Space complexity refers to the amount of additional memory an operation uses as the input size increases. In the case of basic array operations, most operations are performed in-place — meaning they modify the array without allocating new memory. Below, we analyze the space complexity of five fundamental operations: insertion, deletion, traversal, update, and search.
Summary Table
| Operation | Space Complexity | Why? |
|---|---|---|
| Insertion | O(1) | Done in-place without using extra memory |
| Deletion | O(1) | Removes item by shifting elements, no extra storage used |
| Traversal | O(1) | Uses only a loop counter or index variable |
| Update | O(1) | Modifies existing data at a specific index |
| Search | O(1) | No new data structures are used during the search |
1. Insertion
Inserting an element into an array involves placing the new value at a specific index and shifting other elements if necessary. The key point is that this operation doesn't require allocating any new memory.
- Space Complexity:
O(1) - Why: The insertion is done within the existing array block. Only a temporary variable may be used to hold the new value — which is constant in size.
2. Deletion
Deletion removes an element at a given index and shifts the remaining elements to the left to maintain order. It works directly on the original array.
- Space Complexity:
O(1) - Why: The operation doesn’t require any additional space. Elements are just overwritten using the same memory block.
3. Traversal
Traversal means iterating over each element of the array to read or process its contents.
- Space Complexity:
O(1) - Why: Only a loop index variable is needed to move through the array — the memory used does not depend on array size.
4. Update
Updating an array element refers to changing the value at a specific index. This does not create new memory, it simply replaces the old value.
- Space Complexity:
O(1) - Why: The update is done at the existing memory location. There’s no need to allocate or copy anything.
5. Search
Searching (linear or binary) involves scanning the array or dividing it (if sorted), but neither case requires creating new data structures.
- Space Complexity:
O(1) - Why: Whether comparing linearly or via mid-pointers, the algorithm only uses a few constant-size variables (like index or start/end pointers).