Rod Cutting Problem in DSA Java

Mental Model

We want to cut a rod into pieces to get the maximum total price from those pieces. Each piece length has a price, and we try all cuts to find the best total.

Analogy: Imagine you have a chocolate bar and a price list for each possible piece size. You want to break the bar into pieces to sell for the most money, deciding where to break it.

Rod length: 8 units
Prices: [1, 5, 8, 9, 10, 17, 17, 20]
Rod: 8 units -> [cut points?]
No cuts yet: 8 units whole -> price unknown

Dry Run Walkthrough

Input: rod length = 4, prices = [1, 5, 8, 9]

Goal: Find the maximum price by cutting the rod of length 4 into pieces using given prices

Step 1: Try no cut: take whole rod length 4, price = 9

Length 4 -> price 9

Why: Check if selling whole rod is best

Step 2: Try cut at length 1 + remaining 3: price = price(1) + price(3) = 1 + 8 = 9

Cut: 1 + 3 -> prices 1 + 8 = 9

Why: Check if cutting into 1 and 3 is better

Step 3: Try cut at length 2 + remaining 2: price = price(2) + price(2) = 5 + 5 = 10

Cut: 2 + 2 -> prices 5 + 5 = 10

Why: Check if cutting into two pieces of length 2 is better

Step 4: Try cut at length 3 + remaining 1: price = price(3) + price(1) = 8 + 1 = 9

Cut: 3 + 1 -> prices 8 + 1 = 9

Why: Check if cutting into 3 and 1 is better

Step 5: Choose max price among all options: max(9, 9, 10, 9) = 10

Max price = 10

Why: Select best cutting strategy for max profit

Result:

Max price for rod length 4 is 10

Annotated Code

DSA Java

public class RodCutting {

    // prices array: price[i] is price of rod length i+1
    public static int cutRod(int[] price, int n) {
        int[] dp = new int[n + 1]; // dp[i] stores max price for length i
        dp[0] = 0; // no rod, no price

        for (int i = 1; i <= n; i++) {
            int max_val = Integer.MIN_VALUE;
            for (int j = 1; j <= i; j++) {
                int val = price[j - 1] + dp[i - j];
                if (val > max_val) {
                    max_val = val; // update max price for length i
                }
            }
            dp[i] = max_val; // store best price for length i
        }
        return dp[n]; // max price for full rod length n
    }

    public static void main(String[] args) {
        int[] price = {1, 5, 8, 9};
        int n = 4;
        int maxPrice = cutRod(price, n);
        System.out.println("Max price for rod length " + n + " is " + maxPrice);
    }
}

for (int i = 1; i <= n; i++) {

iterate over all rod lengths from 1 to n to find max price

for (int j = 1; j <= i; j++) {

try all possible first cut lengths j for current rod length i

int val = price[j - 1] + dp[i - j];

calculate price if we cut first piece length j plus best price for remaining length

if (val > max_val) { max_val = val; }

update max price for current length i if better found

dp[i] = max_val;

store best price for rod length i

return dp[n];

return max price for full rod length n

Output

Max price for rod length 4 is 10

Complexity Analysis

Time: O(n^2) because for each length i, we try all cuts j from 1 to i

Space: O(n) for dp array storing max prices for all lengths up to n

vs Alternative: Naive recursive approach is exponential due to repeated subproblems; DP reduces it to polynomial time by storing results

Edge Cases

rod length 0

returns 0 price since no rod to cut

DSA Java

dp[0] = 0;

rod length 1

returns price of length 1 piece directly

DSA Java

for (int i = 1; i <= n; i++) { ... }

all prices zero

max price will be 0 regardless of cuts

DSA Java

int max_val = Integer.MIN_VALUE; ... if (val > max_val) { max_val = val; }

When to Use This Pattern

When asked to maximize profit by cutting a rod or dividing a resource into parts with given prices, use the Rod Cutting DP pattern to try all cuts and store best results.

Common Mistakes

Mistake: Not storing intermediate results causing repeated calculations and slow performance
Fix: Use a dp array to store max prices for all lengths to avoid recomputation

Mistake: Off-by-one errors in indexing price array
Fix: Remember price[j-1] corresponds to piece length j

Summary

Find the best way to cut a rod to maximize total price from given piece prices.

Use when you need to optimize cutting or partitioning with known piece values.

Try all cuts for each length and store best prices to avoid repeated work.

Practice

1. What is the main goal of the Rod Cutting Problem in algorithms?

easy

A. To find the shortest rod length possible

B. To maximize the total profit by cutting the rod into pieces

C. To minimize the number of cuts made on the rod

D. To count the total number of ways to cut the rod

Solution

1. Step 1: Understand the problem objective

   The rod cutting problem aims to cut a rod into pieces to get the highest total price from selling those pieces.

2. Step 2: Identify the optimization goal

   The goal is to maximize profit, not minimize cuts or count ways.

3. Final Answer:

   To maximize the total profit by cutting the rod into pieces -> Option B

4. Quick Check:

   Maximize profit = To maximize the total profit by cutting the rod into pieces [OK]

Hint: Focus on maximizing profit, not minimizing cuts [OK]

Common Mistakes:

• Thinking the goal is to minimize cuts
• Confusing with counting ways to cut
• Assuming shortest rod length is the goal

2. Which of the following is the correct way to declare the price array for rod lengths in Java?

easy

A. int[] price = {1, 5, 8, 9};

B. int price[] = (1, 5, 8, 9);

C. int price = {1, 5, 8, 9};

D. int[] price = (1, 5, 8, 9);

Solution

1. Step 1: Recall Java array declaration syntax

   In Java, arrays are declared with square brackets and initialized with curly braces.

2. Step 2: Check each option

   int[] price = {1, 5, 8, 9}; uses correct syntax: int[] price = {1, 5, 8, 9};. Others use parentheses or wrong assignment.

3. Final Answer:

   int[] price = {1, 5, 8, 9}; -> Option A

4. Quick Check:

   Correct Java array syntax = int[] price = {1, 5, 8, 9}; [OK]

Hint: Use curly braces {} to initialize arrays in Java [OK]

Common Mistakes:

• Using parentheses () instead of curly braces {}
• Missing square brackets [] in declaration
• Assigning array to int variable

3. What will be the output of the following Java code snippet for rod length 4?

int[] price = {1, 5, 8, 9};
int n = 4;
int maxVal = 0;
for (int i = 1; i <= n; i++) {
    maxVal = Math.max(maxVal, price[i-1] + price[n - i]);
}
System.out.println(maxVal);

medium

A. 13

B. 9

C. 10

D. 8

Solution

1. Step 1: Trace the loop iterations and calculations

   For n=4, loop i=1 to 4:
   i=1: maxVal = max(0, price[0]+price[3]) = max(0,1+9)=10
   i=2: maxVal = max(10, price[1]+price[2])=max(10,5+8)=13
   i=3: maxVal = max(13, price[2]+price[1])=max(13,8+5)=13
   i=4: maxVal = max(13, price[3]+price[0])=max(13,9+1)=13

2. Step 2: Identify the final maxVal value

   After loop, maxVal is 13, but code prints maxVal after loop ends. The code prints 13.

3. Final Answer:

   13 -> Option A

4. Quick Check:

   Max profit for length 4 = 13 [OK]

Hint: Check all splits and sum prices to find max [OK]

Common Mistakes:

• Not considering all splits i=1 to n
• Off-by-one errors in array indexing
• Printing maxVal before loop ends

4. Identify the error in this Java method for rod cutting:

public int cutRod(int[] price, int n) {
    if (n <= 0) return 0;
    int maxVal = Integer.MIN_VALUE;
    for (int i = 1; i < n; i++) {
        maxVal = Math.max(maxVal, price[i-1] + cutRod(price, n - i));
    }
    return maxVal;
}

medium

A. Recursive call parameters are incorrect

B. Base case should return 1 instead of 0

C. Loop should run from i = 1 to i <= n

D. maxVal should be initialized to 0 instead of Integer.MIN_VALUE

Solution

1. Step 1: Analyze the loop range

   The loop runs from i=1 to i<n, so it misses the case i=n which is needed to consider the whole rod as one piece.

2. Step 2: Correct the loop condition

   The loop should run from i=1 to i<=n to include all possible cuts including the full length.

3. Final Answer:

   Loop should run from i = 1 to i <= n -> Option C

4. Quick Check:

   Include full length cut by looping to i <= n [OK]

Hint: Loop must include full rod length i = n [OK]

Common Mistakes:

• Using i < n instead of i <= n
• Incorrect base case return value
• Wrong initialization of maxVal

5. Given a rod of length 5 and price array int[] price = {2, 5, 7, 8, 10};, what is the maximum profit using dynamic programming approach?

hard

A. 14

B. 10

C. 15

D. 12

Solution

1. Step 1: Calculate max profit for each length up to 5

   Length 1: max = 2
   Length 2: max = max(5, 2+2=4) = 5
   Length 3: max = max(7, 5+2=7, 2+5=7) = 7
   Length 4: max = max(8, 7+2=9, 5+5=10, 2+7=9) = 10
   Length 5: max = max(10, 10+2=12, 7+5=12, 5+7=12, 2+10=12) = 12

2. Step 2: Check if cutting into smaller pieces yields better profit

   Try cutting length 5 into 1+4, 2+3, 3+2, 4+1 etc. The best is 2+3 or 3+2 or 1+4 giving 12, but also check if full length price 10 is less than these.

3. Step 3: Recalculate carefully

   Check all splits:
   1+4: 2 + 10 = 12
   2+3: 5 + 7 = 12
   3+2: 7 + 5 = 12
   4+1: 8 + 2 = 10
   But also check if cutting into 1+1+3 or 1+1+1+2 etc. can yield more.
   Try 1+1+3: 2+2+7=11
   Try 1+1+1+2: 2+2+2+5=11
   Try 1+1+1+1+1: 2*5=10
   So max is 12.

4. Final Answer:

   12 -> Option D

5. Quick Check:

   Max profit for length 5 = 12 [OK]

Hint: Check all splits and pick max sum for length 5 [OK]

Common Mistakes:

• Choosing price of full length only
• Not checking all split combinations
• Confusing price array indices