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String Interleaving in Python

Suppose we have two strings s and t, we have to find two strings interleaved, starting with first string s. If there are leftover characters in a string they will be added to the end.

So, if the input is like s = "abcd", t = "pqrstu", then the output will be "apbqcrdstu"

To solve this, we will follow these steps −

• res:= blank string
• m:= minimum of size of s, size of t
• res := res concatenate s[i] concatenate t[i]
• return res concatenate s[from index i to end] concatenate t [from index i to end]

 Live Demo

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How to Interleave Two Strings of Variable Lengths in Python?

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 :

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:

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 and characters 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:

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 a nice performance analysis. Which function performs best? I don’t want to hold the interesting results back because you may find them valuable, too:

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 itertools solution.

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 your solution!

Being able to quickly understand and write source code is a crucial skill of every single coder. Companies such as Amazon, Google, and Facebook are famously interviewing every applicant — testing their understanding and proficiency with source code. Nowadays, understanding Python code fast is one of the most valuable skills you can have as an ambitious coder.

To help you attain this valuable skill, we’ve created the “ Coffee Break Python ” book series. Check them out!

While working as a researcher in distributed systems, Dr. Christian Mayer found his love for teaching computer science students.

To help students reach higher levels of Python success, he founded the programming education website Finxter.com that has taught exponential skills to millions of coders worldwide. He’s the author of the best-selling programming books Python One-Liners (NoStarch 2020), The Art of Clean Code (NoStarch 2022), and The Book of Dash (NoStarch 2022). Chris also coauthored the Coffee Break Python series of self-published books. He’s a computer science enthusiast, freelancer , and owner of one of the top 10 largest Python blogs worldwide.

His passions are writing, reading, and coding. But his greatest passion is to serve aspiring coders through Finxter and help them to boost their skills. You can join his free email academy here.

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Interleave Strings

Given three strings A, B and C. Determine if C can be created by merging A and B in a way that maintains the relative order of the characters in A and B.

Assumptions

• none of A, B, C is null
• C = "abcde", A = "acd", B = "be", return true
• C = "abcde", A = "adc", B = "be", return false

Solution: 比较a[i]与c[i + j]，b[j]与c[i + j]，所以有2种情况

dp[0][0] = true

或者可以写成：

results matching " "

No results matching " ".

Interleaving Strings

Let's solve the Interleaving Strings problem using Dynamic Programming.

• Try it yourself
• Time complexity
• Space complexity

Given strings s1 , s2 , and s3 , find whether an interleaving of s1 and s2 forms s3 .

An interleaving of two strings a and b is a configuration where a and b splits into n n n and m m m substrings, respectively, such that:

• a = a 1 = a_{1} = a 1 ​ + a 2 a_{2} a 2 ​ + … + a n a_{n} a n ​
• b = b 1 = b_{1} = b 1 ​ + b 2 b_{2} b 2 ​ + … + b m b_{m} b m ​
• ∣ n − m ∣ ⩽ 1 \left | n - m \right | \leqslant 1 ∣ n − m ∣ ⩽ 1
• a 1 + b 1 + a 2 + b 2 + a 3 + b 3 + . . . a_{1} + b_{1} + a_{2} + b_{2} + a_{3} + b_{3} + ... a 1 ​ + b 1 ​ + a 2 ​ + b 2 ​ + a 3 ​ + b 3 ​ + ...
• b 1 + a 1 + b 2 + a 2 + b 3 + a 3 + . . . b_{1} + a_{1} + b_{2} + a_{2} + b_{3} + a_{3} + ... b 1 ​ + a 1 ​ + b 2 ​ + a 2 ​ + b 3 ​ + a 3 ​ + ...
• a x a_{x} a x ​ can be a consecutive sequence of characters in string s1 .
• b x b_{x} b x ​ can be a consecutive sequence of characters in string s2 .

Note: a + b a + b a + b is the concatenation of the strings a a a and b b b .

Let’s say we have two strings, “abc” and “xyz”. We may interleave them in multiple ways. We may alternately pick one character from each string until we have used up all the characters, giving us “axbycz” and “xaybzc”. Or, we may try to pick two characters at a time from each string, giving us “abxycz” and “xyabzc”. If we pick three characters at a time, we get “abcxyz” and “xyzabc”. Another class of variations is to pick a variable number of characters from each string, which leads to many more possibilities, such as “abxcyz”, “xayzbc”, and so on. However, “bxaycz” is not a valid interleaving, as the order of the characters taken from the first string is not preserved, since “b” appears here before “a”. Similarly, “acxybz” is not a valid interleaving.

Constraints:

• 0 ≤ 0 \leq 0 ≤ s1.length , s2.length ≤ 1500 \leq 1500 ≤ 1500
• 0 ≤ 0 \leq 0 ≤ s3.length ≤ 3000 \leq 3000 ≤ 3000
• s1 , s2 , and s3 consist of lowercase English letters.

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Answer to Question #176131 in Python for bhuvanesh

Given two dates D1 and D2, write a program to count the number of Saturdays and Sundays from D1 to D2 (including D1 and D2).

The date in string format is like "8 Feb 2021".Input

The first line of input will contain date D1 in the string format.

The second line of input will contain date D2 in the string format.Output

The output should be a single line containing two integers separated by space.Explanation

For example, if the given dates are "25 Jan 2021" and "14 Feb 2021", the Saturdays and Sundays dates from "25 Jan 2021" to "14 Feb 2021" are

"30 Jan 2021" is a Saturday

"31 Jan 2021" is a Sunday

"6 Feb 2021" is a Saturday

"7 Feb 2021" is a Sunday

"13 Feb 2021" is a Saturday

"14 Feb 2021" is a Sunday

So the output should be

Saturday: 3

Sample Input 1

25 Jan 2021

14 Feb 2021

Sample Output 1

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how do I interleave 2 arrays of different sizes?

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