Prompt and scope
The stick has endpoints 0 and n, and cuts contains distinct internal positions. At each step, choose a cut in the current segment, pay that segment's length, and split it into two segments. Use the scale of LeetCode 1547: n is at most 1,000,000 and the number of cuts m is at most 100. Return only the minimum cost; recovering an order requires storing the decision point.
What the interviewer is testing
- Whether you recognize interval DP instead of greedily following the input order.
- Whether you add 0 and n as sentinels and sort the cut positions.
- Whether you can explain why choosing the first or last cut splits the problem into independent intervals.
- Whether you can state the m-based time, space, and integer-safety bounds.
Clarifying questions
- Can cuts include 0, n, or duplicates? If so, should the API deduplicate or reject them?
- Do we need only the minimum cost, or also one optimal order? The latter needs a choice table.
- What is the upper bound for m? A larger m may rule out cubic time.
- Is every cut cost exactly the current segment length? A weighted cost changes the transition.
30-second answer
I sort the cuts and add 0 and n. Let dp[i][j] be the minimum cost to perform every cut strictly between points i and j; adjacent endpoints have value zero. For each interval, try every internal pivot k as the first cut. That cut costs the interval length, and the left and right intervals are independent, so I add dp[i][k] and dp[k][j]. Filling spans in increasing order gives dp[0][m+1] in O(m cubed) time and O(m squared) space.
Deep-dive answer
1. Establish coordinates and invariants
Sort cuts and build points containing 0, all cut positions, and n. After sorting, the internal cuts of the interval from points[i] to points[j] are exactly indices between i and j. This invariant makes the state depend on endpoints rather than the history of earlier cuts.
2. Define the interval state
dp[i][j] is the minimum cost to perform every cut strictly inside points[i] and points[j]. When j is i plus one, there is no internal cut, so the value is zero. The state does not record order because after the first cut the left and right subproblems do not interact, and each later cost depends only on its current segment.
3. Derive the transition
If the first cut is points[k], with i less than k and k less than j, the current payment is points[j] minus points[i]. The cut creates two independent intervals:
dp[i][j] = min(
dp[i][k] + dp[k][j] + points[j] - points[i]
for k in (i + 1 ... j - 1)
)Compute shorter spans first so both subinterval values are already available.
4. Implement it
function minCost(n: number, cuts: number[]): number {
const points = [0, ...cuts.slice().sort((a, b) => a - b), n];
const m = points.length;
const dp = Array.from({ length: m }, () => Array<number>(m).fill(0));
for (let span = 2; span < m; span += 1) {
for (let left = 0; left + span < m; left += 1) {
const right = left + span;
let best = Number.POSITIVE_INFINITY;
for (let pivot = left + 1; pivot < right; pivot += 1) {
best = Math.min(
best,
dp[left][pivot] + dp[pivot][right] + points[right] - points[left],
);
}
dp[left][right] = best === Number.POSITIVE_INFINITY ? 0 : best;
}
}
return dp[0][m - 1];
}5. Prove the transition
Use induction on the number of internal cuts. With zero cuts the cost is zero. Assume shorter intervals are optimal. Every optimal ordering has a first pivot k. Its cost is fixed as the full interval length; the remaining work splits into left and right intervals, whose optimal values are dp[i][k] and dp[k][j] by induction. Taking the minimum over every possible k covers every first cut, so the transition is optimal.
6. Complexity, numbers, and reconstruction
With m internal cuts there are O(m squared) states and up to O(m) pivots per state, giving O(m cubed) time and O(m squared) space. With n up to 1,000,000 and m at most 100, a JavaScript number covers the stated cost range; for larger weighted costs, use BigInt or explicit safe-integer checks. To reconstruct an order, store the pivot that achieved each minimum and recursively emit the decision tree.
7. Counterexample and tests
For n=7 and cuts=[1,3,4,5], cutting in input order costs 20, while the order 3,5,1,4 costs 16. This disproves input-order and left-to-right greedy choices. Also test one cut, positions near endpoints, unsorted input, duplicate policy, consecutive gaps, maximum m, and the zero-internal-cut base intervals.
Model answer
I sort the cuts and add 0 and n. dp[i][j] is the minimum cost for every cut between the two endpoints, with zero for adjacent endpoints. For each interval I try pivot k as the first cut: pay the interval length, then add the optimal left and right costs. Filling by increasing span returns the outer interval. With m cuts the complexity is O(m cubed) time and O(m squared) space; storing each best pivot reconstructs an order. I test unsorted input, endpoint-near cuts, a single cut, and the n=7, [1,3,4,5] counterexample.
Common mistakes
- Cutting in the input order → That order may be far from optimal → Sort and enumerate the first pivot with interval DP.
- Omitting 0 and n → Boundary lengths and states are incomplete → Include both endpoints as sentinels.
- Defining dp as only one cut's cost → Later cuts are omitted → Define it as the minimum cost for the entire interior set.
- Choosing the shortest or longest segment greedily → Local choices change both subproblem costs → Show the recursive split and induction proof.
- Testing only sorted input → The code may silently assume order → Copy and sort inside the function, then test an unsorted array.
Follow-up questions
How do you return one optimal cut order?
Store the pivot that achieves each interval minimum. Emit that pivot, then recurse on the left and right intervals. If several pivots tie, define a deterministic rule such as the smallest index or lexicographically smallest order.
Is O(m cubed) acceptable when m grows from 100 to 2,000?
Do not promise it without measurement. Estimate the state count and time budget, then look for additional structure, approximation, or offline constraints. A generic O(m squared) claim requires a proven monotonicity or quadrangle-inequality property.
What if cuts contains duplicate positions?
A repeated cut has no second physical effect. Sort and deduplicate, or reject duplicates according to the input contract; document the choice and test it instead of creating zero-length intervals.
What if each cut costs segment length times a weight?
If the weight depends only on the chosen pivot k, replace the interval length term with interval length times weight[k], and the split remains independent. If cost depends on history, cut count, or cross-interval state, the subproblems are no longer independent and the state must be redesigned.