LeetCode #301 — HARD

Remove Invalid Parentheses

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

Given a string s that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid.

Return a list of unique strings that are valid with the minimum number of removals. You may return the answer in any order.

Example 1:

Input: s = "()())()"
Output: ["(())()","()()()"]

Example 2:

Input: s = "(a)())()"
Output: ["(a())()","(a)()()"]

Example 3:

Input: s = ")("
Output: [""]

Constraints:

1 <= s.length <= 25
s consists of lowercase English letters and parentheses '(' and ')'.
There will be at most 20 parentheses in s.

Step 01

Brute Force Baseline

Problem summary: Given a string s that contains parentheses and letters, remove the minimum number of invalid parentheses to make the input string valid. Return a list of unique strings that are valid with the minimum number of removals. You may return the answer in any order.

Baseline thinking

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

Pattern signal: Backtracking

Example 1

"()())()"

Example 2

"(a)())()"

Example 3

")("

Core Insight

What unlocks the optimal approach

Since we do not know which brackets can be removed, we try all the options! We can use recursion.
In the recursion, for each bracket, we can either use it or remove it.
Recursion will generate all the valid parentheses strings but we want the ones with the least number of parentheses deleted.
We can count the number of invalid brackets to be deleted and only generate the valid strings in the recusrion.

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

Largest constraint values

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 #301: Remove Invalid Parentheses
class Solution {
    private String s;
    private int n;
    private Set<String> ans = new HashSet<>();

    public List<String> removeInvalidParentheses(String s) {
        this.s = s;
        this.n = s.length();
        int l = 0, r = 0;
        for (char c : s.toCharArray()) {
            if (c == '(') {
                ++l;
            } else if (c == ')') {
                if (l > 0) {
                    --l;
                } else {
                    ++r;
                }
            }
        }
        dfs(0, l, r, 0, 0, "");
        return new ArrayList<>(ans);
    }

    private void dfs(int i, int l, int r, int lcnt, int rcnt, String t) {
        if (i == n) {
            if (l == 0 && r == 0) {
                ans.add(t);
            }
            return;
        }
        if (n - i < l + r || lcnt < rcnt) {
            return;
        }
        char c = s.charAt(i);
        if (c == '(' && l > 0) {
            dfs(i + 1, l - 1, r, lcnt, rcnt, t);
        }
        if (c == ')' && r > 0) {
            dfs(i + 1, l, r - 1, lcnt, rcnt, t);
        }
        int x = c == '(' ? 1 : 0;
        int y = c == ')' ? 1 : 0;
        dfs(i + 1, l, r, lcnt + x, rcnt + y, t + c);
    }
}

// Accepted solution for LeetCode #301: Remove Invalid Parentheses
func removeInvalidParentheses(s string) []string {
	vis := map[string]bool{}
	l, r, n := 0, 0, len(s)
	for _, c := range s {
		if c == '(' {
			l++
		} else if c == ')' {
			if l > 0 {
				l--
			} else {
				r++
			}
		}
	}
	var dfs func(i, l, r, lcnt, rcnt int, t string)
	dfs = func(i, l, r, lcnt, rcnt int, t string) {
		if i == n {
			if l == 0 && r == 0 {
				vis[t] = true
			}
			return
		}
		if n-i < l+r || lcnt < rcnt {
			return
		}
		if s[i] == '(' && l > 0 {
			dfs(i+1, l-1, r, lcnt, rcnt, t)
		}
		if s[i] == ')' && r > 0 {
			dfs(i+1, l, r-1, lcnt, rcnt, t)
		}
		x, y := 0, 0
		if s[i] == '(' {
			x = 1
		} else if s[i] == ')' {
			y = 1
		}
		dfs(i+1, l, r, lcnt+x, rcnt+y, t+string(s[i]))
	}
	dfs(0, l, r, 0, 0, "")
	ans := make([]string, 0, len(vis))
	for v := range vis {
		ans = append(ans, v)
	}
	return ans
}

# Accepted solution for LeetCode #301: Remove Invalid Parentheses
class Solution:
    def removeInvalidParentheses(self, s: str) -> List[str]:
        def dfs(i, l, r, lcnt, rcnt, t):
            if i == n:
                if l == 0 and r == 0:
                    ans.add(t)
                return
            if n - i < l + r or lcnt < rcnt:
                return
            if s[i] == '(' and l:
                dfs(i + 1, l - 1, r, lcnt, rcnt, t)
            elif s[i] == ')' and r:
                dfs(i + 1, l, r - 1, lcnt, rcnt, t)
            dfs(i + 1, l, r, lcnt + (s[i] == '('), rcnt + (s[i] == ')'), t + s[i])

        l = r = 0
        for c in s:
            if c == '(':
                l += 1
            elif c == ')':
                if l:
                    l -= 1
                else:
                    r += 1
        ans = set()
        n = len(s)
        dfs(0, l, r, 0, 0, '')
        return list(ans)

// Accepted solution for LeetCode #301: Remove Invalid Parentheses
struct Solution;

use std::collections::HashSet;

impl Solution {
    fn remove_invalid_parentheses(s: String) -> Vec<String> {
        let mut cur = vec![];
        let s: Vec<char> = s.chars().collect();
        let n = s.len();
        let mut min = std::usize::MAX;
        let mut res: HashSet<String> = HashSet::new();
        Self::dfs(0, 0, 0, &mut cur, &mut res, &mut min, &s, n);
        res.into_iter().collect()
    }

    fn dfs(
        start: usize,
        left: usize,
        remove: usize,
        cur: &mut Vec<char>,
        all: &mut HashSet<String>,
        min: &mut usize,
        s: &[char],
        n: usize,
    ) {
        if start == n {
            if left != 0 {
                return;
            }
            if remove > *min {
                return;
            }
            if remove < *min {
                *min = remove;
                all.clear();
            }
            let s = cur.iter().copied().collect();
            all.insert(s);
        } else {
            match s[start] {
                '(' => {
                    cur.push('(');
                    Self::dfs(start + 1, left + 1, remove, cur, all, min, s, n);
                    cur.pop();
                    Self::dfs(start + 1, left, remove + 1, cur, all, min, s, n);
                }
                ')' => {
                    if left > 0 {
                        cur.push(')');
                        Self::dfs(start + 1, left - 1, remove, cur, all, min, s, n);
                        cur.pop();
                    }
                    Self::dfs(start + 1, left, remove + 1, cur, all, min, s, n);
                }
                _ => {
                    cur.push(s[start]);
                    Self::dfs(start + 1, left, remove, cur, all, min, s, n);
                    cur.pop();
                }
            }
        }
    }
}

#[test]
fn test() {
    let s = "()())()".to_string();
    let mut res = vec_string!["()()()", "(())()"];
    let mut ans = Solution::remove_invalid_parentheses(s);
    ans.sort();
    res.sort();
    assert_eq!(ans, res);
    let s = "(a)())()".to_string();
    let mut res = vec_string!["(a)()()", "(a())()"];
    let mut ans = Solution::remove_invalid_parentheses(s);
    ans.sort();
    res.sort();
    assert_eq!(ans, res);
    let s = ")(".to_string();
    let mut res = vec_string![""];
    let mut ans = Solution::remove_invalid_parentheses(s);
    ans.sort();
    res.sort();
    assert_eq!(ans, res);
    let s = ")(f".to_string();
    let mut res = vec_string!["f"];
    let mut ans = Solution::remove_invalid_parentheses(s);
    ans.sort();
    res.sort();
    assert_eq!(ans, res);
}

// Accepted solution for LeetCode #301: Remove Invalid Parentheses
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #301: Remove Invalid Parentheses
// class Solution {
//     private String s;
//     private int n;
//     private Set<String> ans = new HashSet<>();
// 
//     public List<String> removeInvalidParentheses(String s) {
//         this.s = s;
//         this.n = s.length();
//         int l = 0, r = 0;
//         for (char c : s.toCharArray()) {
//             if (c == '(') {
//                 ++l;
//             } else if (c == ')') {
//                 if (l > 0) {
//                     --l;
//                 } else {
//                     ++r;
//                 }
//             }
//         }
//         dfs(0, l, r, 0, 0, "");
//         return new ArrayList<>(ans);
//     }
// 
//     private void dfs(int i, int l, int r, int lcnt, int rcnt, String t) {
//         if (i == n) {
//             if (l == 0 && r == 0) {
//                 ans.add(t);
//             }
//             return;
//         }
//         if (n - i < l + r || lcnt < rcnt) {
//             return;
//         }
//         char c = s.charAt(i);
//         if (c == '(' && l > 0) {
//             dfs(i + 1, l - 1, r, lcnt, rcnt, t);
//         }
//         if (c == ')' && r > 0) {
//             dfs(i + 1, l, r - 1, lcnt, rcnt, t);
//         }
//         int x = c == '(' ? 1 : 0;
//         int y = c == ')' ? 1 : 0;
//         dfs(i + 1, l, r, lcnt + x, rcnt + y, t + c);
//     }
// }

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

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.