This content originally appeared on DEV Community and was authored by Kathan Vakharia
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 ❓
- Possible Explanation ?
- Documented C++ Code ?
- Time and Space Complexity Analysis ⌛?
The Question
? 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) → (second, second-last) → (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
? If you are wondering why there's
<instead of≤? It's becausestartwill be equal toendonly 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,startandendboth 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 thanend(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.
This content originally appeared on DEV Community and was authored by Kathan Vakharia
Kathan Vakharia | Sciencx (2021-07-22T11:13:11+00:00) Arrays Questions: Reverse an array. Retrieved from https://www.scien.cx/2021/07/22/arrays-questions-reverse-an-array/
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