How do you solve Extra Characters in a String on LeetCode?

Extra Characters in a String (LeetCode #2707) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Dynamic programming helps break down the problem into manageable parts.

What is the brute force solution for Extra Characters in a String?

The brute force approach for Extra Characters in a String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible substring of the given string to see if it exists in the dictionary. If it doesn't, we count it as an extra character. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Extra Characters in a String?

Extra Characters in a String uses the following concepts: Array, Hash Table, String, Dynamic Programming, Trie. The recommended patterns to study are: Dynamic Programming, Substring Search.

What are the interview tips for Extra Characters in a String?

Always clarify the problem statement and constraints. Think about edge cases, such as empty strings or no valid substrings. Explain your thought process clearly while coding.

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

Not considering all possible substrings. Failing to initialize the DP array correctly.

#2707

Extra Characters in a String

Medium

ArrayHash TableStringDynamic ProgrammingTrieDynamic ProgrammingSubstring Search

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

Using dynamic programming, we can keep track of the minimum extra characters needed to form valid substrings up to each index. This allows us to build on previous results and avoid redundant checks.

⚙️

Algorithm

5 steps

1Step 1: Create a DP array where DP[i] represents the minimum extra characters for the substring s[0:i].
2Step 2: Initialize DP[0] to 0 (no characters, no extras).
3Step 3: For each index i, check all substrings ending at i and update DP[i] based on previous DP values.
4Step 4: If a substring is found in the dictionary, update DP[i] to the minimum of its current value and DP[start] (where start is the beginning of the substring).
5Step 5: If no valid substring is found, increment the extra count from DP[i-1] by 1.

solution.py10 lines

1def minExtraChars(s, dictionary):
2    word_set = set(dictionary)
3    n = len(s)
4    dp = [0] * (n + 1)
5    for i in range(1, n + 1):
6        dp[i] = dp[i - 1] + 1  # Assume the character is extra
7        for j in range(i):
8            if s[j:i] in word_set:
9                dp[i] = min(dp[i], dp[j])
10    return dp[n]

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops for substring checks, but we optimize the process by storing results in a DP array. The space complexity is O(n) for the DP array.

1Dynamic programming helps break down the problem into manageable parts.
2Using a set for the dictionary allows for O(1) average time complexity for lookups.

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