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Kathan Vakharia
Kathan Vakharia

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Arrays Questions: Reverse an array

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,

  1. Question Link โ“
  2. Possible Explanation ๐Ÿ“
  3. Documented C++ Code ๐Ÿงน
  4. Time and Space Complexity Analysis โŒ›๐ŸŒŒ

The Question

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


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
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๐Ÿ’ก If you are wondering why there's < instead of โ‰ค ? It's because start will be equal to end only in the case of odd length arrays and they both will point to the middle element 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



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;
    int arr[n];
    for(int i=0; i<n; i++){

    reverse(arr, 0, n-1);

    for(auto val: arr){

    return 0;
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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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