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Minimum Extra Characters in String Problem - Step by Step Explanation

This README walks through the solution for calculating the minimum number of extra characters required to split a string into valid words from a given dictionary. We explore this in C++, Java, JavaScript, Python, and Go. Each step provides a code snippet followed by detailed explanations.

C++ Solution

  1. Convert Dictionary to Unordered Set:

    • Convert the list of dictionary words into an unordered set for fast lookups (O(1) time complexity).

  2. DP Array Initialization:

    • Create a DP array dp[] where dp[i] represents the minimum number of extra characters required to parse the substring s[0:i].
    • Initialize each entry of dp[] to the maximum possible number of extra characters, which is n (the length of the string s).
    • Set the base case dp[0] = 0 (no extra characters for an empty string).

  3. Iterate Over String:

    • For each index i from 1 to n, check all possible substrings ending at i by looping through each j from 0 to i-1.
    • Extract the substring s[j:i] and check if it exists in the dictionary.

  4. Update DP Array:

    • If a valid substring is found, update dp[i] by considering dp[j] (no extra characters needed for this substring).
    • If no valid substring is found, treat the current character as an extra character and update dp[i] by incrementing dp[i-1].

  5. Return Result:

    • The result is stored in dp[n], which represents the minimum extra characters needed for the entire string.

Java Solution

  1. Create Dictionary Set:

    • Convert the dictionary array into a set for fast O(1) lookups.

  2. Initialize DP Array:

    • Create a dp[] array of size n+1, where dp[i] stores the minimum number of extra characters required to process the first i characters of the string.
    • Set all values in dp[] to n (maximum possible extra characters) and set dp[0] = 0.

  3. Check All Substrings:

    • Loop over each position i in the string.
    • For each i, check all possible substrings s[j:i] where j < i and check if the substring is present in the dictionary.

  4. Update DP Array:

    • If a valid substring is found, update dp[i] with the minimum between dp[i] and dp[j].
    • If no valid substring is found, consider the current character as an extra and update dp[i] by adding 1 to dp[i-1].

  5. Final Output:

    • The minimum extra characters for the entire string is stored in dp[n].

JavaScript Solution

  1. Convert Dictionary to Set:

    • Use a Set to store dictionary words for O(1) lookups.

  2. Initialize DP Array:

    • Create an array dp[] of size n+1 and initialize it with n (the maximum number of extra characters).
    • Set the base case dp[0] = 0.

  3. Check Possible Substrings:

    • Loop through each index i from 1 to n, and for each i, check all possible substrings s[j:i] where j < i.
    • Use Set.has() to check if the substring exists in the dictionary.

  4. Update DP Array:

    • If a valid substring is found, update dp[i] by considering dp[j].
    • Otherwise, treat the current character as extra and update dp[i] to dp[i-1] + 1.

  5. Final Answer:

    • The answer is stored in dp[n].

Python Solution

  1. Convert Dictionary to Set:

    • Convert the dictionary list into a set for O(1) time complexity during lookup.

  2. Initialize DP Array:

    • Create a DP array dp[] where dp[i] represents the minimum extra characters needed for the substring s[0:i].
    • Initialize dp[] with the value n, and set dp[0] = 0 (base case).

  3. Iterate Over Substrings:

    • Loop through each index i from 1 to n and check all substrings s[j:i] where j < i.
    • If the substring exists in the dictionary, update dp[i].

  4. Update for Extra Characters:

    • If no valid substring is found, consider the current character s[i-1] as extra and update dp[i] = dp[i-1] + 1.

  5. Return Result:

    • The final result is found in dp[n].

Go Solution

  1. Create Dictionary Map:

    • Create a map to store dictionary words for O(1) lookups.

  2. Initialize DP Array:

    • Create a DP array dp[] where dp[i] stores the minimum number of extra characters required up to index i.
    • Initialize the DP array to n (maximum possible extra characters) and set dp[0] = 0.

  3. Iterate Over Substrings:

    • Loop through each index i from 1 to n and check all substrings s[j:i].
    • If the substring is in the dictionary, update dp[i] by considering dp[j].

  4. Handle Extra Characters:

    • If no valid substring is found, consider the current character s[i-1] as extra and update dp[i] = dp[i-1] + 1.

  5. Final Result:

    • The result is stored in dp[n].