Minimum Steps on a Circular Character Dial
Reported by candidates from Navan'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 Navan OA, reported in January 2024, is going greedy: grabbing the nearest occurrence of each target character and moving on. It looks right on Example 1 and falls apart the moment the ring has duplicates, like "ABACA". You've got a circular dial, a start index, and a target to hit in order. It's a string problem wearing a DP costume. If you blank on the state definition during the live assessment, StealthCoder is the safety net that stays invisible on screen while you recover. But the pattern is small enough to own tonight.
The problem
You are given a circular array of characters as ring, a starting index startIndex, and a target string. Moving clockwise or counterclockwise to an adjacent ring position costs one step. For this exercise, assume you must visit the target characters in order. Visiting a character costs no extra selection step. The ring may contain duplicate characters, and you may choose any occurrence when visiting the next target character. Return the minimum total movement cost, or -1 if a target character is absent. An empty target costs zero steps. Function calculateMinSteps(ring: String, startIndex: int, target: String) → int Examples Example 1 ring = "ABCDEF" startIndex = 2 target = "FA" return = 4 From C at index 2, reach F in three moves in either direction, then A in one move. The total is four. Example 2 ring = "ABACA" startIndex = 1 target = "AC" return = 2 Choose A at index 2, one step from B, then C at index 3 in one more step. Choosing A at index 0 would cost more before the following C. Example 3 ring = "XYZ" startIndex = 2 target = "A" return = -1 A does not occur on the supplied ring. Constraints 1 <= ring.length <= 100. 0 <= startIndex < ring.length. 0 <= target.length <= 100. For this exercise, assume ring and target characters are ASCII letters and digits.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is dynamic programming over target positions. Let dp[i][j] be the minimum cost to have visited target[0..i] while standing on ring index j, where ring[j] equals target[i]. For each j that matches target[i], take the minimum over every k matching target[i-1] of dp[i-1][k] plus the circular distance between k and j. Circular distance is min(|a-b|, n-|a-b|). Seed the first layer from startIndex. The pitfall is greedy nearest-occurrence, which fails when a farther A sits closer to the next C, exactly Example 2. Also handle the edge cases: empty target returns 0, and any target character missing from the ring returns -1. With ring and target both capped at 100, the O(m * n * n) cost is trivial. If the recurrence slips away mid-OA, StealthCoder can feed you the full solution in real time.
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Minimum Steps on a Circular Character Dial FAQ
What's the trick in the Navan circular dial problem?+
Don't go greedy. Duplicate characters mean the nearest occurrence of the next target letter can lead to a worse path later. Use DP where state is the target index plus which ring position you're standing on, and take the minimum over all previous matching positions.
How hard is this really?+
Medium. The code is short, but you have to spot that greedy fails. Once you see the DP state, it's about 20 lines. The constraints, 100 for both ring and target, let a cubic solution pass easily.
How do I compute distance on a circular ring?+
Take d = abs(a - b), then the answer is min(d, n - d), where n is the ring length. That covers clockwise and counterclockwise in one expression. Forgetting the wraparound is the second most common bug after going greedy.
What edge cases should I test before submitting?+
Empty target returns 0. A target character not on the ring returns -1, like Example 3. Start index already sitting on the first target letter costs zero. Also test a ring with repeated letters, like Example 2, to confirm you're not picking the nearest occurrence blindly.
How do I prepare in 48 hours for a problem like this?+
Write this one from scratch twice, without peeking. Focus on defining the DP state out loud before coding. Then do two or three other layered-choice DP problems where each step picks among duplicates. The pattern repeats more than the specific problem does.