g0801_0900.s0809_expressive_words.Solution Maven / Gradle / Ivy

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Java-based LeetCode algorithm problem solutions, regularly updated

There is a newer version: 1.37

package g0801_0900.s0809_expressive_words;

// #Medium #Array #String #Two_Pointers #2022_03_23_Time_2_ms_(86.09%)_Space_42.9_MB_(22.41%)

public class Solution {
    public int expressiveWords(String s, String[] words) {
        int ans = 0;
        for (String w : words) {
            if (check(s, w)) {
                ans++;
            }
        }
        return ans;
    }

    private boolean check(String s, String w) {
        int i = 0;
        int j = 0;
        /* Logic is to check whether character at same index of S and w are same
          if same,
           1. Find the consecutive number of occurrences of the char in S (say len1) and w ( say len2)
           2. If len1 == len 2 , move to the next char in S and w
           3. If  len1 >= 3 and len2 < len1, means we can make the char in w stretchy to match len1
           4. else, return false, because it's not possible to stretch the char in w
        */
        while (i < s.length() && j < w.length()) {
            char ch1 = s.charAt(i);
            char ch2 = w.charAt(j);

            int len1 = getLen(s, i);
            int len2 = getLen(w, j);
            if (ch1 == ch2) {
                if (len1 == len2 || (len1 >= 3 && len2 < len1)) {
                    i = i + len1;
                    j = j + len2;
                } else {
                    return false;
                }
            } else {
                return false;
            }
        }
        return i == s.length() && j == w.length();
    }

    private int getLen(String value, int i) {
        i = i + 1;
        int count = 1;
        for (int j = i; j < value.length(); j++) {
            if (value.charAt(j) == value.charAt(i - 1)) {
                count++;
            } else {
                break;
            }
        }
        return count;
    }
}