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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.
1using System;
2using System.Collections.Generic;
3
4public class Solution {
5 public int MinExtraChars(string s, IList<string> dictionary) {
6 var dictSet = new HashSet<string>(dictionary);
7 int n = s.Length;
8 var dp = new int[n + 1];
9 Array.Fill(dp, n);
10 dp[0] = 0;
11
12 for (int i = 1; i <= n; i++) {
13 dp[i] = dp[i - 1] + 1;
14 for (int j = 0; j < i; j++) {
15 if (dictSet.Contains(s.Substring(j, i - j))) {
16 dp[i] = Math.Min(dp[i], dp[j]);
17 }
18 }
19 }
20 return dp[n];
21 }
22
23 public static void Main(string[] args) {
24 var solution = new Solution();
25 string s = "leetscode";
26 IList<string> dictionary = new List<string>{"leet", "code", "leetcode"};
27 Console.WriteLine(solution.MinExtraChars(s, dictionary)); // Output: 1
28 }
29}The C# solution employs a HashSet for efficient dictionary lookups and utilizes a dynamic programming approach to minimize the count of extra characters by evaluating substrings.
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 Python solution builds a Trie to store dictionary words and uses a recursive function with memoization to evaluate segments. Each segment is checked in the Trie to ensure minimal unaccounted characters.