Closest Dessert Cost
Time O(m * t) · Space O(t) · Official statement on LeetCode
Solutions
// Time: O(m * max(max_base, target + max_topping / 2)) ~= O(m * t)
// Space: O(max(max_base, target + max_topping / 2)) ~= O(t)
class Solution {
public:
int closestCost(vector<int>& baseCosts, vector<int>& toppingCosts, int target) {
static const int MAX_COUNT = 2;
const auto max_base = *max_element(cbegin(baseCosts), cend(baseCosts));
const auto max_topping = *max_element(cbegin(toppingCosts), cend(toppingCosts));
vector<bool> dp(max(max_base, target + max_topping / 2) + 1);
for (const auto& b : baseCosts) {
dp[b] = true;
}
for (const auto& t : toppingCosts) {
for (int count = 0; count < MAX_COUNT; ++count) {
for (int i = size(dp) - 1 - t; i >= 1; --i) {
if (dp[i]) {
dp[i + t] = true;
}
}
}
}
int result = numeric_limits<int>::max();
for (int i = 1; i <= size(dp) - 1; ++i) {
if (!dp[i]) {
continue;
}
if (abs(i - target) < abs(result - target)) {
result = i;
}
if (i >= target) {
break;
}
}
return result;
}
};
// Time: O(n * 3^m)
// Space: O(m * t)
class Solution2 {
public:
int closestCost(vector<int>& baseCosts, vector<int>& toppingCosts, int target) {
unordered_map<int, unordered_set<int>> lookup;
int result = numeric_limits<int>::max();
for (const auto& b : baseCosts) {
backtracking(toppingCosts, 0, b, target, &lookup, &result);
}
return result;
}
private:
void backtracking(const vector<int>& toppingCosts,
int i, int cost, int target,
unordered_map<int, unordered_set<int>> *lookup,
int *result) {
static const int max_top = 2;
if (lookup->count(i) && (*lookup)[i].count(cost)) {
return;
}
(*lookup)[i].emplace(cost);
if (cost >= target || i == size(toppingCosts)) {
if (pair(abs(cost - target), cost) < pair(abs(*result - target), *result)) {
*result = cost;
}
return;
}
for (int j = 0; j <= max_top; ++j) {
backtracking(toppingCosts, i + 1, cost + j * toppingCosts[i], target, lookup, result);
}
}
};
// Time: O(3^m*log(3^m)) + O(n*log(3^m)) = O(m*(3^m + n))
// Space: O(3^m)
class Solution3 {
public:
int closestCost(vector<int>& baseCosts, vector<int>& toppingCosts, int target) {
static const int MAX_COUNT = 2;
unordered_set<int> combs_set = {0};
for (const auto& t : toppingCosts) {
unordered_set<int> new_combs_set;
for (const auto& c : combs_set) {
for (int i = 0; i <= MAX_COUNT; ++i) {
new_combs_set.emplace(c + i * t);
}
}
combs_set = move(new_combs_set);
}
vector<int> combs(cbegin(combs_set), cend(combs_set));
sort(begin(combs), end(combs));
int result = numeric_limits<int>::max();
for (const auto& b : baseCosts) {
const auto& cit = lower_bound(cbegin(combs), cend(combs), target - b);
if (cit != cend(combs)) {
if (pair(abs(b + *cit - target), b + *cit) < pair(abs(result - target), result)) {
result = b + *cit;
}
}
if (cit != cbegin(combs)) {
if (pair(abs(b + *prev(cit) - target), b + *prev(cit)) < pair(abs(result - target), result)) {
result = b + *prev(cit);
}
}
}
return result;
}
};
// Time: O(n * 3^m)
// Space: O(3^m)
class Solution4 {
public:
int closestCost(vector<int>& baseCosts, vector<int>& toppingCosts, int target) {
static const int MAX_COUNT = 2;
unordered_set<int> combs_set = {0};
for (const auto& t : toppingCosts) {
unordered_set<int> new_combs_set;
for (const auto& c : combs_set) {
for (int i = 0; i <= MAX_COUNT; ++i) {
new_combs_set.emplace(c + i * t);
}
}
combs_set = move(new_combs_set);
}
int result = numeric_limits<int>::max();
for (const auto& b : baseCosts) {
for (const auto& c : combs_set) {
if (pair(abs(b + c - target), b + c) < pair(abs(result - target), result)) {
result = b + c;
}
}
}
return result;
}
};
Beginner Explanation
What is Closest Dessert Cost?
Closest Dessert Cost (LeetCode #1774) 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 dynamic programming.
- 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.
AlgoForge explanations are original teaching notes. Always open the official problem statement on LeetCode for constraints and examples.
Interview Walkthrough
Interview approach for Closest Dessert Cost
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 dynamic programming.
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(m * t)) and space (O(t)) 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(m * t) time and O(t) space.
Pattern focus: dynamic programming
Use the pattern as a checklist:
- dynamic programming — 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(m * t) |
| Space | O(t) |
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 Closest Dessert Cost
- Skipping edge cases — empty collections, single-element inputs, max constraints.
- Wrong invariant for dynamic programming — 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 dynamic programming:
- 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 Closest Dessert Cost 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 dynamic programming 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
Closest Dessert Cost (#1774) — Medium. Pattern: dynamic programming. Complexity: O(m * t) time / O(t) space. Re-derive the invariant before coding.
FAQs
What is the time complexity of Closest Dessert Cost?+
The reference solutions aim for O(m * t) time and O(t) space. Always re-derive complexity from the code you write in the interview.
What pattern does Closest Dessert Cost use?+
It primarily maps to dynamic programming, within the broader topic of dynamic programming.
Is Closest Dessert Cost 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/closest-dessert-cost/