g0801_0900.s0809_expressive_words.Solution Maven / Gradle / Ivy
Go to download
Show more of this group Show more artifacts with this name
Show all versions of leetcode-in-java Show documentation
Show all versions of leetcode-in-java Show documentation
Java-based LeetCode algorithm problem solutions, regularly updated
The newest version!
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;
}
}
© 2015 - 2024 Weber Informatics LLC | Privacy Policy