Formatted question description: https://leetcode.ca/all/1209.html
1209. Remove All Adjacent Duplicates in String II (Medium)
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 <= 10^5
2 <= k <= 10^4
s
only contains lower case English letters.
Related Topics:
Stack
Similar Questions:
Solution 1. Stack
// OJ: https://leetcode.com/problems/remove-all-adjacent-duplicates-in-string-ii/
// Time: O(N)
// Space: O(N)
class Solution {
public:
string removeDuplicates(string s, int k) {
vector<pair<char, int>> st;
for (char c : s) {
if (st.empty() || st.back().first != c) st.emplace_back(c, 1);
else if (++st.back().second == k) st.pop_back();
}
string ans;
for (int i = 0; i < st.size(); ++i) {
for (int j = 0; j < st[i].second; ++j) ans += st[i].first;
}
return ans;
}
};
Java
class Solution {
public String removeDuplicates(String s, int k) {
StringBuffer sb = new StringBuffer();
List<Integer> duplicatesCounts = new ArrayList<Integer>();
int length = s.length();
for (int i = 0; i < length; i++) {
int curLength = sb.length();
char c = s.charAt(i);
int curCount = 1;
if (curLength > 0 && c == sb.charAt(curLength - 1))
curCount = duplicatesCounts.get(curLength - 1) + 1;
if (curCount == k) {
int startIndex = curLength - k + 1;
sb.delete(startIndex, curLength);
for (int j = curLength - 1; j >= startIndex; j--)
duplicatesCounts.remove(j);
} else {
sb.append(c);
duplicatesCounts.add(curCount);
}
}
return sb.toString();
}
}