Minimum Cost for Cutting Cake I
Time O(mlogm + nlogn) · Space O(1) · Official statement on LeetCode
Solutions
// Time: O(mlogm + nlogn)
// Space: O(1)
// sort, greedy
class Solution {
public:
int minimumCost(int m, int n, vector<int>& horizontalCut, vector<int>& verticalCut) {
sort(begin(horizontalCut), end(horizontalCut));
sort(begin(verticalCut), end(verticalCut));
int result = 0;
for (int cnt_h = 1, cnt_v = 1; !empty(horizontalCut) || !empty(verticalCut); ) {
if (empty(verticalCut) || (!empty(horizontalCut) && horizontalCut.back() > verticalCut.back())) {
result += horizontalCut.back() * cnt_h; horizontalCut.pop_back();
++cnt_v;
} else {
result += verticalCut.back() * cnt_v; verticalCut.pop_back();
++cnt_h;
}
}
return result;
}
};
// Time: O(mlogm + nlogn)
// Space: O(1)
// sort, greedy
class Solution2 {
public:
int minimumCost(int m, int n, vector<int>& horizontalCut, vector<int>& verticalCut) {
sort(begin(horizontalCut), end(horizontalCut), greater<int>());
sort(begin(verticalCut), end(verticalCut), greater<int>());
int result = 0;
for (int i = 0, j = 0; i < size(horizontalCut) || j < size(verticalCut); ) {
if (j == size(verticalCut) || (i < size(horizontalCut) && horizontalCut[i] > verticalCut[j])) {
result += horizontalCut[i++] * (j + 1);
} else {
result += verticalCut[j++] * (i + 1);
}
}
return result;
}
};
// Time: O((m + n) * m^2 * n^2)
// Space: O(m^2 * n^2)
// memoization
class Solution3 {
public:
int minimumCost(int m, int n, vector<int>& horizontalCut, vector<int>& verticalCut) {
vector<vector<vector<vector<int>>>> lookup(m, vector<vector<vector<int>>>(n, vector<vector<int>>(m, vector<int>(n, -1))));
const function<int (int, int, int, int)> memoization = [&](int x1, int y1, int x2, int y2) {
static const int INF = numeric_limits<int>::max();
if (x1 == x2 && y1 == y2) {
return 0;
}
if (lookup[x1][y1][x2][y2] == -1) {
int mn = INF;
for (int x = x1; x + 1 <= x2; ++x) {
mn = min(mn, memoization(x1, y1, x, y2) + memoization(x + 1, y1, x2, y2) + horizontalCut[x]);
}
for (int y = y1; y + 1 <= y2; ++y) {
mn = min(mn, memoization(x1, y1, x2, y) + memoization(x1, y + 1, x2, y2) + verticalCut[y]);
}
lookup[x1][y1][x2][y2] = mn;
}
return lookup[x1][y1][x2][y2];
};
return memoization(0, 0, m - 1, n - 1);
}
};
Beginner Explanation
What is Minimum Cost for Cutting Cake I?
Minimum Cost for Cutting Cake I (LeetCode #3218) is a Medium problem that primarily trains greedy.
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 greedy.
- 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, Greedy.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Minimum Cost for Cutting Cake I
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 greedy.
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(mlogm + nlogn)) and space (O(1)) 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(mlogm + nlogn) time and O(1) space.
Pattern focus: greedy
Use the pattern as a checklist:
- greedy — 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(mlogm + nlogn) |
| Space | O(1) |
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 Cost for Cutting Cake I
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for greedy — 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 greedy:
- 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: greedy.
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 Cost for Cutting Cake I in a second language (cpp, python).
- Drill 3–5 more problems tagged greedy.
- 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 greedy 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 Cost for Cutting Cake I (#3218) — Medium. Pattern: greedy. Complexity: O(mlogm + nlogn) time / O(1) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Minimum Cost for Cutting Cake I?+
The reference solutions aim for O(mlogm + nlogn) time and O(1) space. Always re-derive complexity from the code you write in the interview.
What pattern does Minimum Cost for Cutting Cake I use?+
It primarily maps to greedy, within the broader topic of greedy.
Is Minimum Cost for Cutting Cake I 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-cost-for-cutting-cake-i/