Build confidence with an intuition-first walkthrough focused on stack fundamentals.
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 <= 105s consists of lowercase English letters.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack
"abbaca"
"azxxzy"
remove-all-adjacent-duplicates-in-string-ii)removing-stars-from-a-string)minimize-string-length)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1047: Remove All Adjacent Duplicates In String
class Solution {
public String removeDuplicates(String s) {
StringBuilder sb = new StringBuilder();
for (char c : s.toCharArray()) {
if (sb.length() > 0 && sb.charAt(sb.length() - 1) == c) {
sb.deleteCharAt(sb.length() - 1);
} else {
sb.append(c);
}
}
return sb.toString();
}
}
// Accepted solution for LeetCode #1047: Remove All Adjacent Duplicates In String
func removeDuplicates(s string) string {
stk := []rune{}
for _, c := range s {
if len(stk) > 0 && stk[len(stk)-1] == c {
stk = stk[:len(stk)-1]
} else {
stk = append(stk, c)
}
}
return string(stk)
}
# Accepted solution for LeetCode #1047: Remove All Adjacent Duplicates In String
class Solution:
def removeDuplicates(self, s: str) -> str:
stk = []
for c in s:
if stk and stk[-1] == c:
stk.pop()
else:
stk.append(c)
return ''.join(stk)
// Accepted solution for LeetCode #1047: Remove All Adjacent Duplicates In String
impl Solution {
pub fn remove_duplicates(s: String) -> String {
let mut stack = Vec::new();
for c in s.chars() {
if !stack.is_empty() && *stack.last().unwrap() == c {
stack.pop();
} else {
stack.push(c);
}
}
stack.into_iter().collect()
}
}
// Accepted solution for LeetCode #1047: Remove All Adjacent Duplicates In String
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1047: Remove All Adjacent Duplicates In String
// class Solution {
// public String removeDuplicates(String s) {
// StringBuilder sb = new StringBuilder();
// for (char c : s.toCharArray()) {
// if (sb.length() > 0 && sb.charAt(sb.length() - 1) == c) {
// sb.deleteCharAt(sb.length() - 1);
// } else {
// sb.append(c);
// }
// }
// return sb.toString();
// }
// }
Use this to step through a reusable interview workflow for this problem.
For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.
Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.
Review these before coding to avoid predictable interview regressions.
Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.
Usually fails on: Indices point to blocked elements and outputs shift.
Fix: Pop while invariant is violated before pushing current element.