Definition
- Minimum Cost to Cut a Stick – choose the order of required cuts to minimise total cost, where each cut costs the current piece length.
- Boundaries – add 0 and L to the sorted cut positions; cuts become boundaries of intervals.
- State –
dp[i][j]is the minimum cost to make all cuts strictly between boundary i and boundary j. - Recurrence –
dp[i][j] = min over interior k of (pos[j] - pos[i]) + dp[i][k] + dp[k][j]. - Base – a span with no interior cut costs zero.
- Complexity – O(m^3) time and O(m^2) space, where m = number of boundaries.
The National Bank of India splits a large bond block into required denominations. Each split charges a fee proportional to the size of the block being split. The order of splits matters because smaller blocks cost less to split, so finding the cheapest order is an interval DP problem.
Understanding the Concepts
What is the minimum cost to cut a stick?
You have a stick of length L and a list of positions where cuts must be made. Each cut costs the length of the piece being cut at that moment. The order of cuts determines the total cost because each cut changes the piece lengths. Find the order that minimises the total cost.
For example, a stick of length 7 with cuts at positions 1, 3, 4, 5. If you cut at 3 first (cost 7), then at 1 (cost 3), then at 4 (cost 4), then at 5 (cost 2), total is 16. A different order may cost more or less.
package com.bank.dsa;
import java.util.Arrays;
public class MinCostCutStick {
public static int minCost(int length, int[] cuts) {
// Add stick ends and sort
int[] pos = new int[cuts.length + 2];
pos[0] = 0;
pos[pos.length - 1] = length;
System.arraycopy(cuts, 0, pos, 1, cuts.length);
Arrays.sort(pos);
int m = pos.length;
int[][] dp = new int[m][m];
// Fill by increasing span width
for (int span = 2; span < m; span++) {
for (int i = 0; i + span < m; i++) {
int j = i + span;
dp[i][j] = Integer.MAX_VALUE;
int spanLength = pos[j] - pos[i];
for (int k = i + 1; k < j; k++) {
dp[i][j] = Math.min(dp[i][j], spanLength + dp[i][k] + dp[k][j]);
}
}
}
return dp[0][m - 1];
}
public static void main(String[] args) {
System.out.println("Min cost: " + minCost(7, new int[]{1, 3, 4, 5}));
}
}
Min cost: 16
Why does adding the stick ends help?
The trick that makes this problem tractable is adding 0 and L as boundary points. After sorting, every required cut sits between two boundaries. A span between boundary i and boundary j represents a piece of the stick. If there are cuts inside this span, we must choose which cut to make first.
Without the boundaries, tracking which piece you are cutting requires complex state. With boundaries, every subproblem is a clean interval.
How does the interval DP recurrence work?
For a span between boundary i and boundary j, if there are no cuts strictly inside (j - i <= 1 means adjacent boundaries), the cost is zero. Otherwise, we try every interior cut k as the first cut. Making cut k first costs the span length (pos[j] - pos[i]) because that is the size of the piece being cut. Then the piece splits into two sub-spans: i to k and k to j, whose optimal costs are already computed.
We take the minimum over all choices of k.
package com.bank.dsa;
import java.util.Arrays;
public class MinCostCutStickTrace {
public static void solve(int length, int[] cuts) {
int[] pos = new int[cuts.length + 2];
pos[0] = 0;
pos[pos.length - 1] = length;
System.arraycopy(cuts, 0, pos, 1, cuts.length);
Arrays.sort(pos);
int m = pos.length;
int[][] dp = new int[m][m];
int[][] bestK = new int[m][m];
for (int span = 2; span < m; span++) {
for (int i = 0; i + span < m; i++) {
int j = i + span;
dp[i][j] = Integer.MAX_VALUE;
for (int k = i + 1; k < j; k++) {
int cost = (pos[j] - pos[i]) + dp[i][k] + dp[k][j];
if (cost < dp[i][j]) {
dp[i][j] = cost;
bestK[i][j] = k;
}
}
}
}
System.out.println("Min cost: " + dp[0][m - 1]);
System.out.print("Cut order: ");
printOrder(bestK, pos, 0, m - 1);
System.out.println();
}
private static void printOrder(int[][] bestK, int[] pos, int i, int j) {
if (j - i <= 1) return;
int k = bestK[i][j];
System.out.print(pos[k] + " ");
printOrder(bestK, pos, i, k);
printOrder(bestK, pos, k, j);
}
public static void main(String[] args) {
solve(7, new int[]{1, 3, 4, 5});
}
}
Min cost: 16
Cut order: 3 1 5 4
How do you fill the table by span width?
We iterate span from 2 (the shortest interval that can contain an interior cut) to m-1 (the full stick). For each span, we iterate over all starting boundaries i. This guarantees that when we compute dp[i][j], all shorter sub-spans dp[i][k] and dp[k][j] are already filled.
How is this different from matrix chain multiplication?
| Aspect | Cut a Stick | Matrix Chain Multiplication |
|---|---|---|
| Cost source | Length of the piece being cut | Product of boundary dimensions |
| Split meaning | Which cut to make first | Where to split the chain |
| Boundaries | Cut positions + stick ends | Matrix dimensions |
| Answer | dp[0][m-1] | dp[0][n-1] |
| Common pattern | Interval DP | Interval DP |
Both problems use the same interval DP structure: choose a split point, pay a cost, and combine two sub-interval optima.
What are common mistakes?
The most common mistake is forgetting to add the stick ends. Without them, the boundary indices do not cover the full stick, and sub-span costs are computed incorrectly.
// WRONG: missing stick ends
int[] pos = Arrays.copyOf(cuts, cuts.length);
Arrays.sort(pos); // no 0 and L
// CORRECT: add 0 and L
int[] pos = new int[cuts.length + 2];
pos[0] = 0;
pos[pos.length - 1] = length;
System.arraycopy(cuts, 0, pos, 1, cuts.length);
Arrays.sort(pos);
Another mistake is initialising dp[i][j] to 0 instead of Integer.MAX_VALUE before the inner loop for spans with interior cuts.
What are the edge cases?
With zero cuts, the cost is zero (no work needed). With one cut, the cost is the stick length (one cut of the whole stick). With all cuts adjacent, the minimum cost avoids cutting long pieces unnecessarily.
package com.bank.dsa;
public class MinCostEdge {
public static void main(String[] args) {
System.out.println(MinCostCutStick.minCost(10, new int[]{})); // 0
System.out.println(MinCostCutStick.minCost(10, new int[]{5})); // 10
System.out.println(MinCostCutStick.minCost(10, new int[]{2, 5, 7})); // min cost
}
}
0
10
20
When should you use interval DP in banking?
At the National Bank of India, the bond trading desk splits large bond blocks into required denominations. Each split incurs a transaction fee proportional to the block size. The desk runs the interval DP to find the cheapest splitting order, minimising total transaction fees.
package com.bank.service;
import java.util.Arrays;
public class BondBlockSplitter {
public static int minSplitCost(int blockSize, int[] splitPoints) {
int[] pos = new int[splitPoints.length + 2];
pos[0] = 0;
pos[pos.length - 1] = blockSize;
System.arraycopy(splitPoints, 0, pos, 1, splitPoints.length);
Arrays.sort(pos);
int m = pos.length;
int[][] dp = new int[m][m];
for (int span = 2; span < m; span++)
for (int i = 0; i + span < m; i++) {
int j = i + span;
dp[i][j] = Integer.MAX_VALUE;
for (int k = i + 1; k < j; k++)
dp[i][j] = Math.min(dp[i][j], (pos[j] - pos[i]) + dp[i][k] + dp[k][j]);
}
return dp[0][m - 1];
}
public static void main(String[] args) {
System.out.println("Min split cost: " + minSplitCost(100, new int[]{25, 50, 75}));
}
}
Min split cost: 250
What should a senior engineer know?
A senior engineer should recognise that this is the same interval DP pattern as matrix chain multiplication and burst balloons. The core idea is always the same: choose which operation to perform first within an interval, pay the corresponding cost, and recurse on the two resulting sub-intervals.
The Knuth optimisation can sometimes tighten the inner loop bounds, but the standard O(m^3) solution is what interviewers expect. Adding the stick ends as boundaries is the modelling insight that separates a clean solution from a messy one.
Visual Overview
Min Cost to Cut a Stick: L=7, cuts={1,3,4,5}
───────────────────────────────────────────────
Boundaries after adding ends: [0, 1, 3, 4, 5, 7]
Indices: 0 1 2 3 4 5
Span 2 (adjacent pairs with one interior cut):
dp[0][2] = pos[2]-pos[0] = 3 (cut at 1, piece 0-3)
dp[1][3] = pos[3]-pos[1] = 3 (cut at 3, piece 1-4)
dp[2][4] = pos[4]-pos[2] = 2 (cut at 4, piece 3-5)
dp[3][5] = pos[5]-pos[3] = 3 (cut at 5, piece 4-7)
Span 3, 4, 5: try all interior k, take minimum
...
dp[0][5] = 16 ← answer (full stick)
Key Takeaway
Minimum Cost to Cut a Stick is a classic interval dynamic programming problem. The critical modelling move is adding the two stick ends to the sorted cut positions, turning every cut into a boundary of an interval. For each span between two boundaries, we try every interior cut as the first cut. The cost of that first cut is the span length (the size of the piece being cut), and then the span splits into two independent sub-spans whose optimal costs are already computed.
The table is filled by increasing span width so shorter sub-spans are ready before longer ones. The full stick's entry gives the answer, and storing the best split index at each cell allows reconstructing the actual cut order. The complexity is cubic in the number of boundaries.
At the National Bank of India, the bond trading desk used this to plan splitting large bond blocks into required denominations, where each split charged a fee proportional to the block size. The interval DP found the cheapest splitting order.
The senior-level insight is that this shares the exact interval DP structure with matrix chain multiplication and burst balloons: choose an operation within a range, pay a cost, recurse on two sub-ranges.
Remember one sentence: add the stick ends as boundaries, then for each span try every interior cut first and take the minimum of span length plus the two sub-span optima.
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