = 2 ;
j - = 2 ;
printArr(arr, n);
if __name__ = = '__main__' :
arr = [ 1 , 2 , 3 , 4 , 5 , 6 ];
n = len (arr);
UpdateArr(arr, n);
|
C#
using System;
class GFG
{
static void printArr( int []arr, int n)
{
for ( int i = 0; i < n; i++)
Console.Write(arr[i] + " " );
}
static void UpdateArr( int []arr, int n)
{
int i = 0, j = n - 1;
while (i < j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i += 2;
j -= 2;
}
printArr(arr, n);
}
public static void Main (String[] args)
{
int []arr = { 1, 2, 3, 4, 5, 6 };
int n = arr.Length;
UpdateArr(arr, n);
}
}
|
Javascript
<script>
function printArr(arr, n)
{
for (let i = 0; i < n; i++)
document.write(arr[i] + " " );
}
function UpdateArr(arr, n)
{
let i = 0, j = n - 1;
while (i < j) {
let temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
i += 2;
j -= 2;
}
printArr(arr, n);
}
let arr = [ 1, 2, 3, 4, 5, 6 ];
let n = arr.length;;
UpdateArr(arr, n);
</script>
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Time Complexity: O(n)
Auxiliary Space: O(1)
Last Updated :
29 Jun, 2022
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