Leetcode 2707: Extra Characters in a String
You are given a string and a list of words. Your task is to break the string into one or more non-overlapping substrings such that each substring is a word from the given list. However, some characters of the string may not be used in any substring. Your goal is to minimize the number of unused characters.
Problem
Approach
Steps
Complexity
Input: The input consists of a string and a list of distinct words.
Example: Input: s = "codingrocks", dictionary = ["code", "rocks", "dog"]
Constraints:
• 1 <= s.length <= 50
• 1 <= dictionary.length <= 50
• 1 <= dictionary[i].length <= 50
• dictionary[i] and s consists of only lowercase English letters
• dictionary contains distinct words
Output: The output should be the minimum number of unused characters left after breaking the string optimally into substrings from the dictionary.
Example: Output: 3
Constraints:
• The number of unused characters should be minimized.
Goal: The goal is to find the minimum number of extra characters by breaking the string into substrings present in the dictionary.
Steps:
• Step 1: Use dynamic programming to try and break the string into valid substrings from the dictionary.
• Step 2: For each possible break, check if the substring exists in the dictionary.
• Step 3: Minimize the number of unused characters after breaking the string.
Goal: The input string and dictionary should be handled efficiently, and the solution must work within the given constraints.
Steps:
• The string length is between 1 and 50.
• The number of words in the dictionary is between 1 and 50.
Assumptions:
• The dictionary contains distinct words.
• Input: Input: s = "helloworld", dictionary = ["hello", "world"]
• Explanation: We can break the string into two substrings: "hello" and "world", using all characters from the string. There are no extra characters left, so the output is 0.
• Input: Input: s = "codingrocks", dictionary = ["code", "rocks", "dog"]
• Explanation: The string can be split into "code" and "rocks". The leftover characters are 'i' and 'n', so the output is 2.
Approach: To solve this problem, a dynamic programming approach is used where we break the string and attempt to match substrings with the dictionary words, minimizing the leftover characters.
Observations:
• The problem can be approached using dynamic programming (DP) to check all possible splits of the string.
• We should keep track of the minimum extra characters after each possible split.
Steps:
• Step 1: Initialize a DP array where each index represents the minimum leftover characters up to that point.
• Step 2: Iterate through the string and try to match substrings to words in the dictionary.
• Step 3: If a valid match is found, update the DP array with the new minimum leftover characters.
• Step 4: Return the minimum leftover characters after processing the entire string.
Empty Inputs:
• The string will never be empty as per constraints.
Large Inputs:
• The solution should handle strings up to length 50 efficiently.
Special Values:
• If no valid breaks are possible, the solution should return the entire length of the string as leftover.
Constraints:
• Ensure efficient handling of all possible substrings within the given constraints.
set<string> dic;
string str;
vector<int> memo;
int n;
int dp(int idx) {
if(idx == n) return 0;
if(memo[idx] != -1) return memo[idx];
int ans = INT_MAX;
int len = 1;
for(int i = idx; i < n; i++) {
string s = str.substr(idx, len);
if(dic.count(s)) {
ans = min(ans, dp(idx + len));
} else {
ans = min(ans, len + dp(idx + len));
}
len++;
}
return memo[idx] = ans;
}
int minExtraChar(string s, vector<string>& dictionary) {
n = s.size();
str = s;
memo.resize(n, -1);
for(auto x: dictionary) {
dic.insert(x);
}
return dp(0);
}
1 : Set Initialization
set<string> dic;
A set `dic` is initialized to store the dictionary words for efficient look-up.
2 : String Initialization
string str;
A string `str` is initialized to store the input string `s`.
3 : Vector Initialization
vector<int> memo;
A vector `memo` is declared to store the computed results for dynamic programming. This helps to avoid recalculating results for overlapping subproblems.
4 : Integer Initialization
int n;
An integer `n` is declared to store the length of the input string `s`.
5 : Function Declaration
int dp(int idx) {
The `dp` function is declared. It takes an index `idx` and returns the minimum number of extra characters needed to split the substring starting at `idx` into valid words from the dictionary.
6 : Base Case - End of String
if(idx == n) return 0;
Base case: if the index `idx` has reached the end of the string, return 0 as no extra characters are needed.
7 : Memoization Check
if(memo[idx] != -1) return memo[idx];
Check if the result for the current index `idx` has already been computed. If so, return the stored result from the `memo` vector.
8 : Initialization
int ans = INT_MAX;
Initialize a variable `ans` to store the minimum number of extra characters found for the current substring. It is initially set to the maximum possible value.
9 : Length Initialization
int len = 1;
Initialize the variable `len` to 1, representing the length of the current substring being considered.
10 : For Loop
for(int i = idx; i < n; i++) {
Start a for loop to iterate through the string starting at index `idx` and consider substrings of increasing length.
11 : Substring Extraction
string s = str.substr(idx, len);
Extract a substring `s` from the string `str`, starting at index `idx` with length `len`.
12 : Dictionary Lookup
if(dic.count(s)) {
Check if the substring `s` exists in the dictionary `dic`.
13 : Recursive Call - Valid Word
ans = min(ans, dp(idx + len));
If `s` is a valid word in the dictionary, recursively call `dp` with the next index `idx + len` and update the answer `ans` with the minimum result.
14 : Else Block - Invalid Word
} else {
If the substring `s` is not a valid word in the dictionary, handle the case where extra characters are needed.
15 : Recursive Call - Invalid Word
ans = min(ans, len + dp(idx + len));
If `s` is not a valid word, add `len` (the length of the substring) to the result of the recursive call `dp(idx + len)` to account for extra characters.
16 : Increment Length
len++;
Increment the length of the substring `len` for the next iteration.
17 : Return Statement
return memo[idx] = ans;
Store the computed result `ans` in `memo[idx]` and return it.
18 : Function Declaration
int minExtraChar(string s, vector<string>& dictionary) {
The `minExtraChar` function is declared. It takes the input string `s` and the dictionary `dictionary` as parameters and returns the minimum number of extra characters.
19 : Initialize n
n = s.size();
Set `n` to the length of the input string `s`.
20 : Assign String
str = s;
Assign the string `s` to the global variable `str`.
21 : Resize Memo
memo.resize(n, -1);
Resize the `memo` vector to size `n` and initialize all values to `-1` to indicate that no subproblem has been solved yet.
22 : Insert Dictionary Words
for(auto x: dictionary) {
Iterate through the `dictionary` and insert each word into the `dic` set.
23 : Insert Word
dic.insert(x);
Insert each word `x` from the dictionary into the set `dic`.
24 : Return Statement
return dp(0);
Call the `dp` function starting at index `0` to calculate the result and return it.
Best Case: O(n^2)
Average Case: O(n^2)
Worst Case: O(n^2)
Description: The time complexity is O(n^2) due to checking all possible substrings against the dictionary.
Best Case: O(n)
Worst Case: O(n)
Description: The space complexity is O(n) for storing the DP array.
| LeetCode Solutions Library / DSA Sheets / Course Catalog |
|---|
comments powered by Disqus