The O(n) Club: Minimum Remove to Make Valid Parentheses — Java’s Answer to Your Bracket Nightmares

The O(n) Club: Minimum Remove to Make Valid Parentheses — Java’s Answer to Your Bracket Nightmares

⚡ TL;DR

Dev nightmare: You have a string riddled with extra parentheses, and you want the smallest surgery possible to make it valid. The fastest cure? Walk it left-to-right and skip spare ), then go right-to-left and dump any lonely (. All letters survive, only parentheses get evicted. No NP-hard tears. Here’s what not to do:

// This will make your CPU weep
public String minRemove(String s) {
  // Brute-force: Try all removals. But unless you’re writing a recursive horror movie, don’t.
  return "Nope."; 
} 

🧠 How Devs Usually Mess This Up

It’s classic: Devs hack off extra ) on the first pass and forget the poor ( orphans hiding at the end. Or they pile on stack logic like they’re being charged by the paren and wind up juggling indices for sport. News flash: You don’t need to reconstruct some mythical perfect string—any minimal fix is valid. And stacks? That’s overkill for this one, pal.

🔍 What’s Actually Going On

Pretend each ( is trying to swipe into a club, and every ) is the bouncer. Move left-to-right: only openers with chaperones get in. If a closer shows up early, the bouncer (your code) just ignores them. Then you boomerang right-to-left: if you spot an opener whose match already left, just keep walking. Letters? They’re the VIPs—never bounced.

🛠️ PseudoCode

• Create a StringBuilder for round one.
• Go left-to-right:
  • If you hit (, increment a balance counter and append it.
  • If you get ) and balance > 0, decrement balance and append. If not, skip it.
  • Non-parentheses? Keep ’em—they never hurt anyone.
• Round two: right-to-left.
  • Track “closings” remaining.
  • If ), increment close and append.
  • If ( and close > 0, decrement close and append; otherwise, skip.
  • Others? Always in the club.
• Reverse the final StringBuilder and you’re out the door.

// First pass: kick out wild ')'
StringBuilder sb = new StringBuilder();
int open = 0;
for (char c : s.toCharArray()) {
  if (c == '(') { open++; sb.append(c); }
  else if (c == ')') {
    if (open > 0) { open--; sb.append(c); } // admit if opener exists
    // else, skip
  } else { sb.append(c); }
}
 // Second pass: kick out lonely '('
StringBuilder res = new StringBuilder();
int close = 0;
for (int i = sb.length() - 1; i >= 0; i--) {
  char c = sb.charAt(i);
  if (c == ')') { close++; res.append(c); }
  else if (c == '(') {
    if (close > 0) { close--; res.append(c); }
    // else, skip
  } else { res.append(c); }
}
return res.reverse().toString();

💻 The Code

public String minRemoveToMakeValid(String s) {
    StringBuilder sb = new StringBuilder();
    int open = 0;
    for (char ch : s.toCharArray()) {
        if (ch == '(') {
            open++;
            sb.append(ch);
        } else if (ch == ')') {
            if (open > 0) {
                open--;
                sb.append(ch);
            }
        } else {
            sb.append(ch);
        }
    }
     StringBuilder result = new StringBuilder();
    int close = 0;
    for (int i = sb.length() - 1; i >= 0; i--) {
        char ch = sb.charAt(i);
        if (ch == ')') {
            close++;
            result.append(ch);
        } else if (ch == '(') {
            if (close > 0) {
                close--;
                result.append(ch);
            }
        } else {
            result.append(ch);
        }
    }
    return result.reverse().toString();
}

⚠️ Pitfalls, Traps & Runtime Smacks

• If you only remove extra ) up front, you’ll leave dangling openers. Do both passes or enjoy failing test cases.
• Stack approaches teach balancing, but for this you can do even less.
• Leave non-parenthesis chars alone—or the prompt police will find you.
• O(n) time and space—don’t bring this to a zero-copy fight; you’re making buffers both ways.
• Don’t stress about one canonical result—any minimal fix is fair game.

🧠 Interviewer Brain-Tattoo

“String a mess? Two sweeps, no stack, O(n) time—minimal therapy for even maximal bracket chaos. You can quote me, but try not to use this in a regex interview.”