**Array-DS**

### Problem Statement :

An array is a type of data structure that stores elements of the same type in a contiguous block of memory. In an array, A, of size N, each memory location has some unique index, i (where 0<=i<N), that can be referenced as A[i] or Ai. Reverse an array of integers. Note: If you've already solved our C++ domain's Arrays Introduction challenge, you may want to skip this. Example: A=[1,2,3] Return [3,2,1]. Function Description: Complete the function reverseArray in the editor below. reverseArray has the following parameter(s): 1. int A[n]: the array to reverse Returns 1 .int[n]: the reversed array Input Format: The first line contains an integer, N, the number of integers in A. The second line contains N space-separated integers that make up A. Constraints: 1<=N<=10^3 1<=N<=10^4 where A[i] is ith integer in A.

### Solution :

` ````
Solution in C :
In C:
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
int main() {
int n, new_number;
scanf("%d", &n);
int *p_numbers = (int*)malloc(n * sizeof(int));
int i = 0;
while (i < n) {
scanf("%d", &new_number);
*(p_numbers + i) = new_number;
i++;
}
while (i >= 1) {
printf("%d ", *(p_numbers + --i));
}
free(p_numbers);
return 0;
}
In C++:
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int N;
cin>>N;
vector<int> arr(N);
for (int i=0;i<N;i++)
cin>>arr[i];
reverse(arr.begin(),arr.end());
for (int i=0;i<N;i++)
cout<<arr[i]<<" ";
return 0;
}
In Java:
import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int length = sc.nextInt();
int[] array = new int[length];
for (int i = 0; i < length ; i++) {
array[i] = sc.nextInt();
}
for (int i = 0; i < length; i++){
System.out.print(array[length-i-1] + " ");
}
}
}
In Python 3:
n = int(input())
m = input().strip().split(' ')
res = m[::-1]
r=""
for i in range(n):
r = r + str(res[i]) + " "
print(r)
```

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