2707. Extra Characters in a String - Detailed Explanation

Problem Statement

You are given a string s and an array of strings dictionary. You want to partition s into substrings such that each substring appears in dictionary. However, some characters in s might not be used in forming these dictionary words. Such characters are called extra characters. Your goal is to remove extra characters in order to form a valid segmentation of s where every remaining substring is a word from dictionary. Return the minimum number of extra characters left over after such a partitioning.

In other words, you need to choose a set of substrings (taken in order from s) that are in dictionary and minimize the total number of characters in s that are not covered by any chosen substring.

Example 1

Input:

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

Output:

1

Explanation:

• One optimal segmentation is to form "leet" + "s" + "code".
• The substrings "leet" and "code" are valid dictionary words.
• The extra character is "s" which is not part of any dictionary word.
• Thus, the minimum number of extra characters is 1.

Example 2

Input:

s = "sayhelloworld"
dictionary = ["hello", "world"]

Output:

3

Explanation:

• An optimal segmentation is "say" + "hello" + "world".
• Here, "hello" and "world" are valid dictionary words.
• The extra characters are "s", "a", "y" (i.e. the substring "say" is extra since it is not in the dictionary).
• Thus, the answer is 3.

Constraints

• 1 ≤ s.length ≤ 50
• 1 ≤ dictionary.length ≤ 50
• 1 ≤ dictionary[i].length ≤ 50
• s and dictionary[i] consist of only lowercase English letters.

Hints

1. Dynamic Programming (DP) Formulation:
   Think about defining a DP array where dp[i] represents the minimum number of extra characters from index i to the end of the string.

2. Choice at Each Index:
   At any position i in s, you have two choices:

   • Consider the character at s[i] as an extra character (and then add 1 to dp[i+1]).
   • Try to match a dictionary word starting at index i. For every valid word w where s[i:j+1] == w, you can "cover" that part and set dp[i] = min(dp[i], dp[j+1]).

3. Dictionary Lookup:
   Use a hash set for dictionary words so that you can quickly check if a substring is valid.

4. Bottom-Up vs. Top-Down:
   A bottom-up DP solution (iterating from the end of s backwards) works naturally for this problem.

Approaches

Approach 1: Brute Force (Inefficient)

• Idea:
  Try every possible segmentation of s and count the extra characters for each segmentation.

• Downside:
  The number of segmentations is exponential in the length of s, making this approach impractical even for moderate-sized inputs.

Approach 2: Dynamic Programming (Optimal)

• Idea:
  Use a DP array where dp[i] gives the minimum extra characters needed for the substring starting at index i.

• DP Recurrence:
  For each index i from the end of s to the beginning:

  • Initialize dp[i] = dp[i+1] + 1 (treating s[i] as extra).
  • Then, for every index j from i to the end:
    • If the substring s[i:j+1] is in the dictionary, update dp[i] = min(dp[i], dp[j+1]).

• Final Answer:
  The answer is dp[0], representing the minimum extra characters for the entire string.

• Why It Works:
  The DP approach considers both keeping characters as extra and grouping valid dictionary words, ensuring that every possibility is considered in an efficient manner.

Step-by-Step Walkthrough (Example: "leetscode")

Consider s = "leetscode" and dictionary = ["leet", "code", "leetcode"].

1. Initialization:

   • Let n = s.length = 9.
   • Create a DP array dp of size n+1 where dp[n] = 0 (base case: no extra characters beyond the end).

2. Iteration from Right to Left:

   • Index 8 (s[8] = 'e'):
     • Assume 'e' is extra: dp[8] = dp[9] + 1 = 1.
     • Check substrings starting at 8:
       Only "e" is possible; it is not in the dictionary.
   • Index 7 (s[7] = 'd'):
     • Default: dp[7] = dp[8] + 1 = 2.
     • Check "de" (s[7:9]): not in dictionary.
   • Index 6 (s[6] = 'o'):
     • Default: dp[6] = dp[7] + 1 = 3.
     • Check "ode" (s[6:9]): not in dictionary.
   • Index 5 (s[5] = 'c'):
     • Default: dp[5] = dp[6] + 1 = 4.
     • Check "code" (s[5:9]): "code" is in dictionary, so dp[5] = min(4, dp[9]) = 0.
   • Index 4 (s[4] = 's'):
     • Default: dp[4] = dp[5] + 1 = 1.
     • Check "scode": not in dictionary.
   • Index 3 (s[3] = 't'):
     • Default: dp[3] = dp[4] + 1 = 2.
     • Check "tscode": not in dictionary.
   • Index 2 (s[2] = 'e'):
     • Default: dp[2] = dp[3] + 1 = 3.
     • Check "etscode": not in dictionary.
   • Index 1 (s[1] = 'e'):
     • Default: dp[1] = dp[2] + 1 = 4.
     • Check "eetscode": not in dictionary.
   • Index 0 (s[0] = 'l'):
     • Default: dp[0] = dp[1] + 1 = 5.
     • Check "leet" (s[0:4]): "leet" is in dictionary, so dp[0] = min(5, dp[4]) = min(5, 1) = 1.
     • Also, check "leetscode" (s[0:9]): not in dictionary.

3. Result:

   • dp[0] = 1 which is the minimum number of extra characters.

Code Implementation

Python Code