Triangle Flags from Adjacent Elements
Reported by candidates from Luma AI'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 Luma AI question, reported in June 2026, is checking only one triangle inequality and calling it done. The task is simple on paper: for every window of three consecutive values in arr, output 1 if they form a valid triangle and 0 if not. It's a sliding window of size three with a tiny check inside. Easy to read, easy to botch with a sloppy condition. If you blank mid-assessment, StealthCoder runs invisibly as a safety net, but you should be able to write this one yourself.
The problem
For every three consecutive positive integers in arr, return 1 if they can be the side lengths of a triangle and 0 otherwise. The result has length arr.length - 2. Function adjacentTriangleFlags(arr: int[]) → int[] Examples Example 1 arr = [1,2,2,4] return = [1,0] The triple 1,2,2 is valid, while 2+2 is not greater than 4. Constraints 3 <= arr.length <= 100000. 1 <= arr[i] <= 10^9.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is the triangle test. All three inequalities must hold: a+b>c, a+c>b, b+c>a. Sort the three values, then only the two smaller ones summed against the largest matters. So check min+mid > max. The example proves it: 1,2,2 passes since 1+2>2, but 2,2,4 fails because 2+2 is not greater than 4. Strict greater-than, not greater-or-equal. That's the pitfall, since degenerate triangles count as 0. Second pitfall is overflow. Values go up to 10^9, so the sum of two can hit 2*10^9, which overflows a 32-bit int in Java or C++. Use long. Loop i from 0 to n-3, so the output has length n-2, one pass, O(n) time. Don't sort the whole array, only the three elements in each window. If the live OA freezes you, StealthCoder is the hedge, but the logic here is small enough to own.
The honest play: practice the pattern, and have StealthCoder ready for the one you didn't see coming.
You can drill Triangle Flags from Adjacent Elements 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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Triangle Flags from Adjacent Elements FAQ
What's the trick in Triangle Flags from Adjacent Elements?+
Sort each window of three, then check that the two smaller values sum to strictly more than the largest. That single comparison covers all three triangle inequalities. Slide across the array once and write 1 or 0 per window. It's O(n) overall with constant work per window.
How hard is this Luma AI question really?+
Easy. It's a fixed-size window with a math check. The difficulty is in the details: strict inequality, integer overflow at 10^9, and getting the output length right at arr.length - 2. Most people who miss it trip on one of those, not the algorithm.
Do I need to worry about integer overflow?+
Yes. Each value can reach 10^9, so adding two gives up to 2*10^9, which exceeds a signed 32-bit int. In Java, C++, or similar languages, cast to long before adding. Python handles big integers for you, so it's a non-issue there.
Is equal-sum a valid triangle here?+
No. The example shows 2,2,4 returns 0 because 2+2 is not greater than 4. The inequality must be strict. A sum equal to the third side is a degenerate triangle, and this problem treats it as invalid.
How do I prepare for this in 48 hours?+
Practice fixed-size sliding window problems and write the triangle check from memory. Test edge cases: length exactly 3, equal values, and values at 10^9. That covers nearly every way this question breaks. Then move on, since the pattern is simple and doesn't need more time.