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LeetCode Solution: 5. Longest Palindromic Substring

Aryan Subudhi — Mon, 25 May 2026 17:54:40 +0000

5. Longest Palindromic Substring: Unraveling the Palindromic Puzzle 🧩

Hey fellow coders! 👋 Today, we're diving into a LeetCode classic that often stumps beginners but reveals a really elegant pattern once you spot it: the Longest Palindromic Substring problem.

Don't let the fancy name scare you! We'll break it down piece by piece, find an intuitive way to tackle it, and emerge victorious with a neat, efficient solution. Ready? Let's go!

📜 Problem Explanation

The problem asks us to find the longest palindromic substring within a given string s.

"Whoa, hold on!" you might say. "What's a palindrome? And what's a substring?" Great questions!

Palindrome: A string that reads the same forwards and backward. Think words like "madam", "racecar", or even single letters like "a".
Substring: A contiguous sequence of characters within a string. For "babad", "bab", "aba", "bad", "a" are all substrings. "bba" is not a substring because the 'b's aren't contiguous in that order.

So, our mission is to scan through the given string s, find all possible substrings that are also palindromes, and then return the longest one among them.

Let's look at the examples:

Example 1: s = "babad"
- Possible palindromic substrings: "b", "a", "b", "a", "d", "bab", "aba".
- The longest ones are "bab" and "aba". Either is a valid answer.
Example 2: s = "cbbd"
- Possible palindromic substrings: "c", "b", "b", "d", "bb".
- The longest is "bb".

Constraints:

The string s will have between 1 and 1000 characters.
It will only contain digits and English letters. No weird symbols to worry about!

🤔 Intuition: Don't Brute Force, Explore From the Middle!

A common first thought for many string problems is: "Let's check every possible substring!"

If you have a string of length N:

There are N * (N + 1) / 2 possible substrings (e.g., for "abc", "a", "b", "c", "ab", "bc", "abc").
For each substring, checking if it's a palindrome takes O(length_of_substring) time.
This quickly leads to an O(N^3) solution, which for N=1000 is 1000^3 = 1,000,000,000 operations – way too slow! 🐌

We need something smarter.

What makes a palindrome special? It's symmetrical around its center!

Consider "racecar":
r a c e c a r
The e is the center. Everything expands outwards equally.

Consider "abba":
a b b a
The "center" is between the two bs.

This gives us a huge hint! Instead of picking two ends and checking inwards, what if we pick a center and expand outwards?

Every palindrome has a center:

Odd-length palindromes: A single character is the center (e.g., 'a' in "bab", 'e' in "racecar").
Even-length palindromes: The "center" is the space between two identical characters (e.g., the space between the two 'b's in "bb", or "abba").

This means we can iterate through every possible "center" in our string and try to expand outwards to find the longest palindrome centered there. Since there are N characters and N-1 spaces between characters, there are roughly 2N - 1 potential centers.

🚀 Approach: Expand Around Center

Our intuition leads us to a solid plan:

We'll maintain variables start and max_len to keep track of the beginning index and length of the longest palindrome we've found so far. Initialize start = 0 and max_len = 1 (a single character is always a palindrome).
We'll iterate through each character in the string s using an index i. For each i, we consider it as a potential center for a palindrome.
Since palindromes can be odd or even length, we'll need to check two types of expansions for each i:
- Odd length: i itself is the center. We'll try to expand outwards from left = i, right = i.
- Even length: The space after i is the center. We'll try to expand outwards from left = i, right = i + 1.
To avoid repeating code, we'll create a helper function, let's call it expandAroundCenter(s, left, right). This function will take the string s and two pointers, left and right, and expand them outwards as long as:
- left is not less than 0 (we're within string bounds).
- right is not greater than or equal to len(s) (we're within string bounds).
- s[left] is equal to s[right] (the characters match, so it's still a palindrome).
- Once these conditions are no longer met, the function returns the length of the palindrome found (which is right - left - 1).
In our main loop, after calling expandAroundCenter for both odd and even cases, we'll compare the lengths of the palindromes found. If either is longer than our current max_len, we update max_len and calculate the new start index. The start index will be i - (current_palindrome_length - 1) // 2. (This formula looks tricky, but it simply finds the leftmost character of the new longest palindrome given its length and its "center" i).
After checking all possible centers, we return the substring s[start : start + max_len].

Let's trace s = "babad":

Initialize start = 0, max_len = 1 (our initial "b")
i = 0 (char 'b')
- expandAroundCenter(s, 0, 0) -> returns length 1 ("b")
- expandAroundCenter(s, 0, 1) -> 'b' != 'a', returns length 0
i = 1 (char 'a')
- expandAroundCenter(s, 1, 1) -> returns length 1 ("a")
- expandAroundCenter(s, 1, 2) -> 'a' == 'b' NO, returns length 0
- expandAroundCenter(s, 0, 2) for center 'a': s[0]=='b', s[2]=='b'. left becomes -1, right becomes 3. Palindrome is "bab", length 3.
  - max_len = 3, start = 1 - (3-1)//2 = 1 - 1 = 0. Current best: "bab".
i = 2 (char 'b')
- expandAroundCenter(s, 2, 2) -> returns length 1 ("b")
- expandAroundCenter(s, 2, 3) -> 'b' != 'a', returns length 0
- expandAroundCenter(s, 1, 3) for center 'b': s[1]=='a', s[3]=='a'. left becomes 0, right becomes 4. Palindrome is "aba", length 3.
  - Length is 3, same as current max_len. We don't need to update start and max_len if it's not strictly greater, but it's fine if we do (it would become "aba"). Let's say it updates to "aba".
... and so on.

Finally, we return s[start : start + max_len].

💻 Code

class Solution:
    def longestPalindrome(self, s: str) -> str:
        if not s:
            return ""

        # Helper function to expand around a center
        # It takes the string s, and the left and right pointers
        # It expands outwards as long as characters match and stay within bounds
        # Returns the length of the palindrome found
        def expandAroundCenter(left: int, right: int) -> int:
            while left >= 0 and right < len(s) and s[left] == s[right]:
                left -= 1
                right += 1
            # After the loop, left and right are one step *past* the palindrome
            # So, the length is (right - 1) - (left + 1) + 1 = right - left - 1
            return right - left - 1

        start_index = 0  # Stores the starting index of the longest palindrome
        max_length = 0   # Stores the maximum length found so far

        # Loop through each character to consider it as a center
        for i in range(len(s)):
            # Case 1: Odd length palindrome (e.g., "aba", "racecar")
            # The current character s[i] is the center
            len1 = expandAroundCenter(i, i)

            # Case 2: Even length palindrome (e.g., "abba", "bb")
            # The center is between s[i] and s[i+1]
            len2 = expandAroundCenter(i, i + 1)

            # Get the maximum length found from these two expansions
            current_max_len = max(len1, len2)

            # If this palindrome is longer than our current max_length
            if current_max_len > max_length:
                max_length = current_max_len
                # Calculate the new starting index for this longest palindrome
                # For an odd length palindrome, start = i - (length - 1) / 2
                # For an even length palindrome, start = i - (length / 2 - 1)
                # Both can be generalized as: i - (length - 1) // 2
                start_index = i - (max_length - 1) // 2

        # Return the substring from start_index with max_length
        return s[start_index : start_index + max_length]

⏱️ Time & Space Complexity Analysis

Let's break down how efficient our solution is:

Time Complexity: O(N^2)
- The main for loop iterates N times (once for each character in s).
- Inside the loop, we call expandAroundCenter twice.
- The expandAroundCenter function, in the worst case (e.g., s = "aaaaa"), might expand all the way to the boundaries of the string. This takes O(N) time.
- Therefore, the total time complexity is N * O(N) = O(N^2).
- Given N <= 1000, N^2 is 1,000,000, which is perfectly acceptable within typical time limits (usually around 10^8 operations per second).
Space Complexity: O(1)
- We are only using a few constant variables (start_index, max_length, i, len1, len2, left, right). We are not storing any data structures that grow with the input size N.
- The slicing s[start_index : start_index + max_length] creates a new string, but this is part of the output, not auxiliary space used by the algorithm itself.

🌟 Key Takeaways

Don't Jump to Brute Force: Always consider the constraints and think about whether a naive O(N^3) or O(N^2) approach will pass. Often, there's a more elegant pattern waiting to be discovered!
Palindromes and Symmetry: The core idea for many palindrome problems revolves around their symmetrical nature. Thinking about "centers" is a powerful paradigm.
Odd vs. Even Length: Remember that palindromes can have either odd or even lengths, and your solution needs to account for both types of "centers."
Helper Functions: Breaking down complex logic into smaller, reusable helper functions (like expandAroundCenter) makes your code cleaner, more readable, and easier to debug.
Smallest Palindrome is 1: Don't forget that a single character is always a palindrome, which can be useful for initializing max_length.

This problem is a fantastic introduction to dynamic programming and string algorithms, and mastering the "expand around center" technique will serve you well in many other coding challenges!

Happy coding! ✨

Submission Details

Author Account: aryan-subudhi
Publishing Time: 2026-05-25 23:24:13

LeetCode Solution: 1. Two Sum

Aryan Subudhi — Mon, 25 May 2026 17:42:02 +0000

Cracking LeetCode 1: Two Sum - Your First Step into Algorithm Mastery!

Hey everyone! 👋 aryan-subudhi here, ready to dive into what's often the very first problem many of us encounter on our LeetCode journey: Two Sum. Don't let its "easy" tag fool you; this problem is a fantastic introduction to fundamental algorithmic thinking, data structures, and the crucial concept of time complexity. If you're just starting out, you're in for a treat!

2. Problem Explanation: The Mission, Should You Choose To Accept It!

Imagine you're given a list of numbers, let's call it nums, and a specific target number, target. Your mission, should you choose to accept it, is to find two different numbers within that nums list that add up precisely to target. Once you find them, you don't return the numbers themselves, but their positions (indices) in the original list.

Here's the lowdown:

Input:
- nums: An array (or list in Python) of integers.
- target: A single integer.
Output:
- A list (or array) of two integers, representing the indices of the two numbers that sum up to target.
Key Guarantees/Constraints:
- There will always be exactly one valid solution. No need to worry about multiple pairs or no pairs at all!
- You can't use the same number twice (i.e., you need two distinct elements from the array).
- The answer can be returned in any order (e.g., [0, 1] or [1, 0] are both fine).

Let's look at an example to make it crystal clear:

Example 1:

nums = [2, 7, 11, 15]
target = 9

We need to find two numbers in [2, 7, 11, 15] that sum to 9.

2 + 7 = 9 Bingo!
2 is at index 0.
7 is at index 1.
So, the output is [0, 1]. Simple, right?

3. Intuition: Thinking Through the Problem (From Simple to Smart!)

Okay, so we need to find two numbers. How would a human do it? You'd probably start picking numbers and seeing if their partners exist.

The "Brute Force" Idea (Our First Thought):

Let's try every possible pair!

Pick the first number (e.g., 2).
Then, go through all the other numbers to see if any of them, when added to 2, equal target (which is 9).
- 2 + 7 = 9 (Found it!)
- If not, you'd continue: 2 + 11 = 13 (No), 2 + 15 = 17 (No).
If you didn't find it with 2, you'd move to the next number, 7.
Then, you'd go through all the numbers after 7 (because 7 + 2 is the same as 2 + 7, and we don't want to recheck or use the same element twice).
- 7 + 11 = 18 (No)
- 7 + 15 = 22 (No)
And so on...

This approach definitely works! You're guaranteed to find the pair because you check every single combination.

The Catch: While correct, this isn't very efficient for large lists. If nums has N numbers, you're essentially doing N checks for the first number, N-1 for the second, and so on. This looks like N * N operations in the worst case, which we call O(N^2) complexity. For N = 10,000, N^2 is 100,000,000 — that's a lot of operations! Our follow-up question specifically asks for something better than O(N^2).

Towards a Smarter Idea:

Can we be faster? When we pick a number, say num, we know what its "partner" or "complement" should be to reach the target.
complement = target - num

Instead of searching through the rest of the list for this complement (which takes time), what if we could instantly know if we've seen this complement before? This is where a hash map (or dictionary in Python) shines!

4. Approach: The Hash Map Magic! ✨

The optimized approach uses a hash map (often called a dictionary in Python) to store numbers we've seen so far, along with their indices. This allows for very fast lookups.

Here's the game plan:

Initialize an empty hash map (dictionary): Let's call it seen. This seen map will store (number: index) pairs.
Iterate through the nums array: For each num and its index (let's call it i): a. Calculate the complement: complement = target - num. This is the number we need to find to hit our target. b. Check if complement is in seen: * If complement is in seen, it means we've previously encountered the number that, when added to our current num, gives us target. We've found our pair! * The index of the complement is seen[complement], and the index of our current num is i. We return [seen[complement], i]. c. If complement is NOT in seen: This means our current num doesn't complete a pair with any number we've seen so far. So, we add our current num and its index (i) to our seen map: seen[num] = i. This way, if a future number needs this num as its complement, it will find it quickly.
Since the problem guarantees exactly one solution, we know our loop will always find a pair and return.

Let's trace nums = [2, 7, 11, 15], target = 9 with this approach:

seen = {} (empty dictionary)

i = 0, num = 2:
- complement = target - num = 9 - 2 = 7.
- Is 7 in seen? No, seen is empty.
- Add num to seen: seen = {2: 0}.
i = 1, num = 7:
- complement = target - num = 9 - 7 = 2.
- Is 2 in seen? YES! seen[2] is 0.
- We found our pair! The indices are seen[2] (which is 0) and our current i (which is 1).
- Return [0, 1].

And we're done! Notice how much faster that was than checking every single pair. We only made one pass through the list.

5. Code: The Solution in Action

class Solution:
    def twoSum(self, nums, target):
        # Initialize an empty hash map (dictionary in Python)
        # to store numbers we've seen and their indices.
        seen = {} 

        # Iterate through the array `nums` with both index `i` and value `num`.
        for i, num in enumerate(nums):
            # Calculate the complement needed to reach the target.
            complement = target - num

            # Check if this complement already exists in our `seen` map.
            # If it does, we've found our pair!
            if complement in seen:
                # Return the index of the complement (from `seen`) and the current index `i`.
                return [seen[complement], i]

            # If the complement wasn't found, store the current number and its index
            # in the `seen` map for future lookups.
            seen[num] = i

6. Time & Space Complexity Analysis: Getting Technical!

Understanding how efficient your code is, is crucial for competitive programming and real-world applications.

Brute Force (Nested Loops) Approach:

Time Complexity: O(N^2)
- We have two nested loops. The outer loop runs N times (for each number). The inner loop runs up to N times for each iteration of the outer loop. This results in approximately N * N operations.
Space Complexity: O(1)
- We don't use any additional data structures that grow with the input size, just a few variables.

Hash Map (Dictionary) Approach:

Time Complexity: O(N)
- We iterate through the nums array exactly once.
- Inside the loop, calculating the complement is O(1).
- Checking if a key (complement) exists in a hash map (dictionary) is, on average, O(1).
- Adding a key-value pair (num: i) to a hash map is, on average, O(1).
- Since all operations inside the loop are constant time on average, the total time complexity is proportional to the number of elements N, making it O(N). This is significantly better than O(N^2) for large inputs!
Space Complexity: O(N)
- In the worst case, we might need to store all N numbers and their indices in our seen hash map before finding the pair (e.g., if the pair is the last two elements).
- Therefore, the space used grows linearly with the input size N.

The hash map approach effectively trades space complexity (O(N)) for significantly improved time complexity (O(N)). This is a very common and powerful optimization technique in algorithms!

7. Key Takeaways: What We Learned!

Don't Settle for Brute Force (Always!): While often the easiest to think of, the most straightforward approach isn't always the most efficient. Always consider if there's a faster way.
The Power of Hash Maps: Hash maps (dictionaries) are your best friends for problems requiring fast lookups. They allow you to check for the existence of an element in (average) constant time, O(1), making algorithms much faster.
The Complement Strategy: Many "find a pair" problems can be simplified by thinking about what complement you need for a given num to reach the target.
Time-Space Trade-off: Often, you can improve the time complexity of an algorithm by using extra space (like our hash map). This is a fundamental concept in algorithm design.
Understanding enumerate: In Python, enumerate(nums) is a neat way to get both the index (i) and the value (num) when iterating through a list.

8. Submission Details

Author Account: aryan-subudhi
Publishing Time: 2026-05-25 23:11:39
Title: 1. Two Sum

That's it for Two Sum! I hope this deep dive was helpful and clarified why this seemingly simple problem is so foundational. Keep practicing, and happy coding! 🚀

LeetCode Solution: 6. Zigzag Conversion

Aryan Subudhi — Sun, 24 May 2026 13:03:49 +0000

Unraveling the Zigzag: A Beginner's Guide to LeetCode 6 (Zigzag Conversion)

Hey LeetCoders and aspiring competitive programmers! aryan-subudhi here, ready to dive into another fascinating problem that looks tricky on the surface but unlocks a cool pattern-recognition skill. Today, we're tackling LeetCode Problem 6: Zigzag Conversion. Get ready to bend your string into some interesting shapes!

Problem Explanation

Imagine you're writing a message, but instead of straight lines, you write it in a zigzag pattern across a given number of rows. Once you've written it, you then read the message line by line, from left to right. Your task is to implement a function convert that takes a string s and an integer numRows and returns the converted string.

Let's look at an example. If our input string is s = "PAYPALISHIRING" and numRows = 3:

We write it like this:

P   A   H   N
A P L S I I G
Y   I   R

Notice how P is in row 0, A in row 1, Y in row 2, then we go back up with P in row 1, A in row 0, and so on. It forms a 'N' or 'Z' shape.

Once written, we read it line by line:

Row 0: P, A, H, N
Row 1: A, P, L, S, I, I, G
Row 2: Y, I, R

Concatenating these gives us: "PAHNAPLSIIGYIR". This is our expected output!

Another example: s = "PAYPALISHIRING", numRows = 4

P     I    N
A   L S  I G
Y A   H R
P     I

Reading line by line:

Row 0: P, I, N
Row 1: A, L, S, I, G
Row 2: Y, A, H, R
Row 3: P, I

Concatenating: "PINALSIGYAHRPI".

Constraints:

The string length s can be up to 1000 characters.
s consists of English letters (upper/lower-case), ',', and '.'.
numRows can be between 1 and 1000.

An important edge case: if numRows = 1, the zigzag pattern doesn't really form. The string remains unchanged. convert("A", 1) should return "A".

Intuition

At its core, this problem is about simulating the pattern. If we were doing this manually, we'd pick a character from the input string, place it in the correct row, then decide where the next character goes. This suggests we need a way to keep track of:

Which row we're currently placing characters into.
The direction we're moving (downwards or upwards).

The pattern repeats in cycles: go down from row 0 to numRows - 1, then go up from numRows - 2 to row 1, and then repeat. Notice that row 0 and numRows - 1 are only hit when going "down." The "up" phase skips these two rows to avoid double-counting or overshooting the pattern.

Approach: Simulating the Zigzag

Let's formalize our intuition into a concrete algorithm:

Handle the Edge Case: If numRows is 1, no zigzag conversion is needed. Simply return the original string s.
Initialize Rows: We need a way to store characters for each row. A list of lists (or list of strings/string builders) is perfect for this. Let's create numRows empty lists. For example, mat = [[] for _ in range(numRows)].
Traverse and Populate: We'll iterate through the input string s character by character. For each character, we need to append it to the correct row. The key is to manage our current_row index and the direction.

The pattern can be seen as a series of "full columns" going down, followed by "diagonal segments" going up-right.
- Phase 1: Going Down: We start at row 0 and append characters sequentially to mat[0], mat[1], ..., up to mat[numRows - 1].
- Phase 2: Going Up-Right: After reaching the bottom, we move "up and right." This means we append characters to mat[numRows - 2], mat[numRows - 3], ..., up to mat[1]. We stop before row 0 because row 0 is the start of the next "down" phase. Similarly, numRows - 1 is only hit when going down.
We repeat these two phases until all characters from s are processed.
Construct Result: Once all characters have been placed into their respective row lists, concatenate all the lists (or strings) in mat from row 0 to numRows - 1 to form the final result string.

Let's trace s = "PAYPALISHIRING", numRows = 3 with this approach:

mat = [[], [], []]
s_idx = 0

Cycle 1:

Down Phase:
- s[0] ('P') -> mat[0] -> ['P'], s_idx = 1
- s[1] ('A') -> mat[1] -> ['A'], s_idx = 2
- s[2] ('Y') -> mat[2] -> ['Y'], s_idx = 3 (Reached numRows - 1)
Up-Right Phase: (numRows - 2 down to 1 -> 3 - 2 = 1 down to 1)
- s[3] ('P') -> mat[1] -> ['A', 'P'], s_idx = 4 (Reached row 1)

Cycle 2:

Down Phase:
- s[4] ('A') -> mat[0] -> ['P', 'A'], s_idx = 5
- s[5] ('L') -> mat[1] -> ['A', 'P', 'L'], s_idx = 6
- s[6] ('I') -> mat[2] -> ['Y', 'I'], s_idx = 7
Up-Right Phase:
- s[7] ('S') -> mat[1] -> ['A', 'P', 'L', 'S'], s_idx = 8

Cycle 3:

Down Phase:
- s[8] ('H') -> mat[0] -> ['P', 'A', 'H'], s_idx = 9
- s[9] ('I') -> mat[1] -> ['A', 'P', 'L', 'S', 'I'], s_idx = 10
- s[10] ('R') -> mat[2] -> ['Y', 'I', 'R'], s_idx = 11
Up-Right Phase:
- s[11] ('I') -> mat[1] -> ['A', 'P', 'L', 'S', 'I', 'I'], s_idx = 12

Cycle 4:

Down Phase:
- s[12] ('N') -> mat[0] -> ['P', 'A', 'H', 'N'], s_idx = 13
- s[13] ('G') -> mat[1] -> ['A', 'P', 'L', 'S', 'I', 'I', 'G'], s_idx = 14 (String s exhausted)

Final mat:
mat[0] = ['P', 'A', 'H', 'N']
mat[1] = ['A', 'P', 'L', 'S', 'I', 'I', 'G']
mat[2] = ['Y', 'I', 'R']

Join them: "PAHN" + "APLSIIG" + "YIR" = "PAHNAPLSIIGYIR". Perfect!

Code

class Solution:
    def convert(self, s: str, numRows: int) -> str:
        # Edge case: If there's only one row, no zigzag conversion needed.
        if numRows == 1:
            return s

        # Initialize numRows empty lists to represent each row.
        # Each list will store characters that belong to that row.
        mat = [[] for _ in range(numRows)]

        # Pointer for the current character in the input string s.
        i = 0
        # Length of the input string s.
        n = len(s)

        # Loop until all characters from the input string are processed.
        while i < n:
            # Phase 1: Go down (from row 0 to numRows - 1)
            for down in range(numRows):
                if i < n: # Check if there are still characters left in s
                    mat[down].append(s[i])
                    i += 1

            # Phase 2: Go up-right (from numRows - 2 down to row 1)
            # We skip row 0 and numRows - 1 as they are handled exclusively by the 'down' phase
            # or will be the start/end of the next 'down' phase.
            for up in range(numRows - 2, 0, -1):
                if i < n: # Check if there are still characters left in s
                    mat[up].append(s[i])
                    i += 1

        # After populating all rows, join the characters in each row
        # and then concatenate all rows to form the final result string.
        result = []
        for row in mat:
            result.extend(row) # Use extend for efficiency, then join once

        return "".join(result)

Time & Space Complexity Analysis

Time Complexity: O(N)

We iterate through the input string s once using the while i < n loop.
Inside the loop, each character s[i] is appended to a list in mat exactly once.
Finally, we iterate through the mat (which collectively holds N characters) to build the result list and then join them into a string. The extend and join operations are proportional to the total number of characters, which is N.
Therefore, the overall time complexity is linear with respect to the length of the string s.

Space Complexity: O(N)

We create numRows lists in mat. In the worst case (e.g., numRows = N), each list holds one character. In general, all N characters of the string s are stored across these numRows lists.
The result list also temporarily stores all N characters before joining.
Therefore, the space complexity is linear with respect to the length of the string s.

Key Takeaways

Simulation Power: Many string manipulation or pattern-based problems can be solved by directly simulating the process described. Break down the pattern into repeatable steps.
State Management: Keeping track of current_row and direction (or explicit phases like "down" and "up-right") is crucial for controlling the simulation.
Edge Cases First: Always consider simple edge cases like numRows = 1 early on. They often simplify your logic or prevent errors.
List of Lists for Structure: When dealing with grid-like or row-based structures, a list of lists (or 2D array) is a very natural and effective data structure.
Concatenation Efficiency: For building strings from many smaller pieces, appending to a list of characters/strings and then using "".join(list) is generally more efficient in Python than repeated string concatenation (+=).

Hope this breakdown helped you "unzigzag" this problem! Keep practicing, and you'll master these patterns in no time. Happy coding!

Submission Details

Author Account: aryan-subudhi
Publishing Time: 2026-05-24 18:33:26
Title: Unraveling the Zigzag: A Beginner's Guide to LeetCode 6 (Zigzag Conversion)