Longest Substring With Same Letters After Replacement Leetcode

Longest Substring With Same Letters After Replacement Leetcode – Key Insights – Use a sliding window with HashMap to track character counts. A window is valid if the window length minus the most frequent character is less than or equal to the number of allowed substitutions.

One of my first thoughts on this was to create a hashmap of each character to the positions it occupies. Then analyze the list of positions that have the least gap anyway. Perhaps subtracting letter indices can reveal which letters are close together.

Longest Substring With Same Letters After Replacement Leetcode

. It contains the start and end of the string. This sets us up to determine which letter can form the longest repeating sequence by substitution.

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The next step is to create a function that takes these “diff” arrays and calculates the longest repeated substring from them.

This means the function that calculates the maximum length substring for the “diff” array described above. It basically looks at the array and tries to find the longest subset that contains the sum

The problem with this approach is that it is difficult to calculate the substring length because the difference values ​​are mixed with characters. For example, the variation range for “a” for the string “babbab”.

Obviously the maximum substring length for “a” is 4. It is created by substituting two “b”s in the middle. However, coming up with that amount depends on it

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Represents whether the character is present. So the structure for “a” looks like for the string “babbab”.

. Per-character arrays have the same length as the original string. I am going to start over with this method.

It passes all test cases. However, it only beats 5% of submissions in runtime and 10% in memory usage. I’m going to check out the NeetCode video to see if there’s a cleaner solution.

The optimal solution has some improvements over the solution I found. What’s important is that instead of having a hashmap of lists of character positions, the solution has a hashmap of counts for each character. It uses HashMap to maintain a valid sliding window. A valid condition is

Find Longest Substring Without Repeating Characters

It passes all test cases. It beats only 10% of submissions in runtime, but it now outperforms 77% of submissions in memory usage.

I am proud of myself for sticking with it and coming up with my own solution that passed all test cases. There was some room to improve the solution, but it’s still a great feeling to solve a leetcode problem without help. A fellow redditor from /r/cscareerquestions pointed me to this awesome thread in the leetcode discussion that exposes a sliding window pattern for solving multiple string (substring) problems. Today we’ll explore the intuition behind this powerful technique and apply it to some well-known string problems. Another truth I intend to prove is that some problems tagged as hard are actually easy once we grasp the required pattern.

I’m a strong believer in learning by doing, so let’s walk through a simple example to understand how sliding windows work, and then get back to the Leitcode problems. The sliding window method helps us reduce the time complexity of brute force methods in general.

Solving it using brute force means exploring all possible cases. At this point we don’t really care about efficiency but want correct results. So we can run a nested loop that explores all windows of length k in a given range and chooses the max.

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Now after calculating the sum of the 1st window (size k), we need to add the value of the new (right) item leaving the leftmost item value to get the sum of the next overlapping window, so we are basically saving the re-calculation of the sum for the unchanged part of the window.

Voila! We used memory to save time, the most classic trade-off in algorithms, and sliding windows are what we’ll look at. The time complexity is reduced from O(n²) to O(n). Now let’s move on to the actual Leetcode issues.

Note : This first problem is analyzed crudely because it is a tricky concept and forms the basis for solving the next 5. Master this and try the rest yourself, that’s the best way to learn.

In the previous example the window was of fixed size, but here we use a deterministic variable size window

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Marks. A brute force method is to iterate through all possible substrings and determine the minimum window that contains all characters of T. How do you see if a substring contains all characters of T? You can use a frequency table of T that stores the letter for the numbers that occur as follows:

So we basically count the number of times each unique character in T exists in T by maintaining a counter. Well if T has 4 ‘K’s and B has 7 ‘K’s, the table count for ‘K’ is negative, but before that it goes to 0, proving that string B has at least 4 ‘K’s in it, “K ” Satisfying the requirement with respect to , extending the logic to other characters in T if counter = 0, B contains all characters in T .

To the right of examining each new character and updating the count in the table. As soon as we hit counter = 0, it means we have a valid answer so we try to trim it by removing unnecessary characters from the beginning by sliding to the right. We are constantly trying to validate/invalidate the string by manipulating the counter and table counts.

Take a moment, understand this code. Walk through the paper for this example: [ S:

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Intuition: A good substring for an answer is a permutation of T if such a substring exists in S, but otherwise we can have wasted characters sit between the necessary characters that make the substring valid as an answer. Our attempt here is to remove such characters without losing the essential. After trimming as much as possible we resume with sliding

If counter = 0 then we have a valid candidate for our answer, but we update the answer only if it is shorter than the previously recorded minimum length answer.

Exactly the same as above with the added condition that substring must be equal to p and we must return the indices of all such occurrences.

Here’s my proof that not all leetcode “hard” is exactly hard, a slight modification of the above problem tagged “easy”. Now instead of letters in the above question we have words so it is a bit confusing.

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Since the goal is to find the longest substring without any repetitions we construct a frequency table from the substring explored so far. So we slide like above case

340. A long substring with K unique characters that are literally identical with K = 2.

A good exercise at this point is to think about why the sliding window method actually works, at least draw all possible cases for the window substring problem, and you’ll have a deeper intuition about the technique.

The purpose of the articles published here is to decode common patterns used to solve algorithmic problems and gain a clear intuition of how these work. A very short explanation, it opens up a lot of questions that can be asked. A few clarifications I can think of and why:

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Let’s say the answers to the above are “yes”, “we care about all characters” and “treat A and differently”, then the question now becomes:

Given a string s, which may be empty, find the length of the longest substring without repeating characters. Uppercase and lower case of the same characters are treated as different characters and we expect special characters like !@#.

Suddenly the question becomes clearer and we understand what we are dealing with more clearly. Let’s move on.

Let’s start with the brute force method. A simple way to do this is to iterate through each character once, then iterate through the rest of the characters to see if we see any repeated characters. If yes, we keep track of the number of characters between the starting point and the point where we hit the repeated character, then record the number. If the number is greater than the previous high, we change it to the new high. Otherwise, start the next loop.

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Here we need a simple tracker to check if a character has appeared before, so an easy and fast look-up table is used.

Stores unique values ​​only once and the look-up cost is O(1). It should be quite efficient in terms of tracking.

What is the time complexity of this method? We have 2 for-loops with linear complexity, so the time complexity is simply O(n²).

Usually when we want to remove a loop, we need to track more

Leetcode 424. Longest Repeating Character Replacement