Sponsored
Sponsored
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.
using System.Collections.Generic;
public class Solution {
public int MinExtraChars(string s, IList<string> dictionary) {
var dictSet = new HashSet<string>(dictionary);
int n = s.Length;
var dp = new int[n + 1];
Array.Fill(dp, n);
dp[0] = 0;
for (int i = 1; i <= n; i++) {
dp[i] = dp[i - 1] + 1;
for (int j = 0; j < i; j++) {
if (dictSet.Contains(s.Substring(j, i - j))) {
dp[i] = Math.Min(dp[i], dp[j]);
}
}
}
return dp[n];
}
public static void Main(string[] args) {
var solution = new Solution();
string s = "leetscode";
IList<string> dictionary = new List<string>{"leet", "code", "leetcode"};
Console.WriteLine(solution.MinExtraChars(s, dictionary)); // Output: 1
}
}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.
1class TrieNode:
2 def __init__(self):
3 self.children = {}
4 self.isEnd = False
5
6class Trie:
7 def __init__(self):
8 self.root = TrieNode()
9
10 def insert(self, word):
11 node = self.root
12 for char in word:
13 if char not in node.children:
14 node.children[char] = TrieNode()
15 node = node.children[char]
16 node.isEnd = True
17
18def minExtraHelper(s, idx, root, memo):
19 n = len(s)
20 if idx == n:
21 return 0
22 if memo[idx] != -1:
23 return memo[idx]
24 minExtra = n - idx
25 node = root
26 for i in range(idx, n):
27 if s[i] not in node.children:
28 break
29 node = node.children[s[i]]
30 if node.isEnd:
31 minExtra = min(minExtra, minExtraHelper(s, i + 1, root, memo))
32 minExtra = min(minExtra, 1 + minExtraHelper(s, idx + 1, root, memo))
33 memo[idx] = minExtra
34 return minExtra
35
36
37def minExtraChars(s, dictionary):
38 trie = Trie()
39 for word in dictionary:
40 trie.insert(word)
41 n = len(s)
42 memo = [-1] * n
43 return minExtraHelper(s, 0, trie.root, memo)
44
45s = "leetscode"
46dictionary = ["leet", "code", "leetcode"]
47print(minExtraChars(s, dictionary)) # Output: 1This 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.