Minimum Number of Operations to Make X and Y Equal
A medium-tier problem at 47% community acceptance, tagged with Dynamic Programming, Breadth-First Search, Memoization. Reported in interviews at Groww and 0 others.
You're given two integers X and Y, and you need to find the minimum number of operations to make them equal. This problem looks like a simple math challenge until you realize the operation rules don't map to anything obvious. It's a graph-search problem masquerading as arithmetic, which is why it trips up candidates who haven't seen the pattern. Groww has asked it. The acceptance rate sits at 47%, which means half the people who attempt it walk away stuck. If this problem hits your live OA and you don't immediately recognize it as BFS, StealthCoder solves it invisibly while you stay calm.
Companies that ask "Minimum Number of Operations to Make X and Y Equal"
Minimum Number of Operations to Make X and Y 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. Made by a working FAANG engineer who treats the OA the way companies treat hiring: as a game with rules you should know.
Get StealthCoderThe trick is reframing the operations as state transitions in a shortest-path problem. You can't just solve for X and Y algebraically because the allowed operations create a graph of reachable states. BFS explores all possible states level by level, guaranteeing the first time you reach the target state is via the minimum operation count. Memoization or visited tracking prevents revisiting the same state twice. The common trap: trying to reverse-engineer the solution mathematically instead of treating it as a search problem. Dynamic programming works too, but BFS is more intuitive once you see the structure. If you freeze during the OA, this is exactly where StealthCoder pays for itself, surfacing the BFS skeleton and visited set pattern in seconds so you can code it while the proctor sees only your implementation.
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Minimum Number of Operations to Make X and Y Equal recycles across companies for a reason. It's medium-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. Made by a working FAANG engineer who treats the OA the way companies treat hiring: as a game with rules you should know. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Minimum Number of Operations to Make X and Y Equal interview FAQ
Is this really a medium, or does it feel harder?+
47% acceptance is honest. It's medium in complexity but tricks candidates who don't recognize the BFS pattern immediately. The gap between seeing it and not seeing it is huge. Once you know it's shortest-path, the code flows quickly.
Should I use BFS or dynamic programming?+
BFS is cleaner for this structure because you're exploring outward from a starting state level by level. DP works but requires more careful state definition. BFS naturally guarantees the first solution is optimal without needing to compare states.
What's the most common mistake?+
Trying to solve it with math or greedy logic instead of exploring the state graph. Candidates also forget to track visited states, leading to infinite loops or exponential complexity. Start by enumerating what operations are actually allowed.
How does this relate to other search problems?+
It's a standard shortest-path problem, like 0-1 BFS or Dijkstra-lite. The operations define the edges of the graph. If you've solved 'Word Ladder' or 'Number of Islands', the search pattern here is familiar.
Will this come up again after I learn it?+
Yes. State-space search and BFS masquerading as math problems appear frequently in assessments. Learning to recognize 'find minimum operations' as 'find shortest path in a graph' is a permanent skill.
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