1047, Remove All Adjacent Duplicates In String

I Problem

You are given a string s consisting of lowercase English letters. A duplicate removal consists of choosing two adjacent and equal letters and removing them.

We repeatedly make duplicate removals on s until we no longer can.

Return the final string after all such duplicate removals have been made. It can be proven that the answer is unique.

Example 1
Input: s = "abbaca"
Output: "ca"
Explanation: For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move. The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca".

Example 2
Input: s = "azxxzy"
Output: "ay"

Constraints

• 1 <= s.length <= 10⁵
• s consists of lowercase English letters.

Related Topics

• String
• Stack

II Solution

Approach 1: Brute Force

/// Time Complexity: O(n^2)
///
/// Space Complexity: O(1)
pub fn remove_duplicates(s: String) -> String {
    let bytes = unsafe { s.as_mut_vec() };
    let mut curr = 0;
    let mut next = 1;

    while next < bytes.len() {
        if bytes[curr] == bytes[next] {
            loop {
                if curr == 0 || next == bytes.len() - 1 {
                    break;
                }
                if bytes[curr - 1] == bytes[next + 1] {
                    curr -= 1;
                    next += 1;
                } else {
                    break;
                }
            }
            bytes.drain(curr..=next);
            curr = 0;
            next = 1;
        } else {
            curr = next;
            next += 1;
        }
    }

    s
}

// Time Complexity: O(n^2)
//
// Space Complexity: O(n)
public String removeDuplicates(String s) {
    StringBuilder chars = new StringBuilder(s);
    int curr = 0;
    int next = 1;

    while (next < chars.length()) {
        if (chars.charAt(curr) == chars.charAt(next)) {
            while (true) {
                if (curr == 0 || next == chars.length() - 1) {
                    break;
                }
                if (chars.charAt(curr - 1) == chars.charAt(next + 1)) {
                    curr--;
                    next++;
                } else {
                    break;
                }
            }
            chars.delete(curr, next + 1);
            curr = 0;
            next = 1;
        } else {
            curr = next;
            next++;
        }
    }

    return chars.toString();
}

Approach 2: Use Stack

/// Time Complexity: O(n)
///
/// Space Complexity: O(n)
pub fn remove_duplicates(s: String) -> String {
    let mut stack = String::with_capacity(s.len() / 2);

    for ch in s.chars() {
        match stack.pop() {
            None => {
                stack.push(ch);
            }
            Some(top) => {
                if top != ch {
                    stack.push(top);
                    stack.push(ch);
                }
            }
        }
    }

    stack
}

// Time Complexity: O(n)
//
// Space Complexity: O(n)
public String removeDuplicates(String s) {
    StringBuilder stack = new StringBuilder(s.length() / 2);
    int top = -1;

    for (int i = 0, len = s.length(); i < len; i++) {
        char ch = s.charAt(i);
        if (stack.isEmpty() || stack.charAt(top) != ch) {
            stack.append(ch);
            top++;
        } else {
            stack.deleteCharAt(top);
            top--;
        }
    }

    return stack.toString();
}