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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.
1#include <iostream>
2#include <vector>
3#include <unordered_set>
4
5using namespace std;
6
7int minExtraChars(string s, vector<string>& dictionary) {
8 unordered_set<string> dictSet(dictionary.begin(), dictionary.end());
9 int n = s.length();
10 vector<int> dp(n + 1, n);
11 dp[0] = 0;
12
13 for (int i = 1; i <= n; ++i) {
14 dp[i] = dp[i - 1] + 1;
15 for (int j = 0; j < i; ++j) {
16 if (dictSet.find(s.substr(j, i - j)) != dictSet.end()) {
17 dp[i] = min(dp[i], dp[j]);
18 }
19 }
20 }
21 return dp[n];
22}
23
24int main() {
25 string s = "leetscode";
26 vector<string> dictionary = {"leet", "code", "leetcode"};
27 int result = minExtraChars(s, dictionary);
28 cout << result << endl; // Output: 1
29 return 0;
30}The C++ solution leverages an unordered_set for efficient dictionary look-up. We iterate through each possible ending position in the string to calculate the minimum extra characters needed using a dynamic programming approach.
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.
This C solution utilizes a Trie to store dictionary words for fast matching. It also uses memoization to recursively find the minimum extra characters. Each call attempts to split the string from the current index, using the Trie to validate substrings.