Interleave Equal-Length Strings by Column
Reported by candidates from Upstart'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 Upstart OA, reported in July 2026, is concatenating the strings whole instead of reading them column by column. Candidates see three strings and glue them together, then wonder why "abcdefghi" isn't "adgbehcfi". The problem is tiny. Take index 0 from all three, then index 1, then index 2. That's a plain string traversal with one loop. The risk isn't difficulty, it's rushing and getting the order wrong under a timer. If you blank anyway, StealthCoder runs invisibly during the live OA and hands you the loop as a safety net.
The problem
Given three equal-length strings, build a new string by taking the character at index 0 from all three strings, then the character at index 1 from all three strings, and so on. Function interleaveStrings(first: String, second: String, third: String) → String Examples Example 1 first = "abc" second = "def" third = "ghi" return = "adgbehcfi" Reading each column produces adg, then beh, then cfi.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is that all three strings share one length, so a single index i walks every string at once. Loop i from 0 to n-1, and on each pass append first[i], second[i], third[i] in that order. Done. The output length is 3n, and the work is O(n) time. The common pitfall is the order inside the column. Append first, then second, then third, every time. Another pitfall is building the result with repeated string concatenation in a language where that's slow. Use a list or StringBuilder and join at the end. Check your answer against the example: abc, def, ghi gives adg, beh, cfi, so adgbehcfi. Don't overthink edge cases. Empty strings should return an empty string, and your loop handles that for free. If the timer rattles you, StealthCoder is the hedge in the live OA, but this one is a two-minute write if you stay calm.
The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.
You can drill Interleave Equal-Length Strings by Column 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.
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Interleave Equal-Length Strings by Column FAQ
How hard is the Upstart interleave strings question really?+
It's easy. One loop, one index, three appends per pass. The only way to miss is rushing and concatenating whole strings or mixing up the column order. Write the example by hand first, confirm adgbehcfi, then code it.
What's the trick to interleaveStrings?+
Use one shared index since the strings are equal length. For each i, append first[i], second[i], third[i] in that order. There's no clever algorithm. It's careful traversal and clean output building.
What's the time and space complexity?+
Time is O(n), where n is the length of each string, because you visit each index once. Output space is O(3n), which is just O(n). Use a list or StringBuilder to avoid quadratic cost from repeated string concatenation.
What edge cases should I check?+
Empty strings should return an empty string. A length-1 case like a, b, c should return abc. The input says equal length, so you don't need to handle mismatched lengths unless the assessment text says otherwise.
How do I prepare for this in 48 hours?+
Don't grind. Write this one from memory in your language, then do a few similar string traversal problems like merging two strings alternately. Practice building output with a list and joining. Spend the rest of your time on harder problems the OA might also include.