LeetCode 1209: Remove All Adjacent Duplicates in String II Solution
Master LeetCode problem 1209 (Remove All Adjacent Duplicates in String II), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.
Problem Explanation
Explanation
To solve this problem, we can use a stack to keep track of characters and their counts. We iterate through the input string and for each character, we check if it is the same as the character at the top of the stack. If it is, we increment the count. If the count becomes equal to k, we remove the characters from the stack. Otherwise, we push the character and its count onto the stack. At the end, we construct the final string by appending the characters left in the stack.
Algorithm:
- Initialize an empty stack to store pairs of character and count.
- Iterate through each character in the input string.
- For each character:
- If the stack is not empty and the current character is the same as the character at the top of the stack, increment the count.
- If the count becomes equal to k, remove the characters from the stack.
- Otherwise, push the character and its count onto the stack.
- Construct the final string by appending the characters left in the stack.
Time Complexity: O(n) where n is the length of the input string s. Space Complexity: O(n) where n is the length of the input string s.
Solution Code
class Solution {
public String removeDuplicates(String s, int k) {
Stack<Pair<Character, Integer>> stack = new Stack<>();
for (char ch : s.toCharArray()) {
if (!stack.isEmpty() && stack.peek().getKey() == ch) {
int count = stack.peek().getValue() + 1;
if (count == k) {
stack.pop();
} else {
stack.push(new Pair<>(ch, count));
}
} else {
stack.push(new Pair<>(ch, 1));
}
}
StringBuilder sb = new StringBuilder();
for (Pair<Character, Integer> pair : stack) {
for (int i = 0; i < pair.getValue(); i++) {
sb.append(pair.getKey());
}
}
return sb.toString();
}
}Try It Yourself
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