np.savez() for multiple arrays in NumPy - Time & Space Complexity

Time Complexity: np.savez() for multiple arrays

O(n)

Understanding Time Complexity

We want to understand how the time to save multiple arrays with np.savez() changes as the number and size of arrays grow.

How does saving more or bigger arrays affect the time it takes?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

import numpy as np

arr1 = np.arange(1000)
arr2 = np.arange(2000)
arr3 = np.arange(3000)

np.savez('arrays.npz', a=arr1, b=arr2, c=arr3)

This code saves three numpy arrays of different sizes into one uncompressed file.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Writing each element of every array to disk.
• How many times: Once for each element in all arrays combined.

How Execution Grows With Input

As the total number of elements across all arrays grows, the time to save grows roughly the same way.

Input Size (total elements)	Approx. Operations
10	About 10 write operations
100	About 100 write operations
1000	About 1000 write operations

Pattern observation: The time grows roughly in direct proportion to the total number of elements saved.

Final Time Complexity

Time Complexity: O(n)

This means the time to save grows linearly with the total number of elements in all arrays combined.

Common Mistake

[X] Wrong: "Saving multiple arrays with np.savez() takes constant time regardless of array size."

[OK] Correct: The function must write every element to disk, so more data means more time.

Interview Connect

Understanding how saving data scales helps you reason about performance in real projects where data size varies.

Self-Check

What if we compressed the arrays before saving? How would the time complexity change?