Extra Characters in a String

class Solution {
public:
    int minInsertionsRequired(string str, vector<string> words) {
        int len = str.length();
        unordered_set<string> wordsSet(words.begin(), words.end());
        vector<int> dp(len + 1, 0);

        for (int i = len - 1; i >= 0; i--) {
            dp[i] = dp[i + 1] + 1;
            for (int j = i; j < len; j++) {
                auto currentWord = str.substr(i, j - i + 1);
                if (wordsSet.count(currentWord)) {
                    dp[i] = min(dp[i], dp[j + 1]);
                }
            }
        }

        return dp[0];
    }
};

The provided C++ solution calculates the minimum number of insertions required to construct a given string using a set of words. The method utilized involves dynamic programming.

Here's how this solution works:

1. Initialize the length of the string to len.
2. Convert the list of words into an unordered set called wordsSet for quick look-up.
3. Create a dynamic programming array, dp, of size len + 1, and initialize all values to zero. This array keeps track of the minimal insertions needed from any position i to the end of the string.
4. Fill the dp array backwards. For each position i in the string, start by setting dp[i] as dp[i + 1] + 1.
5. For each possible substring starting at i and ending at j, check whether this substring exists in wordsSet.
6. If the substring exists, update dp[i] with the minimum of its current value or dp[j + 1], where j + 1 is the position right after the end of the current substring.
7. After iterating through all substrings and positions in the string, dp[0] holds the minimum number of insertions required for the entire string.

The use of set for quick look-up and dynamic programming strategy makes the solution efficient by reducing redundant calculations. The final result, returned by the function, is dp[0], which represents the minimum insertions for the whole string based on the available words.

class Solution {
    public int minimumInsertions(String inputString, String[] words) {
        int length = inputString.length();
        var validWords = new HashSet<>(Arrays.asList(words));
        var memo = new int[length + 1];

        for (int i = length - 1; i >= 0; i--) {
            memo[i] = memo[i + 1] + 1;
            for (int j = i; j < length; j++) {
                var sub = inputString.substring(i, j + 1);
                if (validWords.contains(sub)) {
                    memo[i] = Math.min(memo[i], memo[j + 1]);
                }
            }
        }

        return memo[0];
    }
}

The Java program provided is designed to compute the minimum number of insertions required to convert an input string into a sequence composed entirely of words from a specified list. Here is a functional breakdown of how the implementation works:

• The function minimumInsertions accepts two parameters: inputString, which is the string to be transformed, and words, an array of strings against which the inputString will be compared.
• A HashSet named validWords is initialized with the contents of the words array to facilitate fast lookup operations.
• An integer array memo is used for dynamic programming purposes. The array is of size length + 1 where length is the number of characters in inputString. This array tracks the minimum number of insertions needed for each substring of the input string starting from any position i to the end of the string.
• The main logic is implemented in a nested loop:
  • The outer loop iterates over the indices of the input string in reverse, starting from the last character.
  • For each character at position i, it initially assumes the worst case where an insertion is needed for each character (memo[i + 1] + 1).
  • The inner loop checks every possible substring that starts at index i and ends at index j. If the substring exists in validWords, it updates memo[i] with the minimum value between its current value and memo[j + 1].
• The function ultimately returns memo[0], representing the minimum insertions needed for the entire string to transform it into sequences of valid words.

This solution effectively utilizes dynamic programming and memoization to optimize the calculation of minimum insertions, ensuring that substrings are checked efficiently and that previously calculated results are reused.

var minExtraCharacters = function(word, dict) {
    const wordLength = word.length;
    const dictSet = new Set(dict);
    const solution = Array(wordLength + 1).fill(0);

    for (let startPos = wordLength - 1; startPos >= 0; startPos--) {
        solution[startPos] = solution[startPos + 1] + 1;
        for (let endPos = startPos; endPos < wordLength; endPos++) {
            const wordFragment = word.substring(startPos, endPos + 1);
            if (dictSet.has(wordFragment)) {
                solution[startPos] = Math.min(solution[startPos], solution[endPos + 1]);
            }
        }
    }

    return solution[0];
};

This article provides an overview of the JavaScript solution for counting the minimum number of extra characters required to transform a given word into meaningful substrings from a specified dictionary.

• Begin by defining the function minExtraCharacters that takes word and dict as parameters.
• Create a variable wordLength to store the length of the word.
• Use a JavaScript Set to store dictionary words for quick lookup.
• Initialize an array solution where each index represents the minimum number of extra characters starting from that position in the word up to the end.
• Iterate backwards through the word, starting from the last character to the first.
• For each starting position startPos, assume by default that adding one character is necessary, by setting solution[startPos] to solution[startPos + 1] + 1.
• Check every possible ending position endPos from startPos to identify substrings. For each substring, check if it is in the dictionary using the Set.
• If the substring exists in the dictionary, update solution[startPos] to be the minimum value between its current value and solution[endPos + 1], implying no extra characters are needed for the substring itself.
• Return the value of solution[0], which represents the minimum number of extra characters needed for the entire word.

With this approach, efficiently determine the minimal insertions required for the word to solely consist of dictionary words, leveraging dynamic programming and optimal substring checks.

class Solution:
    def minimumInserts(self, string: str, vocab: List[str]) -> int:
        length = len(string)
        vocab_set = set(vocab)
        min_inserts = [0] * (length + 1)

        for i in range(length - 1, -1, -1):
            min_inserts[i] = 1 + min_inserts[i + 1]
            for j in range(i, length):
                word = string[i: j + 1]
                if word in vocab_set:
                    min_inserts[i] = min(min_inserts[i], min_inserts[j + 1])

        return min_inserts[0]

This Python solution provides an efficient way to calculate the minimum number of insertions needed into a string to form other valid strings, as defined by a provided list of words (vocab). The function minimumInserts utilizes dynamic programming to keep track of the minimal operations needed.

• Initialize Important Variables:

  • length stores the length of the input string for easy access.
  • Convert vocab into a set vocab_set for O(1) average lookup time.
  • A list min_inserts initialized with zeros, and an extra space to handle edge cases, records the minimum number of insertions required at each position.

• Iterate Over the String:

  1. Loop backwards through the string starting from the second last character to the beginning.
  2. This reverse iteration helps in building the solution opportunistically from the smallest sub-problem (end of the string) to the beginning.

• Check All Substrings:

  1. For each position i, consider every substring that starts at i. It continues iterating until the character just before the end of the string.
  2. For every word built from these substrings, if it exists in the vocab_set, update the min_inserts for position i to the minimal of its current value or the insertions required after considering the current substring.

• Return the Result:

  • Output min_inserts[0], which reveals the minimum insertions necessary for the entire string to be transformed as per the vocabulary provided.

This efficient and robust solution ensures that you get the minimum insertions needed without unnecessary recomputation, storing intermediate results effectively via dynamic programming.