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1023. Camelcase Matching.cpp
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95 lines (78 loc) · 2.72 KB
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//link: https://leetcode.com/problems/camelcase-matching/
/*
A query word matches a given pattern if we can insert lowercase letters to the pattern word so that it equals the query. (We may insert each character at any position, and may insert 0 characters.)
Given a list of queries, and a pattern, return an answer list of booleans, where answer[i] is true if and only if queries[i] matches the pattern.
Example 1:
Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
Output: [true,false,true,true,false]
Explanation:
"FooBar" can be generated like this "F" + "oo" + "B" + "ar".
"FootBall" can be generated like this "F" + "oot" + "B" + "all".
"FrameBuffer" can be generated like this "F" + "rame" + "B" + "uffer".
Example 2:
Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"
Output: [true,false,true,false,false]
Explanation:
"FooBar" can be generated like this "Fo" + "o" + "Ba" + "r".
"FootBall" can be generated like this "Fo" + "ot" + "Ba" + "ll".
Example 3:
Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"
Output: [false,true,false,false,false]
Explanation:
"FooBarTest" can be generated like this "Fo" + "o" + "Ba" + "r" + "T" + "est".
Note:
1 <= queries.length <= 100
1 <= queries[i].length <= 100
1 <= pattern.length <= 100
All strings consists only of lower and upper case English letters.
*/
/*
Result:
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Camelcase Matching.
Memory Usage: 6.3 MB, less than 100.00% of C++ online submissions for Camelcase Matching.
*/
class Solution {
public:
vector<bool> camelMatch(vector<string>& queries, string pattern) {
vector<bool> ans;
for(int i = 0; i < queries.size(); i ++){
if(match(queries[i], pattern)){
ans.push_back(true);
}
else{
ans.push_back(false);
}
}
return ans;
}
bool match(string s, string p){
int i = 0;
int j = 0;
while(i < s.length() && j < p.length()){
if(s[i] == p[j]){
i ++;
j ++;
}
else{
if(s[i] >= 'a' && s[i] <= 'z'){
i ++;
}
else{
return false;
}
}
}
if(j < p.length()){
return false;
}
if(i < s.length()){
while(i < s.length() && s[i] >= 'a' && s[i] <= 'z') {
i ++;
}
}
if(i < s.length()){
return false;
}
return true;
}
};