2707. Extra Characters in a String

Problem Statement

You are given a 0-indexed string s and a dictionary of words dictionary. You have to break s into one or more non-overlapping substrings such that each substring is present in dictionary. There may be some extra characters in s which are not present in any of the substrings.

Return the minimum number of extra characters left over if you break up s optimally.

Example 1:

Input: s = "leetscode", dictionary = ["leet","code","leetcode"]
Output: 1
Explanation: We can break s in two substrings: "leet" from index 0 to 3 and "code" from index 5 to 8. There is only 1 unused character (at index 4), so we return 1.

Example 2:

Input: s = "sayhelloworld", dictionary = ["hello","world"]
Output: 3
Explanation: We can break s in two substrings: "hello" from index 3 to 7 and "world" from index 8 to 12. The characters at indices 0, 1, 2 are not used in any substring and thus are considered as extra characters. Hence, we return 3.

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

Intuition

Approach:

If the string does not match, Most likely
It is the removal characters

Take that string lenght + f(index+1)
Or if matches 0 + f(index+1)

Video Links

Approach 1:

Memoization

C++

class Solution {
public:
    unordered_map<string, bool> mp;
    vector<int> dp;
    int find_ans(string &s, int index){
        if(index == s.size())
            return 0;

        if(dp[index] != -1)
            return dp[index];

        int ans = INT_MAX;
        for(int cur=index; cur<s.size(); cur++){
            string temp = s.substr(index, cur-index+1);
            int temp_ans;

            if(!mp[temp])
                temp_ans = temp.size() + find_ans(s, cur+1);
            else
                temp_ans = find_ans(s, cur+1);

            ans=min(ans, temp_ans);
        }

        return dp[index] = ans;
    }


    int minExtraChar(string s, vector<string>& dict) {
        for(auto &it: dict)
            mp[it] = true;
        
        dp = vector<int>(s.size(),-1);

        return find_ans(s, 0);
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++