Count Unique Characters of All Substrings of a Given String

Identifying Problem Isomorphism

“Count Unique Characters of All Substrings of a Given String” can be approximately mapped to “Longest Substring Without Repeating Characters”.

Reasoning:

Both problems require you to deal with substrings and unique characters within them. In “Count Unique Characters of All Substrings of a Given String”, you have to consider all substrings and count unique characters within each one. In “Longest Substring Without Repeating Characters”, the task is to find the longest substring with all distinct characters. While not exactly the same, the logic of keeping track of unique characters within a substring is a shared aspect in both problems.

“Longest Substring Without Repeating Characters” is simpler as it only requires you to find the longest substring with unique characters, while “Count Unique Characters of All Substrings of a Given String” requires more computation, considering all possible substrings and counting unique characters within each.

“Count Unique Characters of All Substrings of a Given String” is related to string manipulations, substring calculations, and understanding the frequency count of characters. Here are 10 problems to prepare for it:

  1. 3. Longest Substring Without Repeating Characters: This problem helps you understand how to track unique characters in a substring.

  2. 76. Minimum Window Substring: This problem requires understanding of string traversal and frequency counts.

  3. 159. Longest Substring with At Most Two Distinct Characters: This is a sliding window problem where you must track distinct characters.

  4. 438. Find All Anagrams in a String: This problem involves tracking the frequency of characters in substrings.

  5. 567. Permutation in String: This problem also deals with character frequencies in substrings.

  6. 680. Valid Palindrome II: This is a simpler string problem which helps build fundamentals.

  7. 718. Maximum Length of Repeated Subarray: This problem involves comparing substrings.

  8. 819. Most Common Word: This problem involves character frequency count.

  9. 904. Fruit Into Baskets: Although the problem is not about strings, it uses similar sliding window concepts.

  10. 1244. Design a Leaderboard: This problem helps you understand frequency counts, albeit in a different context.

These cover string manipulation, substring calculations, ability to count character frequencies, which will be useful for “Count Unique Characters of All Substrings of a Given String” problem.

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class Solution:
    def uniqueLetterString(self, string):
        last_two_positions = {char: [-1, -1] for char in ascii_uppercase}
        total = 0
        for index, char in enumerate(string):
            second_last, last = last_two_positions[char]
            total += (index - last) * (last - second_last)
            last_two_positions[char] = [last, index]
        for char in last_two_positions:
            second_last, last = last_two_positions[char]
            total += (len(string) - last) * (last - second_last)
        return total % (10**9 + 7)

In this function:

  • string is the input string on which we want to calculate the sum of unique characters.
  • last_two_positions is a dictionary that keeps track of the last two positions of each character in the string. It is initialized with -1 for each uppercase English alphabet letter, indicating that we haven’t seen any character yet.
  • total is the result we will return at the end. It is the sum of all unique character counts in the substrings.
  • index and char are the current index and character of the string in the enumeration.
  • second_last and last are the last two positions of the current character in the string.
  • The modulo operation at the end is to ensure the result fits in a 32-bit integer, as per the problem constraints.

Problem Classification

Domain Classification: The given problem falls under the domain of ‘String Processing’ and ‘Combinatorics’.

What components:

  1. A function countUniqueChars(s) that returns the count of unique characters in a given string s.
  2. A given string s.
  3. Return the sum of countUniqueChars(t) for every substring t of s.

Problem Classification: Based on the ‘What’ components, the problem is a combination of String Processing, as it involves operations on strings and substrings, and Combinatorics, as it involves counting the sum of unique characters for every possible substring of s.

Explanation: The problem involves manipulating and analyzing strings which is a common task in many real-world applications such as text analysis, pattern recognition, and natural language processing. Moreover, the problem involves counting and summing unique characters for every possible substring, which requires understanding of combinatorics and its principles. As such, it is a problem that tests both the practical skills in manipulating strings as well as the theoretical knowledge in combinatorics.

Clarification Questions

What are the clarification questions we can ask about this problem?

Identifying Problem Isomorphism

Can you help me with finding the isomorphism for this problem?

Which problem does this problem map to the corresponding isomorphic problem on Leetcode ?

Constraints

  1. String Length: The problem specifies that the length of the input string ’s’ is between 1 and 105. This range is relatively small and manageable, allowing us to consider solutions that might involve iterating over the string multiple times, such as generating all possible substrings.

  2. Unique Characters: The problem specifies that ’s’ consists of uppercase English letters only. This tells us that there are at most 26 unique characters in ’s’. This could be leveraged in designing a solution, as we can use a fixed-size array or other data structure to keep track of character frequencies.

  3. Count of Unique Characters: The function countUniqueChars(s) counts the number of unique characters in a string. Since the string contains only uppercase English letters, the maximum number of unique characters in a string can be 26.

  4. Summation of Counts: We are asked to find the sum of countUniqueChars(t) for all substrings t of s. Not all substrings are unique, so we will have to count the unique characters for each substring, even if they have been processed before.

  5. All substrings: As we need to compute countUniqueChars(s) for all possible substrings of ’s’, we need to generate all substrings. The total number of possible substrings of a string of length ’n’ is n * (n + 1) / 2, which can be handled given the constraint on the length of ’s’.

Given these characteristics, an efficient solution could involve using a sliding window approach or a prefix-sum based approach to generate the substrings and compute the unique character counts, keeping track of character frequencies using a fixed-size array.

Thought Process

The problem statement provides some important cues that help in the approach to the problem.

  1. Unique Characters: The problem is asking for the number of unique characters in each substring of the given string.

  2. Sum of Counts: The function should return the sum of the counts of unique characters across all substrings.

  3. Counting Repetitions: If the same substring occurs more than once, its count should be added for each occurrence.

Based on these cues, the problem seems to be a combination of string manipulation and counting. This suggests using some kind of data structure, such as a counter or a hash table, to keep track of the characters in each substring. Also, since we are dealing with substrings, it would be efficient to use a sliding window approach to avoid redundant computation.

Solution Approach and Analysis

We can use a sliding window approach to solve this problem. Let’s break it down:

  1. Initializing counters: For each character in the string, we need to maintain a counter that keeps track of how many times it has occurred in the current substring (i.e., window). For this, we can use a dictionary or an array of size 26 (since there are 26 uppercase English letters). Initially, all counters are set to zero.

  2. Sliding window approach: For each position i in the string, we treat it as the end of the substring, and then we expand the start of the substring as far left as we can go without including a repeated character. We keep track of the leftmost index of the substring in a variable left.

  3. Updating counters: For each new character at position i, we add it to our counter. If the count of this character exceeds 1, it means we have a repeated character in our current substring. So we move the left pointer to the right until we remove the repeated character from our substring.

  4. Counting unique characters: The number of unique characters in the substring ending at position i is just the number of characters in the substring, which is i - left + 1. We add this to our total count.

  5. Result: Our final result is the total count, which is the sum of the counts of unique characters in all substrings.

This approach works because we are essentially enumerating all substrings of the string, and for each substring, we are efficiently counting the number of unique characters using our counter. The sliding window ensures that we do not count repeated characters.

The time complexity of this approach is O(n), where n is the length of the string, because each character is processed only once. The space complexity is O(1), because the size of the counter array is fixed at 26.

Example:

Consider the string s = “ABA”. Initially, our counter is [0, 0, …, 0] (size 26), left is 0, and total is 0.

  • For i = 0 (s[i] = “A”), we add “A” to our counter: counter becomes [1, 0, …, 0], the substring is “A”, and total becomes 1.
  • For i = 1 (s[i] = “B”), we add “B” to our counter: counter becomes [1, 1, …, 0], the substring is “AB”, and total becomes 1 + 2 = 3.
  • For i = 2 (s[i] = “A”), we add “A” to our counter: counter becomes [2, 1, …, 0]. Now we have a repeated character “A”, so we move left to the right until we remove the repeated “A”: left becomes 1 and counter becomes [1, 1, …, 0]. The substring is “BA”, and total becomes 3 + 2 = 5.
  • For i = 3 (s[i] = “C”), we add “C” to our counter: counter becomes [1, 1, 1, …, 0], the substring is “BAC”, and total becomes 5 + 3 = 8.

So, the final result is 8.

Let’s think about how a character can be found as a unique character.

Think about string “XAXAXXAX” and focus on making the second “A” a unique character. We can take “XA(XAXX)AX” and between “()” is our substring. We can see here, to make the second “A” counted as a uniq character, we need to:

insert “(” somewhere between the first and second A insert “)” somewhere between the second and third A For step 1 we have “A(XA” and “AX(A”, 2 possibility. For step 2 we have “A)XXA”, “AX)XA” and “AXX)A”, 3 possibilities.

So there are in total 2 * 3 = 6 ways to make the second A a unique character in a substring. In other words, there are only 6 substring, in which this A contribute 1 point as unique string.

Instead of counting all unique characters and struggling with all possible substrings, we can count for every char in S, how many ways to be found as a unique char. We count and sum, and it will be out answer.

The explanation you provided is centered around counting the ways a character can be unique in a substring within the string. The character in question is the second “A” in the string “XAXAXXAX”.

Let’s break down the steps:

  1. Insert “(” somewhere between the first and second A: This signifies the starting point of the substring in which the second “A” is unique. Two positions are possible, either after the first “A” or right before the second “A”.

  2. Insert “)” somewhere between the second and third A: This signifies the ending point of the substring where the second “A” is unique. Three positions are possible, either right after the second “A”, after the first “X”, or after the second “X”.

By combining the starting and ending points, we get a total of 2 * 3 = 6 substrings where the second “A” is unique.

Therefore, this approach is suggesting that, instead of calculating all the unique characters in every possible substring (which can be quite expensive in terms of computation), we instead calculate for every character in the string, the number of substrings in which it would be unique. This is done by looking at the positions of the same character before and after it. The total sum of these values for all characters will be our answer.

This approach significantly reduces the complexity of the problem by focusing on individual characters rather than all possible substrings.

Language Agnostic Coding Drills

Let’s deconstruct the code into the following individual concepts, listed in increasing order of difficulty:

  1. Variable and Data Types: Understanding how to declare and use basic data types in programming languages is the most fundamental concept.

  2. Control Structures: This involves using loops and conditional statements. Here, we have a for loop that iterates over the string.

  3. Data Structures: Particularly dictionaries and lists. Here, last_two_positions is a dictionary that stores lists as values, and string is handled as a list of characters.

  4. String Manipulation: This concept includes handling and manipulating strings, enumerating them, and accessing individual characters.

  5. Arithmetic Operations: Understanding and correctly applying arithmetic operations is crucial. Here, we have addition, subtraction, and the modulo operation.

  6. Functions: Creating and using functions is a fundamental part of structuring code. Here, we have a function countUniqueSubstrings.

To assemble these concepts into a final solution, we start by creating a function and initializing necessary variables and data structures (Concepts 1, 3, and 6). We then iterate over the string (Concept 2), during which we perform string manipulation to access individual characters (Concept 4). For each character, we calculate the contribution to the total count using arithmetic operations (Concept 5). Finally, we perform additional arithmetic operations to calculate the final count outside the loop.

It’s important to note that while these concepts are distinct, they are not entirely independent. For instance, performing arithmetic operations requires understanding variables and data types, and working with control structures requires some knowledge of the data structure being iterated over. The ability to integrate these concepts is essential to solving the problem.

Targeted Drills in Python

  1. Variable and Data Types

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    integer_var = 10
string_var = "Hello, World!"
print(integer_var, string_var)

  2. Control Structures

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    for i in range(5):
    if i%2 == 0:
        print(i, "is even")
    else:
        print(i, "is odd")

  3. Data Structures

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    my_dict = {'apple': 1, 'banana': 2}
my_list = ['apple', 'banana']
print(my_dict['apple'], my_list[0])

  4. String Manipulation

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    s = "Hello, World!"
for i, char in enumerate(s):
    print(i, char)

  5. Arithmetic Operations

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    a = 10
b = 20
print(a+b, a-b, a*b, a/b, a%b, a**b)

  6. Functions

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    def greet(name):
    return f"Hello, {name}!"
print(greet("World"))

For our problem, a problem-specific concept could be how we calculate the unique contribution for each character. This will involve knowledge of data structures (to store the last two positions of a character) and arithmetic operations (to calculate the unique contribution).

Finally, these drills can be assembled together in the following order to solve the initial problem:

  1. Initialize variables and data structures.
  2. Start a loop to iterate over the string.
  3. For each character in the string, perform string manipulations and calculate the unique contribution using arithmetic operations.
  4. Once the loop is finished, perform additional arithmetic operations to calculate the final count.

These drills illustrate the individual skills necessary to assemble the final solution. By understanding each piece, one can better understand how they fit together to solve the problem.

Q&A

Take “ABBB” for example, the substrings of it are

A AB ABB ABBB B BB BBB B BB B

So the sum = 1 + 2 +2 +2 +1 +1 +1 +1 +1 + 1 = 13, which is judged as being incorrect! What’s wrong with this?

In the problem statement, you are required to calculate the sum of unique characters in all substrings, rather than the length of substrings. Therefore, your approach needs to be slightly different.

For the given example “ABBB”, the substrings are:

  • “A”, “B”, “BB”, “BBB”, “B”, “BB”, “B”

These are the substrings where each character contributes uniquely.

Now, let’s count the unique characters in these substrings:

  • countUniqueChars(“A”) = 1
  • countUniqueChars(“B”) = 1
  • countUniqueChars(“BB”) = 1
  • countUniqueChars(“BBB”) = 1
  • countUniqueChars(“B”) = 1
  • countUniqueChars(“BB”) = 1
  • countUniqueChars(“B”) = 1

If you sum up these values, you get 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7, which is the correct output according to the problem statement.

Remember, each character in a substring contributes to the total if it is unique in that substring. In your example “ABBB”, although there are multiple ‘B’s in some substrings, they only count as one unique character for those substrings. That’s why your initial calculation resulted in a higher number.

Input: s = “ABA” Output: 8 Explanation: The same as example 1, except countUniqueChars(“ABA”) = 1.

Why does the result of countUniqueChars(“ABA”) equal to 1? Should it be 2?

It seems there might be some confusion around the definition of countUniqueChars(). As stated in the problem, the function countUniqueChars(s) is meant to return the number of unique characters that only appear once in the string s.

Given the string “ABA”, the only unique character that appears once is “B”. Hence, countUniqueChars(“ABA”) is equal to 1. The character “A” is not considered unique in this string because it appears more than once.

So, in terms of the problem statement, a character is considered unique if it appears only once in a given string (or substring), not just if it is a distinct character.

Everyone has posted a DP solution without showing how they formulated their solution from a recursive standpoint. This doesn’t help others that haven’t solved the problem

Absolutely, it’s a great practice to understand the recursive formulation before jumping into the Dynamic Programming (DP) solution. It’s often through the recursive solution that we identify overlapping subproblems and opportunities for optimization. Let’s dive into the recursive approach for this problem:

  1. Recursive Thought Process: At the heart of this problem, we’re looking for all substrings of the given string and determining the count of unique characters within each of those substrings. Naturally, this suggests a recursive solution where at each step, we would consider either including or not including the current character in our substring. If we include the current character, we must update our count of unique characters accordingly.

  2. Base Case: The base case for our recursion would be when we’ve considered all characters in our string. At this point, we return the count of unique characters in the substring formed.

  3. Recursive Case: For each character in the string, we have two choices:

    • Exclude the current character and recurse on the remaining string.
    • Include the current character, update the count of unique characters, and recurse on the remaining string.

However, this naive recursive approach would have a time complexity of O(2^n), which isn’t efficient for large inputs due to the huge number of repeated computations. This is where DP comes in.

In the DP approach, we try to optimize this process by storing the results of the subproblems in a data structure (usually an array or a matrix). In the context of this problem, this would involve maintaining a data structure to keep track of the number of unique characters for substrings we’ve already processed.

The process of formulating a DP solution often involves identifying the state variables that define your subproblems (in this case, the start and end indices of your substrings), defining a DP array using these state variables, and then writing a recurrence relation that expresses the solution to the problem in terms of smaller subproblems.

This recursive thought process is the fundamental idea behind most DP problems, and understanding this is key to being able to solve such problems efficiently.

The largest answer <= (26 * 26 * 10000), which is smaller than 10^9+7…

That statement refers to a potential optimization for the problem, where 26 is the number of unique letters in the English alphabet.

We can use this information to our advantage because it can help us narrow down our search space.

Here is the logic behind it:

  • There are at most 26 unique letters.
  • Each unique letter in the given string can be found as a unique character in at most n^2 substrings (where n is the length of the string), as a substring’s uniqueness is determined by the positions of its identical letters before and after it.
  • Hence, the sum for all unique characters will be at most 26 * n^2.
  • The constraints in the problem state that n <= 10^5. Thus, n^2 <= 10^10.
  • So, the sum of all unique characters for all substrings in the worst-case scenario will be 26 * 10^10.
  • However, since we’re taking the modulo of the result with 10^9 + 7 as per the problem statement, the maximum result will always be less than 10^9 + 7.

This calculation helps us ensure that our final result will not exceed the integer limit and cause an overflow, and also assists in understanding the range of the result.

in case when input is AABC possible out should be 16 , but output by system is 14? A 1, AA 1, AAB 2, AABC 3, AB 2, ABC 3, B 1, BC 2, C 1 =16

can anyone explain me where i am wrong !!! need help here

It seems there is a misunderstanding about the counting mechanism in this problem. The function countUniqueChars(t) does not return the number of unique characters in a substring t, but instead, it returns the sum of the counts of each unique character in t.

Given the input string “AABC”, let’s break down the calculation of each substring:

  1. “A”: countUniqueChars(“A”) = 1 (There is only one unique character, “A”)
  2. “AA”: countUniqueChars(“AA”) = 1 (The character “A” is not unique in this substring, so it counts only once)
  3. “AAB”: countUniqueChars(“AAB”) = 2 (The characters “A” and “B” are unique in this substring)
  4. “AABC”: countUniqueChars(“AABC”) = 3 (The characters “A”, “B”, and “C” are unique in this substring)
  5. “AB”: countUniqueChars(“AB”) = 2 (The characters “A” and “B” are unique in this substring)
  6. “ABC”: countUniqueChars(“ABC”) = 3 (The characters “A”, “B”, and “C” are unique in this substring)
  7. “B”: countUniqueChars(“B”) = 1 (There is only one unique character, “B”)
  8. “BC”: countUniqueChars(“BC”) = 2 (The characters “B” and “C” are unique in this substring)
  9. “C”: countUniqueChars(“C”) = 1 (There is only one unique character, “C”)

So the sum is 1 + 1 + 2 + 3 + 2 + 3 + 1 + 2 + 1 = 16.

You’re right about the calculation. If the system gives an output of 14, there might be an error in the system’s calculation or implementation. It should be noted that the code should take into account the concept of counting unique characters as defined in the problem statement.