Minimum Number of Operations to Make X and Y Equal
Time O(x) · Space O(x) · Official statement on LeetCode
Solutions
// Time: O(x)
// Space: O(x)
// memoization
class Solution {
public:
int minimumOperationsToMakeEqual(int x, int y) {
unordered_map<int, int> lookup;
const function<int (int)> memoization = [&](int x) {
if (y >= x) {
return y - x;
}
if (!lookup.count(x)) {
lookup[x] = x - y;
for (const auto& d : {5, 11}) {
lookup[x] = min(lookup[x], min(x % d, d - x % d) + memoization(x / d + (d - x % d < x % d ? 1 : 0)) + 1);
}
}
return lookup[x];
};
return memoization(x);
}
};
// Time: O(x)
// Space: O(x)
// bfs
class Solution2 {
public:
int minimumOperationsToMakeEqual(int x, int y) {
if (y >= x) {
return y - x;
}
const int upper_bound = x + (x - y);
unordered_set<int> lookup = {x};
vector<int> q = {x};
for (int d = 0; !empty(q); ++d) {
vector<int> new_q;
for (const auto& x: q) {
if (x == y) {
return d;
}
vector<int> candidates = {x + 1, x - 1};
for (const auto& d : {5, 11}) {
if (x % d == 0) {
candidates.emplace_back(x / d);
}
}
for (const auto& new_x : candidates) {
if (!(0 <= new_x && new_x <= upper_bound && !lookup.count(new_x))) {
continue;
}
lookup.emplace(new_x);
new_q.emplace_back(new_x);
}
}
q = move(new_q);
}
return -1;
}
};
Beginner Explanation
What is Minimum Number of Operations to Make X and Y Equal?
Minimum Number of Operations to Make X and Y Equal (LeetCode #2998) is a Medium problem that primarily trains dynamic programming.
How to think about it
- Restate the goal in your own words before coding.
- Work a tiny example by hand so the invariant becomes obvious.
- Identify the pattern — this problem aligns with queue bfs.
- Only then translate the idea into code.
Why this problem matters
It sits in the sweet spot of interview difficulty: multiple valid approaches, clear trade-offs. Official solution notes mention: Memoization, BFS.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Minimum Number of Operations to Make X and Y Equal
Opening (30–60 seconds)
- Clarify inputs/outputs and edge cases (empty input, single element, duplicates, overflow).
- State a brute force so the interviewer knows you can solve it naively.
- Propose the optimal direction tied to queue bfs.
Core solution narrative
- Define the state you track (pointers, DP cell, set membership, stack top, etc.).
- Explain the transition when you process the next element.
- Call out time (O(x)) and space (O(x)) before coding.
- Code cleanly; narrate variable names.
What interviewers listen for
- Correctness on edge cases
- Complexity honesty
- Ability to discuss trade-offs (e.g., hash map space vs. sort + two pointers)
Follow-up questions they may ask
- Can you solve it with less memory?
- What if the input stream is infinite / doesn't fit in RAM?
- How would tests look for adversarial inputs?
Optimized Approach
Optimized solution notes
The reference solutions on AlgoForge target O(x) time and O(x) space.
Pattern focus: queue bfs
Use the pattern as a checklist:
- queue bfs — confirm the invariant holds after each step
Multiple methods appear in the source solutions — compare them and explain when each is preferable.
Implementation tips
- Prefer readable names over micro-optimizations in interviews.
- Extract helpers only when they clarify (e.g., expand-around-center, DFS visit).
- After AC-level logic, re-scan for off-by-one and null checks.
Complexity Analysis
Complexity
| Measure | Bound |
|---|---|
| Time | O(x) |
| Space | O(x) |
How to justify this in an interview
- Time: count loops, map/set operations, and recursive branching; state average vs worst case if relevant.
- Space: include hash maps, recursion stack, and output allocation when the problem asks for it.
If your implementation differs from the reference, re-derive big-O from your code — never memorize a complexity you cannot defend.
Common Mistakes
Common mistakes on Minimum Number of Operations to Make X and Y Equal
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for queue bfs — updating state too early or too late.
- Mutating input unexpectedly when the problem forbids it.
- Off-by-one in windows, ranges, or binary search bounds.
- Ignoring overflow / precision for integer arithmetic problems.
- Overengineering — jumping to an advanced structure when a simpler approach works.
Alternative Approaches
Alternatives
The source file includes more than one method. Compare:
- Primary optimized path — best complexity for typical interviews.
- Secondary approach — often brute force, sorting-based, or space-optimized variant.
Practice articulating when you would pick each (constraints, readability, follow-ups).
Edge Cases
Edge cases checklist
- Minimum input size
- Maximum input size / time limits
- Duplicates and already-sorted input
- Negative numbers / zeros (if applicable)
- Disconnected structures (graphs/trees)
- Single path vs branching recursion depth
Pattern Recognition
Spotting this pattern
Signal phrases that point to queue bfs:
- Sorted input or ability to sort without changing the answer class
- Need for contiguous subarray / substring → consider sliding window
- Need for O(1) membership → hash set/map
- Optimal substructure + overlapping subproblems → DP
- Connectivity / components → graph DFS/BFS or Union-Find
Primary topics: dynamic programming.
Follow-up Interview Questions
Follow-ups
- How does the solution change if the input is a stream?
- Can you solve it in-place?
- What if duplicates must be handled differently?
- How would you parallelize the approach?
- Design tests that would break a buggy implementation.
Practice Recommendations
What to practice next
- Re-solve Minimum Number of Operations to Make X and Y Equal in a second language (cpp, python).
- Drill 3–5 more problems tagged dynamic programming.
- Teach the solution out loud in under 5 minutes.
- Add this problem to your revision calendar in 3 days and 14 days.
Visualization
Study checklist
- Read the official problem statement on LeetCode
- Solve on paper / whiteboard first
- Implement the queue bfs approach
- Verify edge cases from the checklist
- State time and space complexity aloud
- Compare with the AlgoForge reference solution
- Schedule a revision session
Revision notes
Minimum Number of Operations to Make X and Y Equal (#2998) — Medium. Pattern: queue bfs. Complexity: O(x) time / O(x) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Minimum Number of Operations to Make X and Y Equal?+
The reference solutions aim for O(x) time and O(x) space. Always re-derive complexity from the code you write in the interview.
What pattern does Minimum Number of Operations to Make X and Y Equal use?+
It primarily maps to queue bfs, within the broader topic of dynamic programming.
Is Minimum Number of Operations to Make X and Y Equal good for interviews?+
Yes — as a Medium problem it is a solid practice target. Pair it with related problems in the same pattern family for spaced repetition.
Where can I read the official statement?+
Open the official LeetCode page for constraints and examples: https://leetcode.com/problems/minimum-number-of-operations-to-make-x-and-y-equal/