Given a string s, a k duplicate removal consists of choosing k adjacent and equal letters from s and removing them causing the left and the right side of the deleted substring to concatenate together.
We repeatedly make k duplicate removals on s until we no longer can.
Return the final string after all such duplicate removals have been made.
It is guaranteed that the answer is unique.
Example 1:
Input: s = "abcd", k = 2
Output: "abcd"
Explanation: There's nothing to delete.
Example 2:
Input: s = "deeedbbcccbdaa", k = 3
Output: "aa"
Explanation:
First delete "eee" and "ccc", get "ddbbbdaa"
Then delete "bbb", get "dddaa"
Finally delete "ddd", get "aa"
Example 3:
Input: s = "pbbcggttciiippooaais", k = 2
Output: "ps"
Constraints:
1 <= s.length <= 10^52 <= k <= 10^4sonly contains lower case English letters.
Solution:
Iterate from left to right, maintain a cound of how many consecutive chars we have at position i and a result string. Whenever the count reaches k we remove k elements.
class Solution {
public:
string removeDuplicates(string s, int k) {
string cur = "";
vector<int> dp(s.size()+1, 0);
for(int i = 1; i<=s.size();i++){
if(cur.empty()){
cur += s[i-1];
dp[i] = 1;
}else{
if(cur[cur.size()-1] == s[i-1]){
cur += s[i-1];
dp[cur.size()] = dp[cur.size()-1]+1;
if(dp[cur.size()] == k){
cur = cur.substr(0,cur.size()-k);
}
}else{
cur += s[i-1];
dp[cur.size()] = 1;
}
}
// cout<<cur<<" "<<dp[cur.size()]<<"\n";
}
return cur;
}
};