LeetCode 2707: Extra Characters in a String Solution

Master LeetCode problem 2707 (Extra Characters in a String), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2707. Extra Characters in a String

Medium

Array Hash Table String Dynamic Programming Trie

Problem Explanation

Explanation

To solve this problem, we can use dynamic programming. We will iterate over the input string s and keep track of the minimum number of extra characters left over up to each index. At each index, we will check all possible substrings starting from that index and ending at a point where a word in the dictionary ends. We update the minimum number of extra characters left over at each index based on the minimum of the current value and the value obtained by considering the current substring.

Algorithm:

Initialize an array dp of size s.length() + 1 to store the minimum number of extra characters left over up to each index.
Initialize dp[0] to 0, as there are no extra characters at the beginning.
Iterate over each index i from 0 to s.length().
For each index i, iterate over each word in the dictionary.
If the substring starting at index i can form the current word, update dp[i + word.length()] to the minimum of its current value and dp[i] + (i + word.length() - i - word.length()), which represents the number of extra characters left over after using the current word.
At the end, return dp[s.length()], which represents the minimum number of extra characters left over after breaking up the string optimally.

Time Complexity:

The time complexity of this solution is O(n * m), where n is the length of the input string s and m is the total number of characters in the dictionary.

Space Complexity:

The space complexity of this solution is O(n), where n is the length of the input string s.

Solution Code

class Solution {
    public int minExtraCharacters(String s, String[] dictionary) {
        int n = s.length();
        int[] dp = new int[n + 1];
        for (int i = 0; i < n; i++) {
            for (String word : dictionary) {
                if (i + word.length() <= n && s.substring(i, i + word.length()).equals(word)) {
                    dp[i + word.length()] = Math.min(dp[i + word.length()], dp[i] + (i + word.length() - i - word.length()));
                }
            }
        }
        return dp[n];
    }
}

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