Reverse an array arr[]. Reversing an array means rearranging the elements such that the first element becomes the last, the second element becomes second last and so on.
Examples:
Input: arr[] = [1, 4, 3, 2, 6, 5]
Output: [5, 6, 2, 3, 4, 1]
Explanation: The first element 1 moves to last position, the second element 4 moves to second-last and so on.
Input: arr[] = [4, 5, 1, 2]
Output: [2, 1, 5, 4]
Explanation: The first element 4 moves to last position, the second element 5 moves to second last and so on.
[Naive Approach] Using a temporary array - O(n) Time and O(n) Space
The idea is to use a temporary array to store the reverse of the array.
- Create a temporary array of same size as the original array.
- Now, copy all elements from original array to the temporary array in reverse order.
- Finally, copy all the elements from temporary array back to the original array.
Working:
C++
#include <iostream>
#include <vector>
using namespace std;
void reverseArray(vector<int> &arr) {
int n = arr.size();
// Temporary array to store elements
// in reversed order
vector<int> temp(n);
// Copy elements from original array
// to temp in reverse order
for(int i = 0; i < n; i++)
temp[i] = arr[n - i - 1];
// Copy elements back to original array
for(int i = 0; i < n; i++)
arr[i] = temp[i];
}
int main() {
vector<int> arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for(int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
#include <stdio.h>
#include <stdlib.h>
void reverseArray(int arr[], int n) {
// Temporary array to store elements
// in reversed order
int temp[n];
// Copy elements from original array
// to temp in reverse order
for(int i = 0; i < n; i++)
temp[i] = arr[n - i - 1];
// Copy elements back to original array
for(int i = 0; i < n; i++)
arr[i] = temp[i];
}
int main() {
int arr[] = { 1, 4, 3, 2, 6, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
reverseArray(arr, n);
for(int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
import java.util.Arrays;
class GfG {
static void reverseArray(int[] arr) {
int n = arr.length;
// Temporary array to store elements
// in reversed order
int[] temp = new int[n];
// Copy elements from original array
// to temp in reverse order
for (int i = 0; i < n; i++)
temp[i] = arr[n - i - 1];
// Copy elements back to original array
for (int i = 0; i < n; i++)
arr[i] = temp[i];
}
public static void main(String[] args) {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.length; i++)
System.out.print(arr[i] + " ");
}
}
def reverseArray(arr):
n = len(arr)
# Temporary array to store elements
# in reversed order
temp = [0] * n
# Copy elements from original array
# to temp in reverse order
for i in range(n):
temp[i] = arr[n - i - 1]
# Copy elements back to original array
for i in range(n):
arr[i] = temp[i]
if __name__ == "__main__":
arr = [1, 4, 3, 2, 6, 5]
reverseArray(arr)
for i in range(len(arr)):
print(arr[i], end=" ")
using System;
class GfG {
static void reverseArray(int[] arr) {
int n = arr.Length;
// Temporary array to store elements
// in reversed order
int[] temp = new int[n];
// Copy elements from original array
// to temp in reverse order
for (int i = 0; i < n; i++)
temp[i] = arr[n - i - 1];
// Copy elements back to original array
for (int i = 0; i < n; i++)
arr[i] = temp[i];
}
static void Main() {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
}
function reverseArray(arr) {
let n = arr.length;
// Temporary array to store elements
// in reversed order
let temp = new Array(n);
// Copy elements from original array
// to temp in reverse order
for (let i = 0; i < n; i++)
temp[i] = arr[n - i - 1];
// Copy elements back to original array
for (let i = 0; i < n; i++)
arr[i] = temp[i];
}
// Driver Code
const arr = [1, 4, 3, 2, 6, 5];
reverseArray(arr);
console.log(arr.join(" "));
Time Complexity: O(n), Copying elements to a new array is a linear operation.
Auxiliary Space: O(n), as we are using an extra array to store the reversed array.
[Expected Approach - 1] Using Two Pointers - O(n) Time and O(1) Space
The idea is to maintain two pointers: left and right, such that left points at the beginning of the array and right points to the end of the array.
While left pointer is less than the right pointer, swap the elements at these two positions. After each swap, increment the left pointer and decrement the right pointer to move towards the center of array. This will swap all the elements in the first half with their corresponding element in the second half.
Working:
C++
#include <iostream>
#include <vector>
using namespace std;
void reverseArray(vector<int> &arr) {
// Initialize left to the beginning
// and right to the end
int left = 0, right = arr.size() - 1;
// Iterate till left is less than right
while(left < right) {
// Swap the elements at left
// and right position
swap(arr[left], arr[right]);
// Increment the left pointer
left++;
// Decrement the right pointer
right--;
}
}
int main() {
vector<int> arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for(int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
#include <stdio.h>
// Function to swap two numbers
void swap(int *a, int *b) {
int temp = *a;
*a = *b;
*b = temp;
}
// function to reverse an array
void reverseArray(int arr[], int n) {
// Initialize left to the beginning
// and right to the end
int left = 0, right = n - 1;
// Iterate till left is less than right
while (left < right) {
// Swap the elements at left
// and right position
swap(&arr[left], &arr[right]);
// Increment the left pointer
left++;
// Decrement the right pointer
right--;
}
}
int main() {
int arr[] = { 1, 4, 3, 2, 6, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
reverseArray(arr, n);
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
import java.util.Arrays;
class GfG {
// function to reverse an array
static void reverseArray(int[] arr) {
// Initialize left to the beginning
// and right to the end
int left = 0, right = arr.length - 1;
// Iterate till left is less than right
while (left < right) {
// Swap the elements at left
// and right position
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
// Increment the left pointer
left++;
// Decrement the right pointer
right--;
}
}
public static void main(String[] args) {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.length; i++)
System.out.print(arr[i] + " ");
}
}
def reverseArray(arr):
# Initialize left to the beginning
# and right to the end
left = 0
right = len(arr) - 1
# Iterate till left is less than right
while left < right:
# Swap the elements at left
# and right position
arr[left], arr[right] = arr[right], arr[left]
# Increment the left pointer
left += 1
# Decrement the right pointer
right -= 1
if __name__ == "__main__":
arr = [1, 4, 3, 2, 6, 5]
reverseArray(arr)
for i in range(len(arr)):
print(arr[i], end=" ")
using System;
class GfG {
static void reverseArray(int[] arr) {
// Initialize left to the beginning
// and right to the end
int left = 0, right = arr.Length - 1;
// Iterate till left is less than right
while (left < right) {
// Swap the elements at left
// and right position
int temp = arr[left];
arr[left] = arr[right];
arr[right] = temp;
// Increment the left pointer
left++;
// Decrement the right pointer
right--;
}
}
static void Main() {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
}
function reverseArray(arr) {
// Initialize left to the beginning
// and right to the end
let left = 0, right = arr.length - 1;
// Iterate till left is less than right
while (left < right) {
// Swap the elements at left
// and right position
[arr[left], arr[right]] = [arr[right], arr[left]];
// Increment the left pointer
left++;
// Decrement the right pointer
right--;
}
}
//Driver Code
const arr = [1, 4, 3, 2, 6, 5];
reverseArray(arr);
console.log(arr.join(" "));
[Expected Approach - 2] By Swapping Elements - O(n) Time and O(1) Space
The idea is to iterate over the first half of the array and swap each element with its corresponding element from the end. So, while iterating over the first half, any element at index i is swapped with the element at index (n - i - 1).
Working:
C++
#include <iostream>
#include <vector>
using namespace std;
void reverseArray(vector<int> &arr) {
int n = arr.size();
// Iterate over the first half and
// for every index i,
// swap arr[i] with arr[n - i - 1]
for(int i = 0; i < n/2; i++) {
swap(arr[i], arr[n - i - 1]);
}
}
int main() {
vector<int> arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for(int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
#include <stdio.h>
void reverseArray(int arr[], int n) {
// Iterate over the first half
// and for every index i, swap
// arr[i] with arr[n - i - 1]
for (int i = 0; i < n / 2; i++) {
int temp = arr[i];
arr[i] = arr[n - i - 1];
arr[n - i - 1] = temp;
}
}
int main() {
int arr[] = { 1, 4, 3, 2, 6, 5 };
int n = sizeof(arr) / sizeof(arr[0]);
reverseArray(arr, n);
for (int i = 0; i < n; i++)
printf("%d ", arr[i]);
return 0;
}
import java.util.Arrays;
class GfG {
static void reverseArray(int[] arr) {
int n = arr.length;
// Iterate over the first half
// and for every index i, swap
// arr[i] with arr[n - i - 1]
for (int i = 0; i < n / 2; i++) {
int temp = arr[i];
arr[i] = arr[n - i - 1];
arr[n - i - 1] = temp;
}
}
public static void main(String[] args) {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.length; i++)
System.out.print(arr[i] + " ");
}
}
def reverseArray(arr):
n = len(arr)
# Iterate over the first half
# and for every index i, swap
# arr[i] with arr[n - i - 1]
for i in range(n // 2):
temp = arr[i]
arr[i] = arr[n - i - 1]
arr[n - i - 1] = temp
if __name__ == "__main__":
arr = [1, 4, 3, 2, 6, 5]
reverseArray(arr)
for i in range(len(arr)):
print(arr[i], end=" ")
using System;
class GfG {
static void reverseArray(int[] arr) {
int n = arr.Length;
// Iterate over the first half
// and for every index i, swap
// arr[i] with arr[n - i - 1]
for (int i = 0; i < n / 2; i++) {
int temp = arr[i];
arr[i] = arr[n - i - 1];
arr[n - i - 1] = temp;
}
}
static void Main() {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
}
function reverseArray(arr) {
let n = arr.length;
// Iterate over the first half
// and for every index i, swap
// arr[i] with arr[n - i - 1]
for (let i = 0; i < n / 2; i++) {
let temp = arr[i];
arr[i] = arr[n - i - 1];
arr[n - i - 1] = temp;
}
}
// Driver Code
const arr = [1, 4, 3, 2, 6, 5];
reverseArray(arr);
console.log(arr.join(" "));
Time Complexity: O(n), the loop runs through half of the array, so it's linear with respect to the array size.
Auxiliary Space: O(1), no extra space is required, therefore we are reversing the array in-place.
Using Inbuilt Methods - O(n) Time and O(1) Space
The idea is to use inbuilt reverse methods available across different languages.
C++
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// function to reverse an array
void reverseArray(vector<int> &arr) {
reverse(arr.begin(), arr.end());
}
int main() {
vector<int> arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for(int i = 0; i < arr.size(); i++)
cout << arr[i] << " ";
return 0;
}
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
class GfG {
// function to reverse an array
static void reverseArray(List<Integer> arr) {
Collections.reverse(arr);
}
public static void main(String[] args) {
List<Integer> arr =
new ArrayList<>(Arrays.asList(1, 4, 3, 2, 6, 5));
reverseArray(arr);
for (int i = 0; i < arr.size(); i++)
System.out.print(arr.get(i) + " ");
}
}
# function to reverse an array
def reverse_array(arr):
arr.reverse()
if __name__ == "__main__":
arr = [1, 4, 3, 2, 6, 5]
reverse_array(arr)
print(" ".join(map(str, arr)))
using System;
class GfG {
// function to reverse an array
static void reverseArray(int[] arr) {
Array.Reverse(arr);
}
static void Main() {
int[] arr = { 1, 4, 3, 2, 6, 5 };
reverseArray(arr);
for (int i = 0; i < arr.Length; i++)
Console.Write(arr[i] + " ");
}
}
// function to reverse an array
function reverseArray(arr) {
arr.reverse();
}
// Driver code
const arr = [1, 4, 3, 2, 6, 5];
reverseArray(arr);
console.log(arr.join(" "));
Time Complexity: O(n), the reverse method has linear time complexity.
Auxiliary Space: O(1) Additional space is not used to store the reversed array, as the in-built array method swaps the values in-place.
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