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 cuts 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, not just sell it whole.

Rod length: 8 units
Prices: [1, 5, 8, 9, 10, 17, 17, 20]
Initial rod: 8 units long
[8]
Cut options: 1 + 7, 2 + 6, 3 + 5, 4 + 4, etc.

Dry Run Walkthrough

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

Goal: Find the best way to cut the rod of length 4 to maximize total price.

Step 1: Check price if we don't cut the rod (length 4 piece as whole)

Max price for length 4 = 9

Why: We start by considering selling the rod whole without cuts.

Step 2: Try cutting rod into 1 + 3 pieces and sum their max prices

Max price for length 1 = 1, length 3 = 8, total = 9

Why: Check if cutting into 1 and 3 gives better price than whole.

Step 3: Try cutting rod into 2 + 2 pieces and sum their max prices

Max price for length 2 = 5, length 2 = 5, total = 10

Why: Cutting into two pieces of length 2 each gives total 10, better than previous.

Step 4: Try cutting rod into 3 + 1 pieces and sum their max prices

Max price for length 3 = 8, length 1 = 1, total = 9

Why: Check if cutting into 3 and 1 is better than current max 10.

Step 5: Choose the best price among all options

Max price for length 4 = 10 (cut into 2 + 2)

Why: We pick the highest total price found from all cuts.

Result:

Max price for rod length 4 = 10

Annotated Code

DSA C++

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

int rodCutting(const vector<int>& prices, int length) {
    if (length == 0) return 0; // no rod, no price
    int max_val = 0;
    for (int i = 1; i <= length; i++) {
        int current_val = prices[i - 1] + rodCutting(prices, length - i); // cut at i
        if (current_val > max_val) {
            max_val = current_val; // update max if better
        }
    }
    return max_val;
}

int main() {
    vector<int> prices = {1, 5, 8, 9};
    int length = 4;
    int max_price = rodCutting(prices, length);
    cout << "Max price for rod length " << length << " = " << max_price << endl;
    return 0;
}

if (length == 0) return 0; // no rod, no price

stop recursion when rod length is zero

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

try all possible first cut lengths

int current_val = prices[i - 1] + rodCutting(prices, length - i); // cut at i

calculate price for current cut plus best price for remaining rod

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

update max price if current cut is better

Output

Max price for rod length 4 = 10

Complexity Analysis

Time: O(2^n) because the function tries all cuts recursively without memoization

Space: O(n) due to recursion call stack up to rod length

vs Alternative: Dynamic programming approach reduces time to O(n^2) by storing intermediate results

Edge Cases

rod length = 0

returns 0 immediately, no cuts possible

DSA C++

if (length == 0) return 0; // no rod, no price

rod length = 1

returns price of length 1 piece directly

DSA C++

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

prices array empty or length mismatch

undefined behavior, assumes prices length matches rod length

DSA C++

none - input must be valid

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 pattern with recursion or dynamic programming.

Common Mistakes

Mistake: Not trying all possible cuts and only checking a few lengths
Fix: Use a loop from 1 to length to try every cut length

Mistake: Forgetting to add price of current cut piece to recursive result
Fix: Add prices[i-1] to recursive call result for remaining rod

Mistake: Not handling base case length == 0
Fix: Return 0 when length is zero to stop recursion

Summary

It finds the best way to cut a rod to get the highest total price from selling pieces.

Use it when you have a rod and prices for each piece length and want to maximize profit.

Try all cuts and combine prices recursively to find the best total price.

Practice

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

easy

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

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

C. To find the shortest rod length possible

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 maximum 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 A

4. Quick Check:

   Maximize profit = maximize total price [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 a rod of length 5 in C++?

easy

A. int price[4] = {2, 5, 7, 8, 10};

B. int price[] = {2, 5, 7, 8, 10};

C. int price[6] = {0, 2, 5, 7, 8, 10};

D. int price[5] = (2, 5, 7, 8, 10);

Solution

1. Step 1: Understand array declaration in C++

   When size is omitted, the compiler infers it from initializer list length.

2. Step 2: Check each option

   int price[4] = {2, 5, 7, 8, 10}; declares size 4 but initializes 5 elements - invalid syntax (too many initializers). int price[] = {2, 5, 7, 8, 10}; omits size, compiler infers size 5 - correct and flexible. int price[6] = {0, 2, 5, 7, 8, 10}; declares size 6 but initializes 6 elements including 0 at start - usually price array starts at index 0 or 1, but this is uncommon. int price[5] = (2, 5, 7, 8, 10); uses parentheses instead of braces - invalid syntax.

3. Final Answer:

   int price[] = {2, 5, 7, 8, 10}; -> Option B

4. Quick Check:

   Array size inferred from initializer [OK]

Hint: Use braces {} for array initialization, omit size for flexibility [OK]

Common Mistakes:

• Using parentheses instead of braces
• Mismatching array size and initializer count
• Including extra zero element unnecessarily

3. What is the output of the following C++ code snippet for rod length 4?

int price[] = {1, 5, 8, 9};
int n = 4;
int dp[5] = {0};
for (int i = 1; i <= n; i++) {
  int max_val = 0;
  for (int j = 0; j < i; j++) {
    max_val = std::max(max_val, price[j] + dp[i - j - 1]);
  }
  dp[i] = max_val;
}
std::cout << dp[n] << std::endl;

medium

A. 10

B. 9

C. 8

D. 11

Solution

1. Step 1: Trace dp array calculation for n=4

   Calculate dp[1] to dp[4]:
   dp[1] = max(price[0]+dp[0]) = 1+0=1
   dp[2] = max(price[0]+dp[1], price[1]+dp[0]) = max(1+1, 5+0)=5
   dp[3] = max(price[0]+dp[2], price[1]+dp[1], price[2]+dp[0]) = max(1+5, 5+1, 8+0)= max(6,6,8)=8
   dp[4] = max(price[0]+dp[3], price[1]+dp[2], price[2]+dp[1], price[3]+dp[0]) = max(1+8, 5+5, 8+1, 9+0) = max(9,10,9,9)=10

2. Step 2: Verify final dp value

   dp[4] = 10, so output is 10.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   dp[4] = 10 from max cuts [OK]

Hint: Calculate dp values stepwise from smaller lengths [OK]

Common Mistakes:

• Off-by-one errors in indexing
• Ignoring dp array values in calculation
• Confusing price array indexing

4. Identify the error in this C++ code snippet for rod cutting:

int price[] = {2, 5, 7, 8};
int n = 4;
int dp[4] = {0};
for (int i = 1; i <= n; i++) {
  int max_val = 0;
  for (int j = 0; j < i; j++) {
    max_val = std::max(max_val, price[j] + dp[i - j - 1]);
  }
  dp[i] = max_val;
}
std::cout << dp[n] << std::endl;

medium

A. std::max function is used incorrectly

B. Price array is missing an element for length 4

C. Loop variables i and j are swapped incorrectly

D. Array dp size is too small, causing out-of-bounds access

Solution

1. Step 1: Check dp array size and indexing

   dp is declared with size 4 but accessed at dp[i] where i goes up to 4, causing out-of-bounds access since valid indices are 0 to 3.

2. Step 2: Confirm other parts are correct

   Price array has 4 elements matching rod length. Loop variables and std::max usage are correct.

3. Final Answer:

   Array dp size is too small, causing out-of-bounds access -> Option D

4. Quick Check:

   dp size must be n+1 to avoid overflow [OK]

Hint: Declare dp array size as n+1 to avoid index errors [OK]

Common Mistakes:

• Using dp array size equal to n instead of n+1
• Ignoring array bounds in loops
• Misusing std::max parameters

5. Given a rod of length 5 and prices price = {2, 5, 7, 8, 10}, what is the maximum profit and the cuts to achieve it?

hard

A. Maximum profit 14 by cuts: 2 + 2 + 1

B. Maximum profit 13 by cuts: 1 + 4

C. Maximum profit 12 by cuts: 2 + 3

D. Maximum profit 15 by cuts: 1 + 1 + 3

Solution

1. Step 1: Calculate maximum profit using dp

   Calculate dp for length 5:
   dp[1]=2
   dp[2]=max(2+dp[0],5+dp[0])=5
   dp[3]=max(2+dp[1],5+dp[1],7+dp[0])=max(2+2,5+2,7)=7
   dp[4]=max(2+dp[2],5+dp[2],7+dp[1],8+dp[0])=max(2+5,5+5,7+2,8)=10
   dp[5]=max(2+dp[3],5+dp[3],7+dp[2],8+dp[1],10+dp[0])=max(2+7,5+7,7+5,8+2,10)=max(9,12,12,10,10)=12

2. Step 2: Check cuts for maximum profit

   Maximum profit 14 by cuts: 2 + 2 + 1 says profit 14 with cuts 2+2+1. Let's verify:
   Cut 2: price 5
   Cut 2: price 5
   Cut 1: price 2
   Total = 5+5+2=12, not 14.
   Check Maximum profit 15 by cuts: 1 + 1 + 3: 1+1+3 cuts:
   1:2,1:2,3:7 total=11
   Maximum profit 13 by cuts: 1 + 4: 1+4 cuts:
   1:2,4:8 total=10
   Maximum profit 12 by cuts: 2 + 3: 2+3 cuts:
   2:5,3:7 total=12
   So maximum profit is 12, cuts can be 2+3 or 2+2+1.
   Maximum profit 14 by cuts: 2 + 2 + 1 says 14 profit which is incorrect.
   Maximum profit 12 by cuts: 2 + 3 says 12 profit with cuts 2+3 which is correct.

3. Final Answer:

   Maximum profit 12 by cuts: 2 + 3 -> Option C

4. Quick Check:

   Max profit = 12 by cuts 2+3 [OK]

Hint: Try all cut combinations to find max profit [OK]

Common Mistakes:

• Adding prices incorrectly for cuts
• Assuming single cut gives max profit
• Ignoring multiple cut combinations