Unique Difference Pattern
Reported by candidates from Microsoft'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 Microsoft OA, reported in June 2026, is comparing the letters themselves instead of the gaps between them. "ACB" and "BDC" look nothing alike, but they share the same shape. The task: every string in the list has the same difference pattern except one, and you return the odd one out. It's a hash-table and string problem in disguise, and the logic is short once you see it. If you blank mid-assessment, StealthCoder runs invisibly on your desktop as a safety net, reads the problem, and gives you a working solution in real time.
The problem
You are given a list of strings containing only uppercase English letters. All strings in the list have the same length. For each string, calculate the difference between adjacent letters based on their alphabetical positions. For example, 'B' - 'A' = 1, 'C' - 'A' = 2. These differences form a difference pattern for the string. In the list, all strings except one share the same difference pattern. Identify and return the string with the unique difference pattern. Function findUniqueDifferencePattern(series: String[]) → String Examples Example 1 series = ["ACB", "BDC", "CED", "DEF"] return = "DEF" Calculate the differences between adjacent letters: "ACB": (C - A, B - C) = (+2, -1) "BDC": (D - B, C - D) = (+2, -1) "CED": (E - C, D - E) = (+2, -1) "DEF": (E - D, F - E) = (+1, +1) The first three strings have the same difference pattern (+2, -1), while "DEF" has a different pattern (+1, +1). Constraints 3 <= size of series[] <= 26 2 <= length of series[i] <= 26 All strings are uppercase English letters only. Within a test case, all strings are of equal length.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is to convert each string into its adjacent-difference signature, like a tuple or a joined string such as "2,-1". Then count how often each signature appears. Exactly one signature shows up once, and the string that produced it is your answer. Pitfall one: comparing raw letters or differences from 'A' instead of adjacent differences. Pitfall two: forgetting negatives, so "+2,-1" and "-2,+1" collapse into the same key if you use absolute values. Pitfall three: sloppy delimiters, where "1,12" and "11,2" blur together if you concatenate with no separator. A hash map from signature to list of strings handles all of it in O(n * m). With at least 3 strings, the majority pattern is unambiguous. If the live OA has you second-guessing the keying step, StealthCoder is the hedge that keeps you moving.
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Unique Difference Pattern FAQ
How hard is the Unique Difference Pattern problem really?+
Easy to low-medium. There's no clever algorithm, just a clean transformation and a frequency count. Most failures come from computing the wrong signature or using a bad key format, not from complexity. If you can write a hash map and loop over characters, you can solve it.
What's the trick to solving it?+
Reduce every string to its adjacent differences, then group by that signature. The group with exactly one member holds your answer. Use a delimiter between numbers so different patterns can't produce the same key by accident.
Do I need to compare every pair of strings?+
No. Pairwise comparison works but it's wasteful. Build signatures once, count them in a map, and return the string whose signature count is 1. That's linear in the total number of characters across the input.
Which edge cases should I test before submitting?+
Test the unique string in first, middle, and last positions. Test minimum length 2, where each pattern is a single number. Test negative differences. Test a case where letters differ wildly but patterns match, like the ACB and BDC pair in the example.
How do I prepare for this in 48 hours?+
Practice frequency-map problems where you build a canonical key per item and group by it. Write the signature function from memory twice. Then run the Microsoft example by hand. That covers this problem and its close variants.