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In this approach, we use dynamic programming to solve the problem. We define a DP array where dp[i] represents the minimum number of extra characters from index 0 to i of the string s. Initially, we set dp[0] to 0 because there are no extra characters in an empty prefix. For each index i while iterating through the string s, we check every j from 0 to i-1 and see if substring s[j:i] exists in the dictionary. If yes, we update dp[i] to min(dp[i], dp[j]), otherwise, we set dp[i] to dp[i-1] + 1.
Time Complexity: O(N^2*M), where N is the length of the string s and M is the average length of words in the dictionary.
Space Complexity: O(N) because of the DP array.
1function minExtraChars(s, dictionary) {
2 const dictSet = new Set(dictionary);
3 const n = s.length;
4 const dp = Array(n + 1).fill(n);
5 dp[0] = 0;
6
7 for (let i = 1; i <= n; i++) {
8 dp[i] = dp[i - 1] + 1;
9 for (let j = 0; j < i; j++) {
10 if (dictSet.has(s.substring(j, i))) {
11 dp[i] = Math.min(dp[i], dp[j]);
12 }
13 }
14 }
15 return dp[n];
16}
17
18const s = "leetscode";
19const dictionary = ["leet", "code", "leetcode"];
20console.log(minExtraChars(s, dictionary)); // Output: 1The JavaScript solution uses a Set for fast membership checking in the dictionary and a DP array to compute minimal extra characters required. The solution inspects substrings using nested loops to achieve the optimal state.
To solve the problem using a Trie and memoization, we first build a Trie from the dictionary. We then use a recursive function with memoization to attempt to decompose the string s into valid segments. For each position in the string, we check possible substrings against the Trie, saving calculated results to avoid redundant computations.
Time Complexity: O(N*M), with N being the string length and M the average dictionary word length due to Trie traversal.
Space Complexity: O(N + T), N is for the memo array and T is for Trie storage.
The Java solution utilizes a Trie for dictionary storage and memoization for backtracking efficiently. Each recursive call evaluates further segments, minimizing unparsed characters using the Trie for word matching.