Half an hour ago, my friend and coauthor of the textbook “Coffee Break NumPy” asked me the following question via WhatsApp:

## Problem Formulation

How would you solve the problem of interleaving two strings in Python:

**Input**: String`s1= "AAA"`

and string`s2 = "BBBBB"`

**Output**: String`s="ABABABBB"`

Being obsessed with finding the most Pythonic way of writing any code snippet (preferably within a single line of code), I quickly became frustrated because there doesn’t seem to be a very simple, clean, and concise answer to this question.

However, in this article, you’ll learn a robust and easy-to-understand way of solving this problem (without external library support). So keep on reading.

## Alternative 1: First String s1 is Shorter

Assuming the first string is shorter gives us the opportunity to solve the problem in a Python one-liner using list comprehension:

s1 = "AAA" s2 = "BBBBB" s = "".join([s1[i] + s2[i] for i in range(len(s1))]) + s2[len(s1):] print(s) # ABABABBB

Because of Python’s efficient implementation of list comprehension, this option is extremely fast — I doubt that there is anything faster (which is still equally readable).

We combine every character of the shorter string `s1`

with the character of the longer string `s2`

at the respective position. This results in the partially interleaved string `"ABABAB"`

. Now, we simply concatenate this with the remaining characters of the longer string `s2`

.

However, this solution doesn’t work if string `s1`

can also be longer than string `s2`

.

Why? Because the Python interpreter will raise an **Index Error **as accessing `s2[i]`

is not possible.

## Alternative 2: Any String May Be Longer

If you don’t assume that one of the string is longer than the other, the problem becomes slightly harder. Still, there is a simple and clean solution to this problem (without using external libraries). It’s not in a single line of code, but it’s readable, fast, and it doesn’t need any length assumptions:

s1 = "AAA" s2 = "BBBBB" s = list(s2) for i,c in enumerate(s1): s.insert(i*2,c) print("".join(s)) # ABABABBB

First, we convert the string `s2`

to a list of characters using the `list(...)`

function. This is the basis of our solution.

Second, we insert the characters of the string `s1`

at positions 0, 2, 4, … by iterating over all indices `i`

`c`

of the first string `s1`

. Now we insert the characters into every other position of the list.

## Alternative 3: Using External Libraries

Expert coders heavily use external libraries because it makes their code more readable, more efficient, and shorter. What’s wrong with that? Here is what an expert reader David of my (free) “Coffee Break Python” email course proposed:

import itertools s1 = "AAA" s2 = "BBBBB" s = "".join([ x + y for x, y in itertools.zip_longest(s1, s2, fillvalue="")]) print(s) # ABABABBB

The problem with taking the built-in `zip()`

function is that the number of pairs returned by the `zip()`

function is equal to the shorter iterable.

Here is what my loyal reader David argues:

*[…] zip_longest() vaults the (built-in) zip()‘s ‘limitation’ of cutting-off at the shorter len() […]. It ‘extends’ the shorter iterable with a fillvalue parameter – using [the empty string] rather than the default None, otherwise the subsequent string concatenation will fail!*

Again, if library support is allowed (in other words: you are not in a coding interview), this is my preferred solution.

## Performance Measurements

After publishing this article, my coauthor Lukas (book “Coffee Break NumPy”) came back to me with

import itertools import matplotlib.pyplot as plt plt.xkcd() def interleave_strings_listcomprehension(s1, s2): return "".join([s1[i] + s2[i] for i in range(len(s1))]) + s2[len(s1):] def interleave_strings_enumerate(s1, s2): s = list(s2) for i, c in enumerate(s1): s.insert(i*2, c) return "".join(s) def interleave_strings_slicing(s1, s2): length_s1 = len(s1) length_s2 = len(s2) if length_s1 != length_s2: if length_s1 > length_s2: spaces_count = length_s1 - length_s2 s2 = s2 + spaces_count * ' ' else: spaces_count = length_s2 - length_s1 s1 = s1 + spaces_count * ' ' interleaved = len(s1) * 2 * [''] interleaved[::2] = s1 interleaved[1::2] = s2 return ''.join(interleaved).replace(' ', '') def interleave_strings_zip(s1, s2): length_s1 = len(s1) length_s2 = len(s2) if length_s1 != length_s2: if length_s1 > length_s2: spaces_count = length_s1 - length_s2 s2 = s2 + spaces_count * ' ' else: spaces_count = length_s2 - length_s1 s1 = s1 + spaces_count * ' ' return "".join(i + j for i, j in zip(s1, s2)).replace(' ', '') def interleave_zip_itertools(s1, s2): import itertools return "".join([ x + y for x, y in itertools.zip_longest(s1, s2, fillvalue="")]) import time multiplicator = 1000 s1 = multiplicator * "AAA" s2 = multiplicator * "BBBB" # Test 1 start = time.perf_counter() interleave_strings_listcomprehension(s1, s2) end = time.perf_counter() plt.bar(1,end - start, hatch=" ", label="List comprehension (Alt 1)") # Test 2 start = time.perf_counter() interleave_strings_enumerate(s1, s2) end = time.perf_counter() plt.bar(2,end - start, hatch="o", label="Enumerate (Alt 2)") # Test 3 start = time.perf_counter() interleave_strings_slicing(s1, s2) end = time.perf_counter() plt.bar(3,end - start, hatch="+", label="Slicing") # Test 4 start = time.perf_counter() interleave_strings_zip(s1, s2) end = time.perf_counter() plt.bar(4,end - start, hatch="/", label="Zip") # Test 5 start = time.perf_counter() interleave_zip_itertools(s1, s2) end = time.perf_counter() plt.bar(5,end - start, hatch="-", label="Zip Itertools (Alt 3)") plt.xticks((),()) plt.ylabel("nanosecs") plt.legend() plt.tight_layout() plt.savefig("plot.jpg") plt.show()

Here is the resulting bar plot comparing the runtime of the different functions:

The slicing function outperformed any other function by at least 50%! I knew that slicing is fast but this result blew my mind. I have also tested the result for even larger strings but slicing still seems to be the fastest alternative. It comes at the cost that readability suffers a bit compared to the

## Where to Go From Here?

If you feel like you have a good solution that will be interesting for the readers of this article, leave a comment below with

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