Remove All Occurrences of a Substring
A medium-tier problem at 78% community acceptance, tagged with String, Stack, Simulation. Reported in interviews at X and 3 others.
You're given a string and a substring to remove. Every time the substring appears, you delete it. Then you check the result again because removing one substring might expose a new one. It sounds simple until you realize the naive approach is O(n^2) and fails on large inputs with overlapping removals. X, Zoho, Arista Networks, and Goldman Sachs have all asked this. The trick is knowing when to use a stack instead of string replacement. If you blank on that pattern during the live OA, StealthCoder solves it invisibly in seconds.
Companies that ask "Remove All Occurrences of a Substring"
Remove All Occurrences of a Substring 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. Made for the engineer who has done the work but might still blank with a webcam pointed at him.
Get StealthCoderThe problem tests whether you understand that string operations can cascade. If you just call a built-in replace function or loop with string slicing, you'll time out on adversarial inputs like 'aaaa' with substring 'aa', where each removal exposes a new match. A stack-based simulation avoids this. As you iterate through the string, you push characters onto a stack. Before pushing, check if the top of the stack plus the current character forms the substring. If it does, pop instead of pushing. Continue until the string is consumed. This is O(n) and handles all cascading removals in a single pass. The insight is treating the problem as a state machine where each character either extends the current prefix or triggers a cancellation. Most candidates miss the stack approach and waste time optimizing the wrong algorithm. When this shows up in your assessment, StealthCoder is your hedge for that moment.
Pattern tags
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Remove All Occurrences of a Substring 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. Made for the engineer who has done the work but might still blank with a webcam pointed at him. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Remove All Occurrences of a Substring interview FAQ
Is this really a medium problem?+
The acceptance rate is about 78 percent, which reflects a genuine medium difficulty. It's not a hard algorithmic problem, but the stack insight isn't obvious if you've only done basic string manipulation. Many candidates get partial credit with slow solutions before recognizing the pattern.
What's the biggest trap when solving this?+
Using built-in string functions like replace or slicing in a loop. Each operation creates a new string, and you'll loop multiple times on tricky inputs. The stack approach processes left-to-right in one pass and is dramatically faster. That's the core insight Goldman Sachs and the other companies are testing.
Do I need to handle overlapping substrings?+
Yes. That's what makes the stack solution necessary. For example, removing 'ab' from 'aabb' leaves 'ab' again. A stack catches this automatically because it simulates the result string incrementally. Each removal can immediately expose the next match.
How does this relate to the other string topics?+
String problems often require simulation or state tracking. This one combines string parsing with stack behavior. You're not doing regex or parsing a grammar, just recognizing when a prefix matches and reacting accordingly. It's simulation using a stack instead of brute force.
What if the substring appears millions of times?+
The stack approach is O(n) time and O(n) space where n is the input string length. You iterate once and push or pop each character once. No nested loops, no repeated string creation. On large inputs, this is the only solution that won't time out.
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