Formatted question description: https://leetcode.ca/all/758.html

758. Bold Words in String

Level

Easy

Description

Given a set of keywords words and a string S, make all appearances of all keywords in S bold. Any letters between <b> and </b> 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 "a<b>abc</b>d". Note that returning "a<b>a<b>b</b>c</b>d" would use more tags, so it is incorrect.

Note:

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

Solution

Use an array with type boolean to determine whether each index should become bold. Find indices of all occurrences of keywords in S, and set all the indices where the keywords occur to be bold.

Loop over the string S, and add tags at start and end indices in S.

class Solution {
    public String boldWords(String[] words, String S) {
        int length = S.length();
        boolean[] bold = new boolean[length];
        for (String word : words) {
            int wordLength = word.length();
            int beginIndex = 0;
            while (beginIndex >= 0) {
                beginIndex = S.indexOf(word, beginIndex);
                if (beginIndex >= 0) {
                    for (int i = 0; i < wordLength; i++)
                        bold[beginIndex + i] = true;
                    beginIndex++;
                }
            }
        }
        StringBuffer sb = new StringBuffer(S);
        if (bold[length - 1])
            sb.append("</b>");
        for (int i = length - 1; i > 0; i--) {
            if (bold[i] && !bold[i - 1])
                sb.insert(i, "<b>");
            else if (!bold[i] && bold[i - 1])
                sb.insert(i, "</b>");
        }
        if (bold[0])
            sb.insert(0, "<b>");
        String boldStr = sb.toString();
        return boldStr;
    }
}

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