Minimum Insertions to Balance a Parentheses String
A medium-tier problem at 53% community acceptance, tagged with String, Stack, Greedy. Reported in interviews at J.P. Morgan and 0 others.
You're given a string of parentheses where the rules are stricter than the classic problem: every closing paren must be immediately preceded by two opening parens, and the string must end balanced. It's not a common pattern, but J.P. Morgan has asked it. The trick isn't obvious on first read, and if you freeze on the OA, you need a working solution fast. StealthCoder solves this invisibly during your screen share, surfacing the greedy count in seconds so you can move on.
Companies that ask "Minimum Insertions to Balance a Parentheses String"
Minimum Insertions to Balance a Parentheses String is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built by a senior engineer who knows the OA is theater. This is the script.
Get StealthCoderThe core insight is greedy counting, not a stack. Track open parens and the required closing pairs. When you see an opening paren, increment open count. When you see a closing paren, you need two of them in a row to match a single open. If the current close doesn't pair with the previous one, or if you hit EOF before closing all opens, you insert the missing parens. Many candidates build a full stack thinking this mirrors the classic balanced paren problem. It doesn't. You only need to count open parens and detect unpaired closing ones. One pass, O(n) time. If you haven't drilled the greedy pattern before, StealthCoder is your hedge on the live OA.
Pattern tags
You know the problem.
Make sure you actually pass it.
Minimum Insertions to Balance a Parentheses String recycles across companies for a reason. It's medium-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built by a senior engineer who knows the OA is theater. This is the script. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Minimum Insertions to Balance a Parentheses String interview FAQ
Why can't I just use a stack like classic balanced parens?+
The constraint is unique: two closing parens must come together for every opening paren. A simple stack doesn't capture that. You need to track whether closing parens are consecutive. Greedy counting of opens and a single pass detects unpaired closes efficiently. The trick is pattern-matching, not nested validation.
Is this still asked at J.P. Morgan and similar firms?+
It appears in their OA reports. It's not a top-frequency problem, but it does show up. Banks like J.P. Morgan value string manipulation and greedy logic. Expect it as a medium-difficulty filter question in a technical round.
How do I know when to insert parens versus when the string is already valid?+
Scan left to right. Count opening parens. For every closing paren, check if it pairs with the previous one (double close). If a close is unpaired, insert an opening before it. At EOF, if opens remain, append the required closing pairs. Greedy means insert only when forced.
What's the most common mistake on this problem?+
Overthinking it with a stack or trying to track a complex state machine. Candidates also forget that closing parens must come in pairs, so they miscalculate the insertion count. The problem is simpler than it looks once you see the greedy pattern.
How does this relate to other String and Greedy problems?+
It combines single-pass greedy counting with string validation, similar to problems about minimum additions or removals. The Stack topic is listed but the solution is greedy, not stack-based. Understanding when to count versus when to use data structures is key across this category.
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