In this series of posts, I will discuss coding questions on the `Arrays`

Data structure.

The posts in this series will be organized in the following way,

- Question Link โ
- Possible Explanation ๐
- Documented C++ Code ๐งน
- Time and Space Complexity Analysis โ๐

## The Question

๐ก Give yourself at least 15-20 mins to figure out the solution :)

## Explanation

The idea is to maintain two index variables `start`

and `end`

, initially pointing to the *first* element and *last* element of the array respectively.

And then, we will *swap* values in the following order: **(first, last)** โ **(second, second-last)** โ **(third, third, third-last)** ....until we reach the *middle* element of the array.

Here's the pseudo-code,

```
while start < end:
swap(arr[start], arr[end])
start = start + 1 //move start ahead by one step
end = end - 1 //move end back by one step
```

๐ก If you are wondering why there's

`<`

instead of`โค`

? It's because`start`

will be equal to`end`

only in the case ofoddlength arrays and they both will point to themiddleelement of the array. And it does no good to swap it with itself as the array is already reversed by then.

### Still confused?

Assume index starts from zero.

Think what happens when

`arr.length = 3`

(*odd*), after one iteration,`start`

and`end`

both will point to index=1 and array is already reversed.Think what happens when

`arr.length = 4`

(*even*), after two iteration,`start(2)`

will be greater than`end(1)`

and array will be reversed.

## C++ Code

### Solution

```
#include<iostream>
using namespace std;
void reverse(int* arr, int start, int end){
//untill we reach the middle
while(start < end){
//swap arr[start] and arr[end]
int temp = arr[start];
arr[start] = arr[end];
arr[end] = temp;
start++;//move start ahead
end--;//move end back
}
}
//driver code
int main(){
int n;
cin>>n;
int arr[n];
for(int i=0; i<n; i++){
cin>>arr[i];
}
reverse(arr, 0, n-1);
for(auto val: arr){
cout<<val<<"\n";
}
return 0;
}
```

## Complexity Analysis

`N`

: length of the array

### Time Complexity: O(N)

Since we are iterating nearly N/2 times, thus time will be O(N/2) = O(N).

### Space Complexity: O(1)

We didn't use any extra space.

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