Rod Cutting Problem in DSA C#

Mental Model

We want to cut a rod into pieces to get the most money by selling those pieces. Each piece length has a price, and we try all ways to cut to find the best total price.

Analogy: Imagine you have a chocolate bar and different prices for each piece size. You want to break it into parts to sell and get the most money, like selling candy pieces separately instead of whole.

Rod length: 8 units
Prices: [1=1$, 2=5$, 3=8$, 4=9$, 5=10$, 6=17$, 7=17$, 8=20$]
Rod: 8 units long
[1][2][3][4][5][6][7][8]
↑ Start cutting here

Dry Run Walkthrough

Input: rod length = 4, prices = [1=2$, 2=5$, 3=7$, 4=8$]

Goal: Find the best way to cut the rod of length 4 to get maximum money

Step 1: Check cutting rod of length 1: best price is price[1] = 2

Length 1: max price = 2

Why: Base case, no cuts needed

Step 2: Check cutting rod of length 2: try no cut (price[2]=5) and cut into two pieces (1+1 = 2+2=4)

Length 2: max price = max(5, 2+2) = 5

Why: Compare selling whole vs two pieces

Step 3: Check cutting rod of length 3: try no cut (7), cut into (1+2) = 2+5=7, cut into (2+1) = 5+2=7

Length 3: max price = 7

Why: Try all cuts to find max

Step 4: Check cutting rod of length 4: try no cut (8), cut into (1+3) = 2+7=9, (2+2) = 5+5=10, (3+1) = 7+2=9

Length 4: max price = 10

Why: Try all cuts, best is two pieces of length 2

Result:

Max price for rod length 4 is 10

Annotated Code

DSA C#

using System;

class RodCutting {
    public static int CutRod(int[] prices, int n) {
        int[] dp = new int[n + 1];
        dp[0] = 0; // no rod, no money

        for (int length = 1; length <= n; length++) {
            int max_val = int.MinValue;
            for (int cut = 1; cut <= length; cut++) {
                int current_val = prices[cut - 1] + dp[length - cut];
                if (current_val > max_val) {
                    max_val = current_val; // update max for this length
                }
            }
            dp[length] = max_val; // store best price for length
        }
        return dp[n];
    }

    static void Main() {
        int[] prices = {2, 5, 7, 8};
        int rodLength = 4;
        int maxPrice = CutRod(prices, rodLength);
        Console.WriteLine(maxPrice);
    }
}

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

iterate over all rod lengths from 1 to n

for (int cut = 1; cut <= length; cut++) {

try all possible first cut lengths

int current_val = prices[cut - 1] + dp[length - cut];

calculate price for current cut plus best price for remaining rod

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

update max price for current rod length

dp[length] = max_val;

store best price found for rod length

Output

10

Complexity Analysis

Time: O(n^2) because for each rod length we try all possible cuts

Space: O(n) for the dp array storing best prices for each length

vs Alternative: Naive recursive approach is exponential due to repeated calculations; DP reduces it to polynomial time

Edge Cases

rod length 0

returns 0 because no rod means no money

DSA C#

dp[0] = 0; // no rod, no money

rod length 1

returns price of length 1 piece directly

DSA C#

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

all prices zero

returns 0 because no piece has value

DSA C#

max_val initialized to int.MinValue and updated only if current_val > max_val

When to Use This Pattern

When you see a problem asking to maximize profit by cutting or splitting a resource into parts, think of the Rod Cutting pattern using dynamic programming to try all splits efficiently.

Common Mistakes

Mistake: Not storing intermediate results causing repeated calculations
Fix: Use a dp array to save best prices for smaller lengths

Mistake: Off-by-one errors in indexing prices array
Fix: Remember prices[cut - 1] corresponds to piece length 'cut'

Summary

Find the best way to cut a rod to maximize total selling price.

Use it when you want to split a resource into parts for maximum gain.

Try all cuts and save best results for smaller lengths to avoid repeated work.

Practice

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

easy

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

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

C. To find the shortest rod length possible

D. To minimize the number of cuts made on 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 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 from pieces, not cuts or counts [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 C# method signature for a function solving the rod cutting problem recursively?

easy

A. int CutRod(int prices, int[] n)

B. void CutRod(int prices[], int n)

C. int CutRod(int[] prices, int n)

D. int CutRod(int[] prices, int[] lengths)

Solution

1. Step 1: Check parameter types and order

   The function should take an array of prices and an integer length n.

2. Step 2: Verify return type

   The function returns an integer representing maximum profit.

3. Final Answer:

   int CutRod(int[] prices, int n) -> Option C

4. Quick Check:

   Correct signature returns int and takes int[] and int [OK]

Hint: Return int max profit; input array prices and int length [OK]

Common Mistakes:

• Using void instead of int return type
• Swapping parameter types
• Passing arrays incorrectly

3. What is the output of the following C# code snippet for rod cutting?

int[] prices = {1, 5, 8, 9};
int n = 4;
int result = CutRod(prices, n);
Console.WriteLine(result);

int CutRod(int[] price, int length) {
    if (length == 0) return 0;
    int max_val = int.MinValue;
    for (int i = 1; i <= length; i++) {
        max_val = Math.Max(max_val, price[i - 1] + CutRod(price, length - i));
    }
    return max_val;
}

medium

A. 10

B. 9

C. 8

D. 5

Solution

1. Step 1: Understand recursive calls and prices

   Prices: length 1=1, 2=5, 3=8, 4=9. The function tries all cuts and sums prices recursively.

2. Step 2: Calculate max profit for length 4

   Possible cuts: - 1 + CutRod(3) = 1 + 8 = 9 - 5 + CutRod(2) = 5 + 5 = 10 - 8 + CutRod(1) = 8 + 1 = 9 - 9 + CutRod(0) = 9 + 0 = 9 Max is 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Max profit for length 4 = 10 [OK]

Hint: Try all cuts, pick max sum of price + recursive result [OK]

Common Mistakes:

• Ignoring recursive calls
• Choosing first price only
• Misindexing price array

4. Identify the bug in this C# code for rod cutting and select the fix:

int CutRod(int[] price, int length) {
    if (length == 0) return 0;
    int max_val = 0;
    for (int i = 1; i <= length; i++) {
        max_val = Math.Max(max_val, price[i] + CutRod(price, length - i));
    }
    return max_val;
}

medium

A. Change 'price[i]' to 'price[i - 1]' to fix array index out of bounds

B. Initialize max_val to int.MinValue instead of 0

C. Change loop to 'for (int i = 0; i < length; i++)' to fix indexing

D. Add base case for length == 1 to return price[0]

Solution

1. Step 1: Check array indexing in loop

   Loop uses i from 1 to length, but price array is zero-indexed. Accessing price[i] causes out of bounds when i == length.

2. Step 2: Fix indexing by subtracting 1

   Change price[i] to price[i - 1] to access correct element safely.

3. Final Answer:

   Change 'price[i]' to 'price[i - 1]' to fix array index out of bounds -> Option A

4. Quick Check:

   Index fix = Change 'price[i]' to 'price[i - 1]' to fix array index out of bounds [OK]

Hint: Array index starts at 0, loop starts at 1, adjust index [OK]

Common Mistakes:

• Using price[i] directly causing runtime error
• Changing loop incorrectly breaking logic
• Ignoring base case

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

hard

A. 10

B. 12

C. 15

D. 14

Solution

1. Step 1: Initialize dp array for lengths 0 to 5

   dp[0]=0. Calculate dp[i] for i=1 to 5 using max of price[j-1]+dp[i-j].

2. Step 2: Calculate dp values step-by-step

   dp[1]=2 dp[2]=max(5,2+2)=5 dp[3]=max(7,5+2,2+5)=7 dp[4]=max(8,7+2,5+5,2+7)=10 dp[5]=max(10,8+2,7+5,5+7,2+10)=14

3. Final Answer:

   14 -> Option D

4. Quick Check:

   Bottom-up dp max profit = 14 [OK]

Hint: Build dp array from length 1 to n, pick max at each step [OK]

Common Mistakes:

• Not using dp array, causing repeated calculations
• Incorrect max calculation in loop
• Ignoring smaller cuts combinations