LeetCode #3703 — MEDIUM

Remove K-Balanced Substrings

Move from brute-force thinking to an efficient approach using stack strategy.

Solve on LeetCode
The Problem

Problem Statement

You are given a string s consisting of '(' and ')', and an integer k.

A string is k-balanced if it is exactly k consecutive '(' followed by k consecutive ')', i.e., '(' * k + ')' * k.

For example, if k = 3, k-balanced is "((()))".

You must repeatedly remove all non-overlapping k-balanced substrings from s, and then join the remaining parts. Continue this process until no k-balanced substring exists.

Return the final string after all possible removals.

​​​​​​​Example 1:

Input: s = "(())", k = 1

Output: ""

Explanation:

k-balanced substring is "()"

Step Current s k-balanced Result s
1 (()) (()) ()
2 () () Empty

Thus, the final string is "".

Example 2:

Input: s = "(()(", k = 1

Output: "(("

Explanation:

k-balanced substring is "()"

Step Current s k-balanced Result s
1 (()( (()( ((
2 (( - ((

Thus, the final string is "((".

Example 3:

Input: s = "((()))()()()", k = 3

Output: "()()()"

Explanation:

k-balanced substring is "((()))"

Step Current s k-balanced Result s
1 ((()))()()() ((()))()()() ()()()
2 ()()() - ()()()

Thus, the final string is "()()()".

Constraints:

  • 2 <= s.length <= 105
  • s consists only of '(' and ')'.
  • 1 <= k <= s.length / 2
Patterns Used
📚Monotonic Stack→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: You are given a string s consisting of '(' and ')', and an integer k. A string is k-balanced if it is exactly k consecutive '(' followed by k consecutive ')', i.e., '(' * k + ')' * k. For example, if k = 3, k-balanced is "((()))". You must repeatedly remove all non-overlapping k-balanced substrings from s, and then join the remaining parts. Continue this process until no k-balanced substring exists. Return the final string after all possible removals. ​​​​​​​

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Stack

Example 1

"(())"
1

Example 2

"(()("
1

Example 3

"((()))()()()"
3
Step 02

Core Insight

What unlocks the optimal approach

  • Use a stack
  • Try run-length encoding; operations only happen at boundaries of '(' and ')' runs
  • When adjacent runs are '(' then ')', you can cancel in blocks of k
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3703: Remove K-Balanced Substrings
class Solution {
    public String removeSubstring(String s, int k) {
        List<int[]> stk = new ArrayList<>();
        for (char c : s.toCharArray()) {
            if (!stk.isEmpty() && stk.get(stk.size() - 1)[0] == c) {
                stk.get(stk.size() - 1)[1] += 1;
            } else {
                stk.add(new int[] {c, 1});
            }
            if (c == ')' && stk.size() > 1) {
                int[] top = stk.get(stk.size() - 1);
                int[] prev = stk.get(stk.size() - 2);
                if (top[1] == k && prev[1] >= k) {
                    stk.remove(stk.size() - 1);
                    prev[1] -= k;
                    if (prev[1] == 0) {
                        stk.remove(stk.size() - 1);
                    }
                }
            }
        }
        StringBuilder sb = new StringBuilder();
        for (int[] pair : stk) {
            for (int i = 0; i < pair[1]; i++) {
                sb.append((char) pair[0]);
            }
        }
        return sb.toString();
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(n)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

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.

MONOTONIC STACK
O(n) time
O(n) space

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.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Breaking monotonic invariant

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.

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