Move from brute-force thinking to an efficient approach using stack strategy.
You are given a string s and an integer k, 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 <= 1052 <= k <= 104s only contains lowercase English letters.Problem summary: You are given a string s and an integer k, 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack
"abcd" 2
"deeedbbcccbdaa" 3
"pbbcggttciiippooaais" 2
remove-all-adjacent-duplicates-in-string)replace-non-coprime-numbers-in-array)minimize-string-length)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #1209: Remove All Adjacent Duplicates in String II
class Solution {
public String removeDuplicates(String s, int k) {
Deque<int[]> stk = new ArrayDeque<>();
for (int i = 0; i < s.length(); ++i) {
int j = s.charAt(i) - 'a';
if (!stk.isEmpty() && stk.peek()[0] == j) {
stk.peek()[1] = (stk.peek()[1] + 1) % k;
if (stk.peek()[1] == 0) {
stk.pop();
}
} else {
stk.push(new int[] {j, 1});
}
}
StringBuilder ans = new StringBuilder();
for (var e : stk) {
char c = (char) (e[0] + 'a');
for (int i = 0; i < e[1]; ++i) {
ans.append(c);
}
}
ans.reverse();
return ans.toString();
}
}
// Accepted solution for LeetCode #1209: Remove All Adjacent Duplicates in String II
func removeDuplicates(s string, k int) string {
stk := []pair{}
for _, c := range s {
if len(stk) > 0 && stk[len(stk)-1].c == c {
stk[len(stk)-1].v = (stk[len(stk)-1].v + 1) % k
if stk[len(stk)-1].v == 0 {
stk = stk[:len(stk)-1]
}
} else {
stk = append(stk, pair{c, 1})
}
}
ans := []rune{}
for _, e := range stk {
for i := 0; i < e.v; i++ {
ans = append(ans, e.c)
}
}
return string(ans)
}
type pair struct {
c rune
v int
}
# Accepted solution for LeetCode #1209: Remove All Adjacent Duplicates in String II
class Solution:
def removeDuplicates(self, s: str, k: int) -> str:
t = []
i, n = 0, len(s)
while i < n:
j = i
while j < n and s[j] == s[i]:
j += 1
cnt = j - i
cnt %= k
if t and t[-1][0] == s[i]:
t[-1][1] = (t[-1][1] + cnt) % k
if t[-1][1] == 0:
t.pop()
elif cnt:
t.append([s[i], cnt])
i = j
ans = [c * v for c, v in t]
return "".join(ans)
// Accepted solution for LeetCode #1209: Remove All Adjacent Duplicates in String II
impl Solution {
pub fn remove_duplicates(s: String, k: i32) -> String {
let mut stack: Vec<(char, usize)> = vec![];
for ch in s.chars() {
if !stack.is_empty() && stack.last().unwrap().0 == ch {
let mut last = stack.pop().unwrap();
last.1 += 1;
stack.push(last);
} else {
stack.push((ch, 1));
}
if stack.last().unwrap().1 == k as usize {
stack.pop();
}
}
stack.iter().fold(String::new(), |acc, &(ch, count)| {
acc + &ch.to_string().repeat(count)
})
}
}
// Accepted solution for LeetCode #1209: Remove All Adjacent Duplicates in String II
function removeDuplicates(s: string, k: number): string {
const stack: { char: string; count: number }[] = []; // [char, count];
for (const c of s) {
if (stack.length !== 0 && stack[stack.length - 1].char === c) {
stack[stack.length - 1].count++;
} else {
stack.push({ char: c, count: 1 });
}
if (stack[stack.length - 1].count === k) {
stack.pop();
}
}
return stack.reduce(
(acc, { char, count }) => (acc += char.repeat(count)),
''
);
}
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.