Given an array arr[] and a position in array, k. Write a function name reverse (a[], k) such that it reverses subarray arr[0..k-1]. Extra space used should be O(1) and time complexity should be O(k).
Example:
Input:
arr[] = {1, 2, 3, 4, 5, 6}
k = 4
Output:
arr[] = {4, 3, 2, 1, 5, 6}
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Below is the implementation for the same.
// C++ program to reverse a subarray arr[0..k-1]
#include <bits/stdc++.h>
using namespace std;
// Reverse subarray a[0..k-1]
void reverse(int a[], int n, int k)
{
if (k > n)
{
cout << "Invalid k";
return;
}
// One by one reverse first and last elements of a[0..k-1]
for (int i = 0; i < k/2; i++)
swap(a[i], a[k-i-1]);
}
// Driver program
int main()
{
int a[] = {1, 2, 3, 4, 5, 6};
int n = sizeof(a) / sizeof(int), k = 4;
reverse(a, n, k);
for (int i = 0; i < n; ++i)
printf("%d ", a[i]);
return 0;
}
// java program to reverse a
// subarray arr[0..k-1]
public class GFG {
// Reverse subarray a[0..k-1]
static void reverse(int []a, int n, int k)
{
if (k > n)
{
System.out.println( "Invalid k");
return;
}
// One by one reverse first
// and last elements of a[0..k-1]
for (int i = 0; i < k / 2; i++)
{
int tempswap = a[i];
a[i] = a[k - i - 1];
a[k - i - 1] = tempswap;
}
}
// Driver code
public static void main(String args[])
{
int []a = {1, 2, 3, 4, 5, 6};
int n = a.length, k = 4;
reverse(a, n, k);
for (int i = 0; i < n; ++i)
System.out.print(a[i] + " ");
}
}
// This code is contributed by Sam007.
# python program to reverse a subarray
# arr[0..k-1]
from __future__ import print_function
# Reverse subarray a[0..k-1]
def reverse(a, n, k):
if (k > n):
print( "Invalid k")
return
# One by one reverse first and
# last elements of a[0..k-1]
for i in range(0, (int)(k/2)):
temp = a[i]
a[i] = a[k-i-1]
a[k-i-1] = temp
# Driver program
a = [1, 2, 3, 4, 5, 6]
n = len(a)
k = 4
reverse(a, n, k);
for i in range(0, n):
print(a[i], end=" ")
# This code is contributed by Sam007.
// C# program to reverse a
// subarray arr[0..k-1]
using System;
class GFG {
static void SwapNum(ref int x, ref int y)
{
int tempswap = x;
x = y;
y = tempswap;
}
// Reverse subarray a[0..k-1]
static void reverse(int []a, int n,
int k)
{
if (k > n)
{
Console.Write( "Invalid k");
return;
}
// One by one reverse first
// and last elements of a[0..k-1]
for (int i = 0; i < k / 2; i++)
SwapNum(ref a[i], ref a[k - i - 1]);
}
// Driver Code
public static void Main()
{
int []a = {1, 2, 3, 4, 5, 6};
int n = a.Length, k = 4;
reverse(a, n, k);
for (int i = 0; i < n; ++i)
Console.Write(a[i] + " ");
}
}
// This code is contributed by Sam007
<script>
// Javascript program to reverse
// a subarray arr[0..k-1]
// Reverse subarray a[0..k-1]
function reverse( a, n, k)
{
if (k > n)
{
document.write("Invalid k");
return;
}
// One by one reverse first
// and last elements of a[0..k-1]
for (let i = 0; i < Math.floor(k/2); i++)
{
let temp = a[i] ;
a[i] = a[k-i-1] ;
a[k-i-1] = temp ;
}
}
// driver code
let a = [1, 2, 3, 4, 5, 6];
let n = a.length, k = 4;
reverse(a, n, k);
for (let i = 0; i < n; ++i)
document.write(a[i] + " ");
</script>
Output
4 3 2 1 5 6
Time complexity: O(k)
Auxiliary Space: O(1) ,since extra space is used.
Method 2 (using STL):
In this method we will use an in-built C++ STL function named reverse. This function completes the task of reversing K elements of array in O(K) time and also doesn't use extra space.
implementation of this method is below.
// C++ program to reverse the first K
// elements using in-built function
#include <bits/stdc++.h>
using namespace std;
int main()
{
int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8 };
int k = 4;
// STL function to reverse element
// from 0 index to K-1 index.
reverse(arr, arr + k);
// printing the array after reversing
// first K elements.
for (int i = 0; i < 8; i++) {
cout << arr[i] << " ";
}
return 0;
}
// this code is contributed by Machhaliya Muhammad
// Java program to reverse the first K
// elements using in-built function
import java.io.*;
import java.util.*;
import java.util.Arrays;
class GFG {
public static void main (String[] args) {
Integer[] arr = { 1, 2, 3, 4, 5, 6, 7, 8 };
int k = 4;
// Java Library function to reverse element
// from 0 index to K-1 index.
Integer[] arr1 = Arrays.copyOfRange(arr, 0, k);
Collections.reverse(Arrays.asList(arr1));
System.arraycopy(arr1, 0, arr, 0, k);
// printing the array after reversing
// first K elements.
for (int i = 0; i < 8; i++) {
System.out.print(arr[i] + " ");
}
}
}
// This code is contributed by Aman Kumar.
# Python3 program to reverse the first K
# elements using in-built function
arr = [ 1, 2, 3, 4, 5, 6, 7, 8 ]
k = 4
# Using list slicing to reverse the array
# from 0 index to K-1 index.
arr[:k] = arr[:k][::-1]
# printing the array after reversing
# first K elements.
print(*arr)
# This code is contributed by phasing17
// C# program to reverse the first K
// elements using in-built function
using System;
using System.Collections.Generic;
class GFG
{
public static void Main(string[] args)
{
int[] arr = { 1, 2, 3, 4, 5, 6, 7, 8 };
int k = 4;
// C# Library function to reverse element
// from 0 index to K-1 index.
Array.Reverse(arr, 0, k);
// printing the array after reversing
// first K elements.
for (int i = 0; i < 8; i++) {
Console.Write(arr[i] + " ");
}
}
}
// this code is contributed by phasing17
// JavaScript program to reverse the first K
// elements using in-built functions
let arr = [ 1, 2, 3, 4, 5, 6, 7, 8 ];
let k = 4;
// Library function to reverse element
// from 0 index to K-1 index.
let arr1 = arr.slice(0, k);
arr1.reverse();
arr.splice(0, k, ...arr1);
// printing the array after reversing
// first K elements.
for (var i = 0; i < 8; i++)
process.stdout.write(arr[i] + " ");
// This code is contributed by phasing17
Output
4 3 2 1 5 6 7 8
Time Complexity :O(K) , as complexity of reverse() function is O(number of elements of array to be sorted).
Auxiliary Space :O(1), as no extra space used.