#2707

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Extra Characters in a String

The problem asks for the minimum number of extra characters left after optimally breaking a string into substrings found in a dictionary.

array hash table string dynamic programming trie

Problem Statement

You are given a 0-indexed string s and a dictionary of words dictionary. The goal is to break s into one or more non-overlapping substrings such that each substring appears in dictionary. Some characters in s may not be part of any valid substring, and these are considered extra characters.

Return the minimum number of extra characters that remain in s after optimally breaking it into valid substrings from the dictionary.

Examples

Example 1

Input: s = "leetscode", dictionary = ["leet","code","leetcode"]

Output: 1

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

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

Solution Approach

Dynamic Programming with Hash Lookup

A dynamic programming approach is used where we iterate through the string and check for possible substrings from the dictionary. A hash table stores the dictionary words, enabling efficient lookups during the scanning of s.

Array Scanning

We perform a scan of the string s from left to right. For each position, we check if a substring from that position exists in the dictionary. If it does, we continue; if not, we count the character as an extra.

Optimal Break Decision

For each substring found in s, we recursively decide whether to break s at that point. If we can break s optimally, we minimize the number of extra characters left over.

Complexity Analysis

Metric	Value
Time	O(N^2 + M \cdot K)
Space	O(N + M \cdot K)

The time complexity is O(N^2 + M * K), where N is the length of string s, M is the size of the dictionary, and K is the average length of words in the dictionary. The space complexity is O(N + M * K), accounting for the dynamic programming table and the hash table used for dictionary lookups.

What Interviewers Usually Probe

Can the candidate explain the reasoning behind the use of dynamic programming here?
Does the candidate demonstrate a good understanding of the pattern of breaking down a string into dictionary words?
Is the candidate able to optimize the solution by minimizing extra characters efficiently?

Common Pitfalls or Variants

Common pitfalls

Failing to optimize the solution, leading to unnecessary extra characters left over.
Overcomplicating the problem by not recognizing the value of hashing dictionary words for efficient lookups.
Not managing dynamic programming states correctly, leading to incorrect results.

Follow-up variants

Handling a dictionary with very large numbers of words.
Optimizing the solution for longer strings with many possible splits.
Implementing the solution without relying on dynamic programming for a more brute force approach.

FAQ

What is the optimal strategy to solve the "Extra Characters in a String" problem?

The optimal strategy is using dynamic programming combined with a hash table for efficient dictionary lookups and array scanning to minimize extra characters.

How does dynamic programming apply to this problem?

Dynamic programming helps by storing the minimum number of extra characters at each position in the string, allowing for optimal breaks at each step.

How can hash tables help in solving this problem?

Hash tables allow for efficient lookups of dictionary words during the array scanning, significantly reducing the time complexity of the solution.

What are common mistakes when solving "Extra Characters in a String"?

Common mistakes include failing to optimize the solution for extra characters, improper management of dynamic programming states, and not utilizing hash tables for efficient lookups.

Can this problem be solved without dynamic programming?

While possible, solving the problem without dynamic programming would typically result in a less efficient solution, especially for larger strings and dictionaries.

terminal

Solution

Solution 1: Hash Table + Dynamic Programming

We can use a hash table $ss$ to record all words in the dictionary, which allows us to quickly determine whether a string is in the dictionary.

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class Solution:
    def minExtraChar(self, s: str, dictionary: List[str]) -> int:
        ss = set(dictionary)
        n = len(s)
        f = [0] * (n + 1)
        for i in range(1, n + 1):
            f[i] = f[i - 1] + 1
            for j in range(i):
                if s[j:i] in ss and f[j] < f[i]:
                    f[i] = f[j]
        return f[n]

Solution 2: Trie + Dynamic Programming

We can use a trie to optimize the time complexity of Solution 1.

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class Solution:
    def minExtraChar(self, s: str, dictionary: List[str]) -> int:
        ss = set(dictionary)
        n = len(s)
        f = [0] * (n + 1)
        for i in range(1, n + 1):
            f[i] = f[i - 1] + 1
            for j in range(i):
                if s[j:i] in ss and f[j] < f[i]:
                    f[i] = f[j]
        return f[n]

Continue Practicing

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