Extra Characters in a String

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Dynamic ProgrammingStrings

You are given a 0-indexed string s and a dictionary of words dictionary. You have to break s into one or more non-overlapping substrings such that each substring is present in dictionary. There may be some extra characters in s which are not present in any of the substrings.

Return the minimum number of extra characters left over if you break up s optimally.

Example 1:

Input: s = "leetscode", dictionary = ["leet","code","leetcode"]
Output: 1
Explanation: We can break s in two substrings: "leet" from index 0 to 3 and "code" from index 5 to 8. There is only 1 unused character (at index 4), so we return 1.

Example 2:

Input: s = "sayhelloworld", dictionary = ["hello","world"]
Output: 3
Explanation: We can break s in two substrings: "hello" from index 3 to 7 and "world" from index 8 to 12. The characters at indices 0, 1, 2 are not used in any substring and thus are considered as extra characters. Hence, we return 3.

Constraints:

1 <= s.length <= 50
1 <= dictionary.length <= 50
1 <= dictionary[i].length <= 50
dictionary[i] and s consists of only lowercase English letters
dictionary contains distinct words

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the maximum lengths of the string `s` and the strings in the `dictionary`?
Can the `dictionary` be empty, and if so, what should the return value be?
Can `s` be an empty string?
Does the `dictionary` contain duplicate words? If a word appears multiple times, does it offer any advantages?
Are the characters in `s` and the words in `dictionary` limited to lowercase English letters, or can they include other characters (e.g., uppercase, digits, special symbols)?

Brute Force Solution

Approach

The brute force strategy tries every possible way to break down the input string into valid words from the dictionary. We explore all possible combinations of word segmentations to find the one that leaves the fewest leftover characters. It's like trying all the possible arrangements until we find the best one.

Here's how the algorithm would work step-by-step:

Start by considering the beginning of the input string.
Try to match a single word from the dictionary at the beginning.
If you find a match, remove that word from the string and repeat the process with the remaining part.
If you don't find a match, try matching a longer word from the dictionary at the beginning.
Keep trying all possible words from the dictionary to see if any of them match at the beginning of the (remaining) string.
For each successful sequence of word matches, keep track of how many characters are left over and weren't part of any word from the dictionary.
Repeat this process, trying all possible combinations of words from the dictionary to cover the input string.
Finally, compare all of the combinations you've tried and pick the one that has the smallest number of leftover characters. That's your answer!

Code Implementation

def minExtraCharBruteForce(input_string, dictionary):    def solve(start_index):        # Base case: If we've reached the end of the string,
        # there are no extra characters.        if start_index == len(input_string):            return 0        min_extra = len(input_string) - start_index        
        # Iterate through all possible substrings starting
        # from the current index.        for end_index in range(start_index, len(input_string)):            substring = input_string[start_index:end_index + 1]

            # If the substring is in the dictionary            if substring in dictionary:                # Recursively calculate the minimum extra
                # characters for the remaining string.                min_extra = min(min_extra, solve(end_index + 1))            else:
                extra_characters = (end_index - start_index + 1)

        return min_extra
    # Initiate the recursive calculation from the beginning
    # of the input string.
    return solve(0)

Big(O) Analysis

Time Complexity

O(2^n) – The brute force approach explores all possible segmentations of the input string of length n. For each position in the string, we have a choice: either include it in a word from the dictionary or consider it an extra character. This branching decision at each position leads to 2 possible choices at each index of the n length string. Consequently, the total number of possible segmentations grows exponentially with the input size n, resulting in a time complexity of O(2^n).

Space Complexity

O(N) – The brute force approach, as described, relies on recursion to explore all possible combinations of word segmentations. In the worst-case scenario, where no words from the dictionary match any prefix of the input string, the recursion depth could reach N, where N is the length of the input string. Each recursive call adds a new frame to the call stack, requiring space to store the function's local variables and return address. Therefore, the auxiliary space complexity is O(N) due to the recursion depth.

Optimal Solution

Approach

The best way to solve this is to think about the problem from the end of the string backwards. We'll use a process to remember the best results we've seen so far to avoid recalculating the same things over and over.

Here's how the algorithm would work step-by-step:

Start by creating a way to remember the best solution we've found for each part of the string.
Begin at the end of the main string.
Check if the last part of the main string is a valid word in the provided word list. If it is, there are no extra characters.
If it is not, or we're dealing with a shorter part of the main string, consider all the ways we could divide the string: Is the last letter an extra character? Are the last two letters an extra character? And so on...
For each of those possibilities, calculate the number of extra characters *plus* the best solution we have for the string *before* the extra characters.
Keep track of the smallest number of extra characters required across all those splits.
Store the best number of extra characters to match the starting position in our memory to reuse for future calculations.
Move backwards one letter at a time in the main string, repeating these calculations, and storing our best results.
Finally, when we reach the very beginning of the string, we'll have the best solution for the entire string.

Code Implementation

def extra_characters(phrase, dictionary):
    string_length = len(phrase)
    # Memoization table to store results for subproblems
    minimum_extra_characters = [0] * (string_length + 1)

    for i in range(string_length - 1, -1, -1):
        minimum_extra_characters[i] = minimum_extra_characters[i + 1] + 1
        for word in dictionary:
            # Check for words that start at the current position.
            if phrase.startswith(word, i):

                minimum_extra_characters[i] = min(
                    minimum_extra_characters[i],
                    minimum_extra_characters[i + len(word)],
                )

    # The result contains the fewest extra chars for the entire string
    return minimum_extra_characters[0]

Big(O) Analysis

Time Complexity

O(n*m) – The algorithm iterates backwards through the input string s of length n. For each position in s, it checks all possible substrings ending at that position to see if they are present in the dictionary word list of average length m. This substring search within the word list contributes a factor of m to the time complexity for each of the n positions in s. Therefore the overall time complexity is O(n*m).

Space Complexity

O(N) – The auxiliary space complexity is primarily determined by the memory used to store the best solutions for each part of the string. The explanation describes storing the best number of extra characters for each starting position in memory to reuse for future calculations. Since there are N possible starting positions in the main string (where N is the length of the string), this memory usage grows linearly with the size of the input string, leading to O(N) space complexity.

Edge Cases

Case	How to Handle
Empty string s	Return 0, as there are no characters to be extra.
Empty dictionary	Return the length of s, since no words can be formed.
Null string s or dictionary	Throw an IllegalArgumentException or return -1 to indicate invalid input.
Dictionary contains an empty string	The empty string matches any prefix of s and results in 0 extra chars if only it is chosen.
s contains a very long sequence with no dictionary matches	The dynamic programming approach handles this by iterating through the string and recording the cost of unmatched characters.
Dictionary contains duplicate words	The algorithm should naturally handle duplicates in the dictionary without issue.
s is very long and the dictionary is very large, leading to potential memory issues in memoization	Consider a top-down dynamic programming approach with memoization to avoid recalculations.
s and dictionary words contain Unicode characters	Ensure the string comparison and substring operations correctly handle Unicode characters.

Empty string s

How to Handle:

Return 0, as there are no characters to be extra.

Empty dictionary

How to Handle:

Return the length of s, since no words can be formed.

Null string s or dictionary

How to Handle:

Throw an IllegalArgumentException or return -1 to indicate invalid input.

Dictionary contains an empty string

How to Handle:

The empty string matches any prefix of s and results in 0 extra chars if only it is chosen.

s contains a very long sequence with no dictionary matches

How to Handle:

The dynamic programming approach handles this by iterating through the string and recording the cost of unmatched characters.

Dictionary contains duplicate words

How to Handle:

The algorithm should naturally handle duplicates in the dictionary without issue.

s is very long and the dictionary is very large, leading to potential memory issues in memoization

How to Handle:

Consider a top-down dynamic programming approach with memoization to avoid recalculations.

s and dictionary words contain Unicode characters

How to Handle:

Ensure the string comparison and substring operations correctly handle Unicode characters.