Given a set of keywords words and a string S, make all appearances of all keywords in S bold. Any letters between and tags become bold.

The returned string should use the least number of tags possible, and of course the tags should form a valid combination.

For example, given that words = [“ab”, “bc”] and S = “aabcd”, we should return “aabcd”. Note that returning “aabcd” would use more tags, so it is incorrect.

Note:

• words has length in range [0, 50].
• words[i] has length in range [1, 10].
• S has length in range [0, 500].
• All characters in words[i] and S are lowercase letters.

Make String Bold using Bruteforce Algorithm

Without the requirement of the shortest, we can make bold on every occurrences of the words in the list. As we are given the original HTML string, we can mark bold for each character that appears in the bold words.

So, with O(N) space, and O(NM) where N is the size of the string and M is the total length of the bold string, the following C++ bruteforce algorithm will insert the least HTML bold tags that satisfy the requirement.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

class Solution {
public:
    string boldWords(vector<string>& words, string S) {
        int n = S.size();
        vector<bool> bold(n, false);
        for (const auto &s: words) {
            for (int i = 0; i + s.size() - 1 < n; ++ i) {
                bool ok = true;
                for (int k = 0; k < s.size(); ++ k) {
                    if (s[k] != S[i + k]) { // bold string s not found in the string
                        ok = false;
                        break;
                    }
                }
                if (ok) { // it is bold
                    for (int k = 0; k < s.size(); ++ k) {
                        bold[i + k] = true; // mark the character bold
                    }
                }
            }
        }
        string ans = "";
        for (int i = 0; i < n; ++ i) {
            // bold start boundary
            if (bold[i] && (i == 0 || !bold[i - 1])) ans += "<b>"; 
            ans += S[i];
            // bold end boundary
            if (bold[i] && (i == n - 1 || !bold[i + 1])) ans += "</b>";
        }
        return ans;
    }
};

class Solution {
public:
    string boldWords(vector<string>& words, string S) {
        int n = S.size();
        vector<bool> bold(n, false);
        for (const auto &s: words) {
            for (int i = 0; i + s.size() - 1 < n; ++ i) {
                bool ok = true;
                for (int k = 0; k < s.size(); ++ k) {
                    if (s[k] != S[i + k]) { // bold string s not found in the string
                        ok = false;
                        break;
                    }
                }
                if (ok) { // it is bold
                    for (int k = 0; k < s.size(); ++ k) {
                        bold[i + k] = true; // mark the character bold
                    }
                }
            }
        }
        string ans = "";
        for (int i = 0; i < n; ++ i) {
            // bold start boundary
            if (bold[i] && (i == 0 || !bold[i - 1])) ans += "<b>"; 
            ans += S[i];
            // bold end boundary
            if (bold[i] && (i == n - 1 || !bold[i + 1])) ans += "</b>";
        }
        return ans;
    }
};

After the bold characters are marked, we re-scan the string and look for the boundarys, and insert the tags accordingly.

–EOF (The Ultimate Computing & Technology Blog) —