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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.
1def min_extra_chars(s, dictionary):
2 dict_set = set(dictionary)
3 n = len(s)
4 dp = [n] * (n + 1)
5 dp[0] = 0
6
7 for i in range(1, n + 1):
8 dp[i] = dp[i - 1] + 1
9 for j in range(i):
10 if s[j:i] in dict_set:
11 dp[i] = min(dp[i], dp[j])
12 return dp[n]
13
14s = "leetscode"
15dictionary = ["leet", "code", "leetcode"]
16print(min_extra_chars(s, dictionary)) # Output: 1The Python solution employs a set for the dictionary and a dynamic programming array to calculate the minimum extra characters for each prefix length, checking substrings against the dictionary.
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.