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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.
The JavaScript solution uses a Trie to store the dictionary and a recursive memoization strategy to determine the minimum extra characters needed by exploring feasible segments stored in the Trie.