Build confidence with an intuition-first walkthrough focused on stack fundamentals.
You are given a string s consisting only of uppercase English letters.
You can apply some operations to this string where, in one operation, you can remove any occurrence of one of the substrings "AB" or "CD" from s.
Return the minimum possible length of the resulting string that you can obtain.
Note that the string concatenates after removing the substring and could produce new "AB" or "CD" substrings.
Example 1:
Input: s = "ABFCACDB" Output: 2 Explanation: We can do the following operations: - Remove the substring "ABFCACDB", so s = "FCACDB". - Remove the substring "FCACDB", so s = "FCAB". - Remove the substring "FCAB", so s = "FC". So the resulting length of the string is 2. It can be shown that it is the minimum length that we can obtain.
Example 2:
Input: s = "ACBBD" Output: 5 Explanation: We cannot do any operations on the string so the length remains the same.
Constraints:
1 <= s.length <= 100s consists only of uppercase English letters.Problem summary: You are given a string s consisting only of uppercase English letters. You can apply some operations to this string where, in one operation, you can remove any occurrence of one of the substrings "AB" or "CD" from s. Return the minimum possible length of the resulting string that you can obtain. Note that the string concatenates after removing the substring and could produce new "AB" or "CD" substrings.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack
"ABFCACDB"
"ACBBD"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2696: Minimum String Length After Removing Substrings
class Solution {
public int minLength(String s) {
Deque<Character> stk = new ArrayDeque<>();
stk.push(' ');
for (char c : s.toCharArray()) {
if ((c == 'B' && stk.peek() == 'A') || (c == 'D' && stk.peek() == 'C')) {
stk.pop();
} else {
stk.push(c);
}
}
return stk.size() - 1;
}
}
// Accepted solution for LeetCode #2696: Minimum String Length After Removing Substrings
func minLength(s string) int {
stk := []byte{' '}
for _, c := range s {
if (c == 'B' && stk[len(stk)-1] == 'A') || (c == 'D' && stk[len(stk)-1] == 'C') {
stk = stk[:len(stk)-1]
} else {
stk = append(stk, byte(c))
}
}
return len(stk) - 1
}
# Accepted solution for LeetCode #2696: Minimum String Length After Removing Substrings
class Solution:
def minLength(self, s: str) -> int:
stk = [""]
for c in s:
if (c == "B" and stk[-1] == "A") or (c == "D" and stk[-1] == "C"):
stk.pop()
else:
stk.append(c)
return len(stk) - 1
// Accepted solution for LeetCode #2696: Minimum String Length After Removing Substrings
impl Solution {
pub fn min_length(s: String) -> i32 {
let mut ans: Vec<u8> = Vec::new();
for c in s.bytes() {
if let Some(last) = ans.last() {
if *last == b'A' && c == b'B' {
ans.pop();
} else if *last == b'C' && c == b'D' {
ans.pop();
} else {
ans.push(c);
}
} else {
ans.push(c);
}
}
ans.len() as i32
}
}
// Accepted solution for LeetCode #2696: Minimum String Length After Removing Substrings
function minLength(s: string): number {
const stk: string[] = [];
for (const c of s) {
if ((stk.at(-1) === 'A' && c === 'B') || (stk.at(-1) === 'C' && c === 'D')) {
stk.pop();
} else {
stk.push(c);
}
}
return stk.length;
}
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.