EASYasked at 1 company

Redistribute Characters to Make All Strings Equal

A easy-tier problem at 67% community acceptance, tagged with Hash Table, String, Counting. Reported in interviews at Moengage and 0 others.

Founder's read

You've got n strings and need to redistribute all characters across them so each string ends up identical. It's an easy problem by rating but the trick is non-obvious if you haven't seen it before. Moengage has asked it. The catch: if character counts don't divide evenly by n, it's impossible. Most candidates waste time on complex redistribution logic when the real solution is a single validation check followed by character frequency math. If you blank on the pattern during the live OA, StealthCoder surfaces the approach in seconds.

Companies asking
1
Difficulty
EASY
Acceptance
67%

Companies that ask "Redistribute Characters to Make All Strings Equal"

If this hits your live OA

Redistribute Characters to Make All Strings Equal is the kind of problem that decides whether you pass. StealthCoder reads the problem on screen and surfaces a working solution in under 2 seconds. Invisible to screen share. The proctor sees nothing. Built because the OA filter rejects engineers who'd pass the on-site. That's a broken filter. This is the workaround.

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What this means

The problem looks like a complex rebalancing task until you realize it's actually a frequency validation problem. Count total occurrences of each character across all n strings. If any character count isn't divisible by n, return false immediately. Otherwise, every character must appear exactly (total_count / n) times in each final string, making equality guaranteed. The common mistake is trying to simulate the redistribution process or overthinking the mechanics of moving characters. Hash tables or frequency maps solve this in one pass. When this hits your assessment and the obvious greedy approach fails, remember: validate divisibility first, then confirm uniform distribution is mathematically possible. StealthCoder runs invisible during screen share and delivers the frequency-validation insight on demand.

Pattern tags

The honest play

You know the problem. Make sure you actually pass it.

Redistribute Characters to Make All Strings Equal recycles across companies for a reason. It's easy-tier, and most candidates blank under the timer. StealthCoder is the hedge: an AI overlay invisible during screen share. It reads the problem and surfaces a working solution in under 2 seconds. Built because the OA filter rejects engineers who'd pass the on-site. That's a broken filter. This is the workaround. Works on HackerRank, CodeSignal, CoderPad, and Karat.

Redistribute Characters to Make All Strings Equal interview FAQ

Is this really an easy problem or are acceptance rates misleading?+

Acceptance sits around 67%, which for easy-rated problems usually signals a conceptual trick rather than implementation difficulty. The gotcha is recognizing this as a frequency validation problem, not a simulation problem. Once you see that pattern, code is trivial.

What's the core trick that solves this?+

Count character frequencies across all strings and check if each frequency divides evenly by n. If yes, equal distribution is possible. If no, it's impossible. No actual redistribution logic needed. The math guarantees the answer.

Which data structures do I need?+

Hash tables or frequency maps are the standard tools. Count each character's total occurrences, then validate divisibility. Some languages use Counter (Python) or similar built-ins. String and hash table topics confirm this approach.

What mistake kills most candidates on this problem?+

Trying to simulate character movement or greedy redistribution. The problem doesn't ask how to redistribute, just whether it's possible. Many waste time on complex logic when a single divisibility check plus frequency count answers everything.

Does Moengage ask harder variants of this problem?+

Only one reported ask from Moengage, and it's the standard version. If they ask it again, expect the same core logic. The edge cases are usually around empty strings or n=1. Focus on the divisibility validation.

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