Merge Strings Alternately
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Here’s the breakdown:
- You’ve first determined the lengths of
word1
andword2
and stored them inm
andn
respectively. i
andj
are initialized to 0 and serve as indices forword1
andword2
respectively.- An empty list
result
is initialized to hold the merged string. - A while loop runs as long as
i
is less thanm
orj
is less thann
. - Inside the while loop, you first check if
i
is less thanm
. If it is, you appendword1[i]
toresult
and incrementi
by 1. This handles the characters fromword1
. - You then check if
j
is less thann
. If it is, you appendword2[j]
toresult
and incrementj
by 1. This handles the characters fromword2
. - After the loop, you join the list
result
into a string and return it.
This solution ensures that the characters from word1
and word2
are added to result
in alternating order and handles cases where word1
and word2
are of different lengths.
What are basic building blocks of this solution?
The basic building blocks of this solution are:
Iteration: The solution iterates over both strings simultaneously, examining each character. The indices
i
andj
are used to track the current position inword1
andword2
respectively.Condition Checking: The solution includes conditions to check whether the end of each string has been reached. This is achieved by comparing
i
andj
withm
andn
, the lengths ofword1
andword2
respectively.List Operations: The solution uses a list to store the characters of the merged string. It uses the
+=
operator to append characters to the list and thejoin()
method to combine all the characters into a single string at the end.String Indexing: The solution uses string indexing to access individual characters from the input strings. This is done with
word1[i]
andword2[j]
.
These are the fundamental operations that the solution uses to solve the problem. The overall strategy is to perform an alternating selection from each string, making sure to handle the case where the strings are of different lengths.
Slightly above this level of abstraction is Two Read Pointers, can you identify more?
In addition to the Two Read Pointers concept you’ve already identified, here are two more abstract principles or strategies at work in this solution:
Merge Operations: This solution involves a merge operation, a common technique used in problems involving multiple ordered or unordered sequences. In this case, the sequences are the individual characters in two strings. The merging happens in an alternating manner between the two strings.
Handling Unequal Lengths: The problem requires that we handle strings of unequal lengths. This is addressed by making the iteration continue as long as there are elements remaining in either string, which is a strategy often employed when dealing with sequences of potentially unequal lengths.
These two concepts, along with the use of two read pointers, form the basis of the strategy employed in this solution.
What is the invariant in this problem?
An invariant in this context refers to a condition that remains unchanged throughout the execution of the program.
In this problem, the invariant is that the resulting merged string at any point during the iteration, when separated back into two parts, would be prefix segments of the original word1
and word2
respectively, in their original order.
More concretely, if we stop at any point during the iteration, and separate the elements contributed by word1
and word2
in result
, they would look like word1[:i]
and word2[:j]
respectively, where i
and j
are the current positions in word1
and word2
.
This condition holds true throughout the execution, making it the invariant of this problem.
What is the loop invariant in this problem?
A loop invariant in this problem is a condition or property that remains true before and after each iteration of the loop.
In this case, the loop invariant is similar to the invariant we discussed before:
Before and after each iteration of the loop, if we break down the characters in the result
list into parts contributed by word1
and word2
, they would be prefix segments of the original word1
and word2
in their original order.
So, before the loop begins and after each iteration, the characters from word1
in the result
list are word1[:i]
and those from word2
are word2[:j]
, where i
and j
are the current positions in word1
and word2
respectively.
Therefore, we can say that the loop maintains the property of result
being an interlaced version of the prefixes of word1
and word2
in their original order, which is our loop invariant.
Is invariant and loop invariant the same for this problem?
Yes, in this specific problem, the invariant and the loop invariant are the same.
This is not always the case in all problems. Invariant refers to a property that remains unchanged throughout the entire execution of the program, while loop invariant is a condition or property that holds true before and after each iteration of a loop.
In this problem, the condition that result
being an interlaced version of the prefixes of word1
and word2
in their original order holds true throughout the program and before and after each loop iteration. Therefore, it serves as both an invariant and a loop invariant in this context.