1662. Check If Two String Arrays are Equivalent
Problem Description
This problem presents two arrays, word1
and word2
, each containing strings. The task is to determine if these two arrays represent the same string when their contents are concatenated together. In other words, if we join all the elements of word1
end-to-end to make a single string and do the same with word2
, and those two resulting strings are identical, we should return true
. Otherwise, we will return false
. It is essential to concatenate the elements in the order they appear in their respective arrays.
Intuition
To solve this problem, we rely on the simple property that strings are equal if they contain the same sequence of characters in the same order. Therefore, the solution approach is straightforward:
- Concatenate all elements in
word1
to form a single string. - Concatenate all elements in
word2
to form another single string. - Compare these two strings for equality.
If both strings are equal, it means the arrays word1
and word2
represent the same string; thus, we return true
. If they differ, the arrays represent different strings, so we return false
. This is achieved with minimal code by using the join
method in Python, which concatenates the elements of a list into a string, separated by the string provided to join
(in this case, an empty string as we don't want any characters between the elements).
Solution Approach
The implementation of the solution is straightforward and elegant, thanks to Python's high-level string handling capabilities. The algorithm does not rely on complex data structures or patterns; it primarily uses the built-in string functionality provided by Python.
Here's a step-by-step walk-through of the arrayStringsAreEqual
function within the Solution
class:
-
The
join
method is called on an empty string (''
) withword1
as the argument. Thejoin
method takes all elements inword1
, which are strings themselves, and concatenates them in the order they appear in the array. This results in a single string that represents the concatenation of all individual strings inword1
.joined_word1 = ''.join(word1)
-
The same process is applied to
word2
:joined_word2 = ''.join(word2)
-
Now, we have two strings represented by
joined_word1
andjoined_word2
. All that remains is to check whether these two strings are identical.result = joined_word1 == joined_word2
-
The result of this comparison is a boolean (
True
orFalse
). The function directly returns this result, completing the check with a single line of code:return joined_word1 == joined_word2
This approach takes full advantage of Python's ability to handle strings and perform operations on lists. It effectively reduces what could be a more complex algorithm involving manual iteration and concatenation to a simple one-liner that is easy to understand and maintain.
Ready to land your dream job?
Unlock your dream job with a 2-minute evaluator for a personalized learning plan!
Start EvaluatorExample Walkthrough
Let's consider an example where word1 = ["ab", "c"]
and word2 = ["a", "bc"]
. We need to follow the described solution approach to determine if these two arrays represent the same string when concatenated.
-
First, we concatenate the elements of
word1
using thejoin
method on an empty string.joined_word1 = ''.join(["ab", "c"]) # This evaluates to "abc"
After applying the
join
method, we end up with the string"abc"
. -
Next, we concatenate the elements of
word2
.joined_word2 = ''.join(["a", "bc"]) # This evaluates to "abc"
Similarly, the
join
method results in the string"abc"
forword2
. -
Now we have both strings obtained from
word1
andword2
respectively. We will now compare these two strings to check if they are identical:result = joined_word1 == joined_word2 # This evaluates to True
In this case,
joined_word1
is"abc"
andjoined_word2
is also"abc"
. The comparison yieldsTrue
. -
Finally, we will directly return the result of the comparison:
return result # This returns True
Since
result
wasTrue
, the function would returnTrue
, indicating thatword1
andword2
do indeed represent the same string when concatenated.
Therefore, given the inputs word1 = ["ab", "c"]
and word2 = ["a", "bc"]
, the function arrayStringsAreEqual
will return True
, as the concatenated strings from both arrays are identical.
Solution Implementation
1from typing import List # Import the List type from typing module for type annotations
2
3class Solution:
4 def array_strings_are_equal(self, word1: List[str], word2: List[str]) -> bool:
5 """
6 Checks if the strings from two lists are equal when concatenated.
7
8 Parameters:
9 word1 (List[str]): First list of strings.
10 word2 (List[str]): Second list of strings.
11
12 Returns:
13 bool: True if the concatenated strings are equal, False otherwise.
14 """
15 # Concatenate all strings in the first list and compare with the concatenation of the second list
16 return ''.join(word1) == ''.join(word2)
17
1class Solution {
2
3 /**
4 * Checks if two string arrays are equal when their elements are concatenated.
5 * @param wordArray1 The first array of strings.
6 * @param wordArray2 The second array of strings.
7 * @return true if the concatenated strings are equal, false otherwise.
8 */
9 public boolean arrayStringsAreEqual(String[] wordArray1, String[] wordArray2) {
10 // Join the elements of the first array into a single string
11 String concatenatedWord1 = String.join("", wordArray1);
12
13 // Join the elements of the second array into a single string
14 String concatenatedWord2 = String.join("", wordArray2);
15
16 // Compare the two concatenated strings for equality
17 return concatenatedWord1.equals(concatenatedWord2);
18 }
19}
20
1#include <vector>
2#include <string>
3#include <numeric> // Required for std::accumulate
4
5class Solution {
6public:
7 bool arrayStringsAreEqual(vector<string>& word1, vector<string>& word2) {
8 // Convert the first vector of strings into a single string
9 std::string concatenatedWord1 = std::accumulate(word1.begin(), word1.end(), std::string(""));
10
11 // Convert the second vector of strings into a single string
12 std::string concatenatedWord2 = std::accumulate(word2.begin(), word2.end(), std::string(""));
13
14 // Compare the two concatenated strings and return whether they are equal
15 return concatenatedWord1 == concatenatedWord2;
16 }
17};
18
1// This function checks if two arrays of strings are equivalent after combining their elements.
2// @param {string[]} firstWordArray - The first array of strings to be compared.
3// @param {string[]} secondWordArray - The second array of strings to be compared.
4// @return {boolean} - Returns true if the concatenated strings are equal, otherwise false.
5function arrayStringsAreEqual(firstWordArray: string[], secondWordArray: string[]): boolean {
6 // Concatenate the elements of the first array into a single string
7 const firstCombinedString = firstWordArray.join('');
8
9 // Concatenate the elements of the second array into a single string
10 const secondCombinedString = secondWordArray.join('');
11
12 // Compare the concatenated strings for equality and return the result
13 return firstCombinedString === secondCombinedString;
14}
15
Time and Space Complexity
The time complexity of the code is O(m + n)
where m
is the total number of characters in word1
and n
is the total number of characters in word2
. This is because the join operations for word1
and word2
each iterate over their respective arrays to build a single string, which takes linear time relative to the number of characters in each array.
The space complexity of the code is O(m + n)
as well, since it needs to allocate space for the new strings generated by ''.join(word1)
and ''.join(word2)
. No additional space is required beyond the strings themselves.
Learn more about how to find time and space complexity quickly using problem constraints.
Which technique can we use to find the middle of a linked list?
Recommended Readings
LeetCode Patterns Your Personal Dijkstra's Algorithm to Landing Your Dream Job The goal of AlgoMonster is to help you get a job in the shortest amount of time possible in a data driven way We compiled datasets of tech interview problems and broke them down by patterns This way we
Recursion Recursion is one of the most important concepts in computer science Simply speaking recursion is the process of a function calling itself Using a real life analogy imagine a scenario where you invite your friends to lunch https algomonster s3 us east 2 amazonaws com recursion jpg You first
Runtime Overview When learning about algorithms and data structures you'll frequently encounter the term time complexity This concept is fundamental in computer science and offers insights into how long an algorithm takes to complete given a certain input size What is Time Complexity Time complexity represents the amount of time
Want a Structured Path to Master System Design Too? Don’t Miss This!