Expressive Words Problem
Description
LeetCode Problem 809.
Sometimes people repeat letters to represent extra feeling. For example:
- “hello” -> “heeellooo”
- “hi” -> “hiiii”
In these strings like “heeellooo”, we have groups of adjacent letters that are all the same: “h”, “eee”, “ll”, “ooo”.
You are given a string s and an array of query strings words. A query word is stretchy if it can be made to be equal to s by any number of applications of the following extension operation: choose a group consisting of characters c, and add some number of characters c to the group so that the size of the group is three or more.
- For example, starting with “hello”, we could do an extension on the group “o” to get “hellooo”, but we cannot get “helloo” since the group “oo” has a size less than three. Also, we could do another extension like “ll” -> “lllll” to get “helllllooo”. If s = “helllllooo”, then the query word “hello” would be stretchy because of these two extension operations: query = “hello” -> “hellooo” -> “helllllooo” = s.
Return the number of query strings that are stretchy.
Example 1:
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Input: s = "heeellooo", words = ["hello", "hi", "helo"]
Output: 1
Explanation:
We can extend "e" and "o" in the word "hello" to get "heeellooo".
We can't extend "helo" to get "heeellooo" because the group "ll" is not size 3 or more.
Example 2:
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Input: s = "zzzzzyyyyy", words = ["zzyy","zy","zyy"]
Output: 3
Constraints:
- 1 <= s.length, words.length <= 100
- 1 <= words[i].length <= 100
- s and words[i] consist of lowercase letters.
Sample C++ Code
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class Solution {
public:
void cntChars(string s, vector<int> &charVec,
vector<int> &cntVec) {
charVec.push_back(s[0]-'a');
int cnt = 1;
for (int i = 1;i < s.size(); i ++) {
if ((s[i] - 'a') == charVec.back()) {
cnt ++;
} else {
cntVec.push_back(cnt);
charVec.push_back(s[i] - 'a');
cnt = 1;
}
}
cntVec.push_back(cnt);
}
int expressiveWords(string S, vector<string>& words) {
int count = 0;
vector<int> charSeq, cntSeq;
vector<int> charW, cntW;
if (S.size() == 0)
return 0;
cntChars(S, charSeq, cntSeq);
for (int i = 0; i < words.size(); i ++) {
string e = words[i];
charW.clear();
cntW.clear();
if (e.size() == 0)
continue;
cntChars(e, charW, cntW);
if (charW.size() != charSeq.size())
continue;
bool is_stretchy = true;
for (int j = 0; j < charW.size(); j ++) {
if (charW[j] != charSeq[j]) {
is_stretchy = false;
break;
}
if (cntW[j] != cntSeq[j]) {
if ((cntSeq[j] < 3) || (cntSeq[j] < cntW[j])) {
is_stretchy = false;
break;
}
}
}
if (is_stretchy)
count ++;
}
return count;
}
};