Move from brute-force thinking to an efficient approach using stack strategy.
You are given a string s consisting only of the characters 'a' and 'b'.
You are allowed to repeatedly remove any substring where the number of 'a' characters is equal to the number of 'b' characters. After each removal, the remaining parts of the string are concatenated together without gaps.
Return an integer denoting the minimum possible length of the string after performing any number of such operations.
Example 1:
Input: s = "aabbab"
Output: 0
Explanation:
The substring "aabbab" has three 'a' and three 'b'. Since their counts are equal, we can remove the entire string directly. The minimum length is 0.
Example 2:
Input: s = "aaaa"
Output: 4
Explanation:
Every substring of "aaaa" contains only 'a' characters. No substring can be removed as a result, so the minimum length remains 4.
Example 3:
Input: s = "aaabb"
Output: 1
Explanation:
First, remove the substring "ab", leaving "aab". Next, remove the new substring "ab", leaving "a". No further removals are possible, so the minimum length is 1.
Constraints:
1 <= s.length <= 105s[i] is either 'a' or 'b'.Problem summary: You are given a string s consisting only of the characters 'a' and 'b'. You are allowed to repeatedly remove any substring where the number of 'a' characters is equal to the number of 'b' characters. After each removal, the remaining parts of the string are concatenated together without gaps. Return an integer denoting the minimum possible length of the string after performing any number of such operations.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Stack
"aabbab"
"aaaa"
"aaabb"
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3746: Minimum String Length After Balanced Removals
class Solution {
public int minLengthAfterRemovals(String s) {
int n = s.length();
int a = 0;
for (int i = 0; i < n; ++i) {
if (s.charAt(i) == 'a') {
++a;
}
}
int b = n - a;
return Math.abs(a - b);
}
}
// Accepted solution for LeetCode #3746: Minimum String Length After Balanced Removals
func minLengthAfterRemovals(s string) int {
a := strings.Count(s, "a")
b := len(s) - a
return abs(a - b)
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
# Accepted solution for LeetCode #3746: Minimum String Length After Balanced Removals
class Solution:
def minLengthAfterRemovals(self, s: str) -> int:
a = s.count("a")
b = len(s) - a
return abs(a - b)
// Accepted solution for LeetCode #3746: Minimum String Length After Balanced Removals
fn min_length_after_removals(s: String) -> i32 {
let (a, b) = s.chars().fold((0i32, 0i32), |(a, b), c| {
if c == 'a' {
(a + 1, b)
} else {
(a, b + 1)
}
});
(a - b).abs()
}
fn main() {
let ret = min_length_after_removals("aabbab".to_string());
println!("ret={ret}");
}
#[test]
fn test() {
assert_eq!(min_length_after_removals("aabbab".to_string()), 0);
assert_eq!(min_length_after_removals("aaaa".to_string()), 4);
assert_eq!(min_length_after_removals("aaabb".to_string()), 1);
}
// Accepted solution for LeetCode #3746: Minimum String Length After Balanced Removals
function minLengthAfterRemovals(s: string): number {
let a = 0;
for (const c of s) {
if (c === 'a') {
++a;
}
}
const b = s.length - a;
return Math.abs(a - b);
}
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.