Expressive Words LeetCode solution in Java

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class Solution {
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  private String getFreqString(String s) {
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    int len = s.length();
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    StringBuilder freqString = new StringBuilder();
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    int currFreq = 1;
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    char prevChar = s.charAt(0);
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    freqString.append(s.charAt(0));
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    for (int i = 1; i < len; i++) {
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      if (s.charAt(i) == prevChar) {
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        currFreq++;
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      } else {
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        freqString.append(currFreq);
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        freqString.append(s.charAt(i));
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        currFreq = 1;
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      }
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      prevChar = s.charAt(i);
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    }
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    if (currFreq > 0) {
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      freqString.append(currFreq);
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    }
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    return freqString.toString();
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  }
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  private boolean isGreaterButLessThanThree(char sChar, char wChar) {
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    return sChar > wChar && sChar < '3';
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  }
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  private boolean isStretchy(String s, String word) {
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    int sLen = s.length();
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    int wordLen = word.length();
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    if (sLen != wordLen) {
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      return false;
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    }
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    for (int i = 0; i < sLen; i++) {
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      char sChar = s.charAt(i);
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      char wChar = word.charAt(i);
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      if (i % 2 != 0) {
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        if (sChar < wChar) {
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          return false;
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        }
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        if (isGreaterButLessThanThree(sChar, wChar)) {
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          return false;
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        }
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      } else if (sChar != wChar) {
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        return false;
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      }
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    }
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    return true;
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  }
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  public int expressiveWords(String s, String[] words) {
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    int wordLen = words.length;
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    if (wordLen < 1 || s.length() < 1) {
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      return 0;
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    }
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    int stretchyWords = 0;
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    String freqStringS = getFreqString(s);
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    for (String word : words) {
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      String freqStringWord = getFreqString(word);
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      if (isStretchy(freqStringS, freqStringWord)) {
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        stretchyWords++;
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      }
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    }
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    return stretchyWords;
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  }
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}