LeetCode #921 — MEDIUM

Minimum Add to Make Parentheses Valid

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

Solve on LeetCode

The Problem

Problem Statement

A parentheses string is valid if and only if:

It is the empty string,
It can be written as AB (A concatenated with B), where A and B are valid strings, or
It can be written as (A), where A is a valid string.

You are given a parentheses string s. In one move, you can insert a parenthesis at any position of the string.

For example, if s = "()))", you can insert an opening parenthesis to be "(()))" or a closing parenthesis to be "())))".

Return the minimum number of moves required to make s valid.

Example 1:

Input: s = "())"
Output: 1

Example 2:

Input: s = "((("
Output: 3

Constraints:

1 <= s.length <= 1000
s[i] is either '(' or ')'.

Step 01

Brute Force Baseline

Problem summary: A parentheses string is valid if and only if: It is the empty string, It can be written as AB (A concatenated with B), where A and B are valid strings, or It can be written as (A), where A is a valid string. You are given a parentheses string s. In one move, you can insert a parenthesis at any position of the string. For example, if s = "()))", you can insert an opening parenthesis to be "(()))" or a closing parenthesis to be "())))". Return the minimum number of moves required to make s valid.

Baseline thinking

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

Pattern signal: Stack · Greedy

Example 1

"())"

Example 2

"((("

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

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 #921: Minimum Add to Make Parentheses Valid
class Solution {
    public int minAddToMakeValid(String s) {
        Deque<Character> stk = new ArrayDeque<>();
        for (char c : s.toCharArray()) {
            if (c == ')' && !stk.isEmpty() && stk.peek() == '(') {
                stk.pop();
            } else {
                stk.push(c);
            }
        }
        return stk.size();
    }
}

// Accepted solution for LeetCode #921: Minimum Add to Make Parentheses Valid
func minAddToMakeValid(s string) int {
	stk := []rune{}
	for _, c := range s {
		if c == ')' && len(stk) > 0 && stk[len(stk)-1] == '(' {
			stk = stk[:len(stk)-1]
		} else {
			stk = append(stk, c)
		}
	}
	return len(stk)
}

# Accepted solution for LeetCode #921: Minimum Add to Make Parentheses Valid
class Solution:
    def minAddToMakeValid(self, s: str) -> int:
        stk = []
        for c in s:
            if c == ')' and stk and stk[-1] == '(':
                stk.pop()
            else:
                stk.append(c)
        return len(stk)

// Accepted solution for LeetCode #921: Minimum Add to Make Parentheses Valid
struct Solution;

impl Solution {
    fn min_add_to_make_valid(s: String) -> i32 {
        let mut stack: Vec<char> = vec![];
        let mut res = 0;
        for c in s.chars() {
            match c {
                '(' => {
                    stack.push(c);
                }
                ')' => {
                    if stack.is_empty() {
                        res += 1;
                    } else {
                        stack.pop();
                    }
                }
                _ => {}
            }
        }
        res + stack.len() as i32
    }
}

#[test]
fn test() {
    let s = "())".to_string();
    let res = 1;
    assert_eq!(Solution::min_add_to_make_valid(s), res);
    let s = "(((".to_string();
    let res = 3;
    assert_eq!(Solution::min_add_to_make_valid(s), res);
    let s = "()".to_string();
    let res = 0;
    assert_eq!(Solution::min_add_to_make_valid(s), res);
    let s = "()))((".to_string();
    let res = 4;
    assert_eq!(Solution::min_add_to_make_valid(s), res);
}

// Accepted solution for LeetCode #921: Minimum Add to Make Parentheses Valid
function minAddToMakeValid(s: string): number {
    const stk: string[] = [];
    for (const c of s) {
        if (c === ')' && stk.length > 0 && stk.at(-1)! === '(') {
            stk.pop();
        } else {
            stk.push(c);
        }
    }
    return stk.length;
}

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.