LeetCode #1541 — MEDIUM

Minimum Insertions to Balance a Parentheses String

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

Solve on LeetCode

The Problem

Problem Statement

Given a parentheses string s containing only the characters '(' and ')'. A parentheses string is balanced if:

Any left parenthesis '(' must have a corresponding two consecutive right parenthesis '))'.
Left parenthesis '(' must go before the corresponding two consecutive right parenthesis '))'.

In other words, we treat '(' as an opening parenthesis and '))' as a closing parenthesis.

For example, "())", "())(())))" and "(())())))" are balanced, ")()", "()))" and "(()))" are not balanced.

You can insert the characters '(' and ')' at any position of the string to balance it if needed.

Return the minimum number of insertions needed to make s balanced.

Example 1:

Input: s = "(()))"
Output: 1
Explanation: The second '(' has two matching '))', but the first '(' has only ')' matching. We need to add one more ')' at the end of the string to be "(())))" which is balanced.

Example 2:

Input: s = "())"
Output: 0
Explanation: The string is already balanced.

Example 3:

Input: s = "))())("
Output: 3
Explanation: Add '(' to match the first '))', Add '))' to match the last '('.

Constraints:

1 <= s.length <= 10⁵
s consists of '(' and ')' only.

Step 01

Brute Force Baseline

Problem summary: Given a parentheses string s containing only the characters '(' and ')'. A parentheses string is balanced if: Any left parenthesis '(' must have a corresponding two consecutive right parenthesis '))'. Left parenthesis '(' must go before the corresponding two consecutive right parenthesis '))'. In other words, we treat '(' as an opening parenthesis and '))' as a closing parenthesis. For example, "())", "())(())))" and "(())())))" are balanced, ")()", "()))" and "(()))" are not balanced. You can insert the characters '(' and ')' at any position of the string to balance it if needed. Return the minimum number of insertions needed to make s balanced.

Baseline thinking

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

Pattern signal: Stack · Greedy

Example 1

"(()))"

Example 2

"())"

Example 3

"))())("

Core Insight

What unlocks the optimal approach

Use a stack to keep opening brackets. If you face single closing ')' add 1 to the answer and consider it as '))'.
If you have '))' with empty stack, add 1 to the answer, If after finishing you have x opening remaining in the stack, add 2x to the answer.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
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 #1541: Minimum Insertions to Balance a Parentheses String
class Solution {
    public int minInsertions(String s) {
        int ans = 0, x = 0;
        int n = s.length();
        for (int i = 0; i < n; ++i) {
            if (s.charAt(i) == '(') {
                ++x;
            } else {
                if (i < n - 1 && s.charAt(i + 1) == ')') {
                    ++i;
                } else {
                    ++ans;
                }
                if (x == 0) {
                    ++ans;
                } else {
                    --x;
                }
            }
        }
        ans += x << 1;
        return ans;
    }
}

// Accepted solution for LeetCode #1541: Minimum Insertions to Balance a Parentheses String
func minInsertions(s string) int {
	ans, x, n := 0, 0, len(s)
	for i := 0; i < n; i++ {
		if s[i] == '(' {
			x++
		} else {
			if i < n-1 && s[i+1] == ')' {
				i++
			} else {
				ans++
			}
			if x == 0 {
				ans++
			} else {
				x--
			}
		}
	}
	ans += x << 1
	return ans
}

# Accepted solution for LeetCode #1541: Minimum Insertions to Balance a Parentheses String
class Solution:
    def minInsertions(self, s: str) -> int:
        ans = x = 0
        i, n = 0, len(s)
        while i < n:
            if s[i] == '(':
                # 待匹配的左括号加 1
                x += 1
            else:
                if i < n - 1 and s[i + 1] == ')':
                    # 有连续两个右括号，i 往后移动
                    i += 1
                else:
                    # 只有一个右括号，插入一个
                    ans += 1
                if x == 0:
                    # 无待匹配的左括号，插入一个
                    ans += 1
                else:
                    # 待匹配的左括号减 1
                    x -= 1
            i += 1
        # 遍历结束，仍有待匹配的左括号，说明右括号不足，插入 x << 1 个
        ans += x << 1
        return ans

// Accepted solution for LeetCode #1541: Minimum Insertions to Balance a Parentheses String
struct Solution;

impl Solution {
    fn min_insertions(s: String) -> i32 {
        let mut right = 0;
        let mut res = 0;
        for c in s.chars().rev() {
            if c == ')' {
                right += 1;
            } else {
                if right >= 2 {
                    right -= 2;
                    if right % 2 == 1 {
                        right += 1;
                        res += 1;
                    }
                } else {
                    if right == 0 {
                        res += 2;
                    } else {
                        res += 1;
                        right -= 1;
                    }
                }
            }
        }
        res += right / 2 + (right % 2) * 2;
        res
    }
}

#[test]
fn test() {
    let s = "())".to_string();
    let res = 0;
    assert_eq!(Solution::min_insertions(s), res);
    let s = "))())(".to_string();
    let res = 3;
    assert_eq!(Solution::min_insertions(s), res);
    let s = "((((((".to_string();
    let res = 12;
    assert_eq!(Solution::min_insertions(s), res);
    let s = ")))))))".to_string();
    let res = 5;
    assert_eq!(Solution::min_insertions(s), res);
    let s = "(()))(()))()())))".to_string();
    let res = 4;
    assert_eq!(Solution::min_insertions(s), res);
}

// Accepted solution for LeetCode #1541: Minimum Insertions to Balance a Parentheses String
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1541: Minimum Insertions to Balance a Parentheses String
// class Solution {
//     public int minInsertions(String s) {
//         int ans = 0, x = 0;
//         int n = s.length();
//         for (int i = 0; i < n; ++i) {
//             if (s.charAt(i) == '(') {
//                 ++x;
//             } else {
//                 if (i < n - 1 && s.charAt(i + 1) == ')') {
//                     ++i;
//                 } else {
//                     ++ans;
//                 }
//                 if (x == 0) {
//                     ++ans;
//                 } else {
//                     --x;
//                 }
//             }
//         }
//         ans += x << 1;
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

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

Custom checkpoints (one per line)

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.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.