How do you solve Total Appeal of A String on LeetCode?

Total Appeal of A String (LeetCode #2262) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how the last occurrence of characters affects substring contributions is crucial.

What is the brute force solution for Total Appeal of A String?

The brute force approach for Total Appeal of A String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible substrings of the given string and calculating the appeal for each substring. This method is straightforward but inefficient for longer strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Total Appeal of A String?

Total Appeal of A String uses the following concepts: Hash Table, String, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Total Appeal of A String?

Always consider edge cases, such as empty strings or strings with all identical characters. Explain your thought process clearly while coding; it helps interviewers understand your approach. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Total Appeal of A String?

Failing to account for overlapping substrings and their contributions correctly. Not using a data structure like a map to efficiently track last occurrences.

#2262

Total Appeal of A String

Hard

Hash TableStringDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages the last occurrence of each character to efficiently calculate the contribution of each character to the total appeal. By tracking the last seen index of each character, we can determine how many substrings end with that character and contribute to the appeal.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to track the total appeal and a map to store the last occurrence of each character.
2Step 2: Iterate through the string, and for each character, calculate its contribution based on its last occurrence.
3Step 3: Update the last occurrence of the character and sum the contributions to the total appeal.

solution.py14 lines

1# Full working Python code
2
3def total_appeal(s):
4    total = 0
5    last_occurrence = {}
6    n = len(s)
7    for i in range(n):
8        char = s[i]
9        last_index = last_occurrence.get(char, -1)
10        total += (i - last_index) * (n - i)
11        last_occurrence[char] = i
12    return total
13
14print(total_appeal('abbca'))  # Output: 28

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once, and each character's contribution is calculated in constant time. The space complexity is O(n) due to the storage of the last occurrence of characters.

1Understanding how the last occurrence of characters affects substring contributions is crucial.
2Using a map to track last occurrences allows for efficient calculations without generating all substrings.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.