Interleave Two Strings
Reported by candidates from ZipRecruiter's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The mistake that sinks a first attempt on this ZipRecruiter OA, reported in April 2022, is stopping the loop when the shorter string runs out and forgetting the leftover tail. The task is simple: alternate characters starting with first, then append whatever's left. It's a two-pointer string build, and the edge cases are where people lose points. Empty strings, unequal lengths, 100000 characters per side. If you blank on the details live, StealthCoder runs invisibly during the assessment and gives you a working solution as a safety net. But you won't need it for this one.
The problem
You are given two strings, first and second. Build a result by taking one character from first, then one from second, and repeating. The first result character comes from first. When one string is exhausted, append the remaining suffix of the other string. Function interleaveStrings(first: String, second: String) → String Examples Example 1 first = "aaaaa" second = "bbb" return = "abababaa" The first six characters alternate, then the two unused characters from the first string are appended. Example 2 first = "" second = "abc" return = "abc" The first string is empty, so the complete second string is appended. Constraints 0 <= first.length,second.length <= 100000 Both strings contain lowercase English letters.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is to loop to the max of the two lengths, or run two pointers until both are exhausted. At each index i, append first[i] if i < first.length, then append second[i] if i < second.length. That handles the suffix automatically, with no separate tail step. The common pitfall is looping only to min(len) and then forgetting the remainder, or hardcoding the remainder logic for only one string. The second pitfall is performance. With up to 100000 characters each, repeated string concatenation in some languages is quadratic, so use a list or StringBuilder and join once at the end. Check example 1: aaaaa and bbb gives ababab then aa. Check example 2: empty first returns second untouched. Time is O(n+m), space is O(n+m). If the live OA rattles you, StealthCoder is the hedge, but the logic fits in ten lines.
The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.
You can drill Interleave Two Strings cold, or you can hedge it. StealthCoder runs invisibly during screen share and surfaces a working solution in under 2 seconds. The proctor sees the IDE. They don't see what's behind it. Built for the candidate who saw this exact problem leak two days before his OA and wondered if anyone had a play.
Get StealthCoderRelated leaked OAs
This OA pattern shows up on LeetCode as merge strings alternately. If you have time before the OA, drill that.
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Make sure you actually pass ZipRecruiter's OA.
ZipRecruiter reuses patterns across OAs. Built for the candidate who saw this exact problem leak two days before his OA and wondered if anyone had a play. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Interleave Two Strings FAQ
How hard is the Interleave Two Strings OA really?+
Easy. It's a two-pointer string build with one real trap, the leftover suffix. If you can write a loop and handle unequal lengths, you're done. Most failures come from skipping edge cases like an empty first string, not from the algorithm.
What's the trick to this problem?+
Iterate to the max of both lengths and guard each append with a bounds check. That way the longer string's remaining characters get added naturally, with no special tail code. Always append from first before second at each index.
What edge cases should I test before submitting?+
Test an empty first string, an empty second string, both empty, equal lengths, and first longer than second like aaaaa and bbb. Also confirm the output starts with a character from first whenever first is non-empty.
Will string concatenation be too slow at 100000 characters?+
It can be in languages where strings are immutable and you rebuild them each step. Push characters into a list or use a StringBuilder, then join once at the end. That keeps the whole thing linear in total length.
How do I prepare for this in 48 hours?+
Write it once from scratch in your OA language, then run the two examples plus the empty-string cases by hand. Spend the rest of your time on other two-pointer and string-building patterns, since this one takes minutes to learn.