Check Monotonic Triples
Reported by candidates from TikTok's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The first thing you notice in this TikTok OA, reported in July 2025, is the output length: arr.length - 2, one flag per window of three. That's the whole problem. It's a fixed-size window scan, not a stack problem despite the hint tag. If you've got an invite for the next day or two, this one is a warm-up you can't afford to overthink. The risk is sloppy comparisons, not hard algorithms. If you blank under the clock, StealthCoder sits invisibly on your screen and hands you the loop so you can finish clean.
The problem
You are given an integer array arr. Determine whether every sequence of three consecutive elements, arr[i], arr[i + 1], and arr[i + 2], is monotonic. Three consecutive elements are monotonic when their values are in strictly increasing or strictly decreasing order. Return an integer array of length arr.length - 2. For every valid index i, set the i-th result to 1 if either arr[i] < arr[i + 1] < arr[i + 2] or arr[i] > arr[i + 1] > arr[i + 2]. Otherwise, set it to 0. A solution with time complexity no worse than O(arr.length^2) will fit within the execution time limit. Function solution(arr: int[]) → int[] Examples Example 1 arr = [1,2,1,-4,5,10] return = [0,1,0,1] [1, 2, 1] is not monotonic. [2, 1, -4] is strictly decreasing. [1, -4, 5] is not monotonic. [-4, 5, 10] is strictly increasing. Example 2 arr = [10,10,10,10,10] return = [0,0,0] All values are equal, so no sequence of three elements is strictly increasing or strictly decreasing.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is that there isn't one. Walk i from 0 to n-3, compare arr[i], arr[i+1], arr[i+2], and write 1 if a < b < c or a > b > c, else 0. That's O(n) time and O(n) output. The pitfall is equality. Example 2 is all tens, and the answer is all zeros because the order must be strict. Using <= or >= will fail that case. The second pitfall is array length. If arr has fewer than 3 elements, the result should be empty, so make sure your loop bound doesn't produce a negative-size array. Don't reach for a stack, a sort, or nested loops. The statement even allows O(n^2), which tells you the checker is forgiving. StealthCoder is your hedge on the live OA if nerves make you second-guess something this simple, but you should have this written in about two minutes.
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Check Monotonic Triples FAQ
How hard is the TikTok Check Monotonic Triples problem really?+
Easy. It's a single pass over the array with a three-element comparison per window. The only real difficulty is reading carefully. Strictly increasing or strictly decreasing means equal neighbors fail. Most candidates finish it in a few minutes if they avoid overengineering.
What's the trick to solving it?+
There's no trick beyond strictness. Loop i from 0 to length minus 3, check arr[i] < arr[i+1] < arr[i+2] or arr[i] > arr[i+1] > arr[i+2], and push 1 or 0. Use strict operators so equal values return 0, as in the all-tens example.
Do I need a stack even though it was tagged stack?+
No. The tag is misleading. Each window depends only on three adjacent values, so no state carries between windows. A plain for loop and a result array is the cleanest answer and runs in linear time.
What edge cases should I test?+
Test arrays with equal values like [10,10,10,10,10], arrays of exactly three elements, and arrays shorter than three. Also test a window where only two of the three are ordered, like [1,2,1]. That should return 0, not 1.
How do I prepare in 48 hours for an OA like this?+
Practice writing sliding-window and array-scan loops quickly and cleanly, paying attention to loop bounds and strict versus non-strict comparisons. Read the output spec twice before coding. Expect a TikTok OA to mix easy scans like this with something harder, so budget your time.