Reorganize String Leetcode Problem 767 [Python Solution]
Are you ready to tackle the Reorganize String LeetCode problem?
This is a medium-level challenge that falls under the category of Heap/Priority Queue.
In this blog post, we will break down the problem, explore the efficient Python solution, discuss time and space complexity, and provide a thorough understanding of the constraints.
By the end, you'll have a solid grasp of how to solve this problem on your own.
Problem Overview
The problem statement is as follows: Given a string s, you need to rearrange its characters so that no two adjacent characters are the same.
If such a rearrangement is possible, return any valid rearrangement; otherwise, return an empty string.
Let's dive into this problem with an example:
Example 1:
Input: s = "aab"
Output: "aba"
In this example, we have two 'a's and one 'b.' We can rearrange them as "aba" so that no adjacent characters are the same.
Example 2:
Input: s = "aaab"
Output: ""
In this case, it's not possible to rearrange the characters to meet the conditions.
So, the output is an empty string.
Understanding Constraints
Before we explore the solution, let's discuss the constraints of this problem:
- The input string
shas a length between 1 and 500. sconsists of lowercase English letters.
Now that we have a clear understanding of the problem, let's dive into the solution.
Efficient Python Solution
To efficiently solve this problem, we'll use a max-heap (a priority queue) to keep track of the most frequent characters.
We'll follow these steps:
-
Count the occurrences of each character in the input string
susing a hashmap. -
Create a max-heap (implemented as a min-heap with negative counts for simplicity).
The heap will store pairs of characters and their counts, sorted by counts in descending order.
-
Initialize variables to keep track of the previous character (
prev) and the result string (res). -
In a loop, repeatedly pop the most frequent character from the max-heap.
Append this character to the result string and update its count.
- If the
prevcharacter is not null (indicating we used it in the previous step), push it back into the max-heap with the updated count.
Reset prev to null.
- Continue this process until the max-heap is empty or until
previs not null.
If we reach a point where the max-heap is empty and prev is not null, it means we cannot find a valid rearrangement, and we return an empty string.
- If the loop completes without any issues, we return the valid rearrangement stored in the
resvariable.
Now, let's provide the Python code for this efficient solution:
from collections import Counter
import heapq
class Solution:
def reorganizeString(self, s: str) -> str:
# Step 1: Count the occurrences of each character
count = Counter(s)
# Step 2: Create a max-heap (implemented as a min-heap with negative counts)
maxHeap = [[-cnt, char] for char, cnt in count.items()]
heapq.heapify(maxHeap)
prev = None
res = ""
# Step 4: Pop the most frequent character and append it to the result string
while maxHeap or prev:
if prev and not maxHeap:
return "" # Step 6: Return an empty string if not possible
# Pop the most frequent character from the max-heap
cnt, char = heapq.heappop(maxHeap)
res += char
cnt += 1
if prev:
heapq.heappush(maxHeap, prev)
prev = None
# Update prev if count is not zero
if cnt != 0:
prev = [cnt, char]
return res # Step 7: Return the valid rearrangement
solution = Solution()
input_str = "aab"
output_str = solution.reorganizeString(input_str)
print(output_str) # Output: "aba"
This Python code efficiently solves the Reorganize String problem using a max-heap for character frequency tracking.
Time and Space Complexity
Let's analyze the time and space complexity of our solution:
Time Complexity:
- Counting the occurrences of each character in
susingCountertakesO(n), where n is the length of the input string. - Building the max-heap using the character counts takes
O(26 * log n)in the worst case (where 26 is the number of lowercase English letters).
In general, this step has a time complexity of O(n * log n).
– The while loop runs in O(n) time since, in the worst case, we may go through the loop n times.
– Overall, the time complexity of the solution is O(n * log n).
Space Complexity:
- The space complexity is determined by the data structures used.
We use O(n) space for the count hashmap and O(n) space for the max-heap.
Therefore, the space complexity is O(n).
Reasoning Behind Our Approach
Our approach takes advantage of the max-heap data structure to efficiently handle the character frequency tracking and ensure that we start with the most frequent character each time.
By using a max-heap, we can easily pop the most frequent character and push it back with an updated count when necessary.
This approach optimizes the rearrangement process and ensures that no two adjacent characters are the same.
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Conclusion
In this blog post, we've explored the Reorganize String LeetCode problem, providing a detailed Python solution that efficiently rearranges the characters to meet the specified conditions.
We've discussed the time and space complexity, constraints, and the reasoning behind our approach.
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Solving coding challenges like this one is a great way to improve your programming skills.
Good luck with your coding journey!
