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, not just sell the whole bar as one piece.

Rod: [1][2][3][4][5][6][7][8]
Prices: p1 p2 p3 p4 p5 p6 p7 p8
Goal: Maximize sum of prices by cutting rod into pieces

Dry Run Walkthrough

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

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

Step 1: Check price if we sell rod as whole length 4

Max price for length 4 = 9 (no cuts)

Why: Start with no cuts to have a baseline price

Step 2: Try cutting rod into length 1 + length 3

Price = price[1] + max price for length 3 = 1 + 8 = 9

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

Step 3: Try cutting rod into length 2 + length 2

Price = price[2] + max price for length 2 = 5 + 5 = 10

Why: Cutting into two pieces of length 2 gives better price

Step 4: Try cutting rod into length 3 + length 1

Price = price[3] + max price for length 1 = 8 + 1 = 9

Why: Check other cuts to confirm max price

Step 5: Choose max price among all options

Max price = 10 by cutting into two pieces of length 2

Why: This is the best way to cut rod length 4

Result:

Max price for rod length 4 = 10

Annotated Code

DSA C

#include <stdio.h>

int max(int a, int b) {
    return (a > b) ? a : b;
}

int rodCutting(int prices[], int n) {
    int dp[n + 1];
    dp[0] = 0; // max price for length 0 is 0

    for (int i = 1; i <= n; i++) {
        dp[i] = 0;
        for (int j = 1; j <= i; j++) {
            dp[i] = max(dp[i], prices[j - 1] + dp[i - j]);
        }
    }
    return dp[n];
}

int main() {
    int prices[] = {1, 5, 8, 9};
    int length = 4;
    int maxPrice = rodCutting(prices, length);
    printf("Max price for rod length %d = %d\n", length, maxPrice);
    return 0;
}

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

Calculate max price for each rod length from 1 to n

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

Try all possible first cut lengths j for rod length i

dp[i] = max(dp[i], prices[j - 1] + dp[i - j]);

Update max price for length i by comparing current max and price of first cut plus best for remaining length

Output

Max price for rod length 4 = 10

Complexity Analysis

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

Space: O(n) because we store max prices for all lengths up to n in dp array

vs Alternative: Naive recursive approach has exponential time due to repeated calculations; DP avoids this by storing results

Edge Cases

rod length 0

max price is 0 since no rod to cut

DSA C

dp[0] = 0; // max price for length 0 is 0

rod length 1

returns price of length 1 piece directly

DSA C

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

all prices zero

max price will be 0 regardless of cuts

DSA C

dp[i] = max(dp[i], prices[j - 1] + dp[i - j]);

When to Use This Pattern

When you see a problem about cutting or splitting to maximize value, think of the Rod Cutting pattern which uses dynamic programming to try all cuts efficiently.

Common Mistakes

Mistake: Not storing intermediate results and recalculating subproblems repeatedly
Fix: Use a dp array to store max prices for all lengths to avoid repeated work

Mistake: Using wrong indices for prices array causing off-by-one errors
Fix: Remember prices[j-1] corresponds to piece length j

Summary

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

Use when you want to optimize profit by splitting a resource into parts with known values.

Try all cuts and store best results for smaller lengths to build up solution efficiently.

Practice

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

easy

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

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

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 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 = maximize total price [OK]

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

Common Mistakes:

• Thinking the goal is to minimize cuts
• Confusing the problem with counting combinations
• Assuming the rod length changes

2. Which of the following is the correct way to declare the price array for a rod of length 4 in C?

easy

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

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

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

D. int price[4] = (2, 5, 7, 8);

Solution

1. Step 1: Understand array indexing for rod prices

   Prices are usually stored with index representing length, so for length 1 to 4, index 0 is unused or zero.

2. Step 2: Check array declaration correctness

   int price[5] = {0, 2, 5, 7, 8}; declares array of size 5 with price[0]=0, then prices for lengths 1 to 4, which is correct.

3. Final Answer:

   int price[5] = {0, 2, 5, 7, 8}; -> Option C

4. Quick Check:

   Index 0 unused, prices start at 1 [OK]

Hint: Use size n+1 with zero at index 0 for convenience [OK]

Common Mistakes:

• Using parentheses instead of braces for array initialization
• Declaring array size smaller than needed
• Not including zero price at index 0

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

int price[5] = {0, 2, 5, 7, 8};
int n = 4;
int dp[5] = {0};
for (int i = 1; i <= n; i++) {
  int max_val = 0;
  for (int j = 1; j <= i; j++) {
    int val = price[j] + dp[i - j];
    if (val > max_val) max_val = val;
  }
  dp[i] = max_val;
}
printf("%d", dp[n]);

medium

A. 10

B. 9

C. 8

D. 7

Solution

1. Step 1: Trace dp array computation for each length

   Calculate dp[1] to dp[4]: dp[1] = max(price[1]+dp[0]) = 2 dp[2] = max(price[1]+dp[1], price[2]+dp[0]) = max(2+2,5+0)=5 dp[3] = max(price[1]+dp[2], price[2]+dp[1], price[3]+dp[0]) = max(2+5,5+2,7+0)=7 dp[4] = max(price[1]+dp[3], price[2]+dp[2], price[3]+dp[1], price[4]+dp[0]) = max(2+7,5+5,7+2,8+0)=10

2. Step 2: Identify final dp[n] value

   dp[4] = 10 is printed.

3. Final Answer:

   10 -> Option A

4. Quick Check:

   Max profit for length 4 = 10 [OK]

Hint: Calculate dp bottom-up, check all cuts for max profit [OK]

Common Mistakes:

• Forgetting to consider all cut lengths j
• Using wrong indices for price or dp
• Not initializing dp array properly

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

int price[5] = {0, 2, 5, 7, 8};
int n = 4;
int dp[5] = {0};
for (int i = 1; i <= n; i++) {
  int max_val = 0;
  for (int j = 1; j < i; j++) {
    int val = price[j] + dp[i - j];
    if (val > max_val) max_val = val;
  }
  dp[i] = max_val;
}
printf("%d", dp[n]);

medium

A. max_val should be initialized to -1

B. dp array size is too small

C. price array initialization is incorrect

D. Inner loop should run till j <= i, not j < i

Solution

1. Step 1: Analyze inner loop range

   The inner loop runs from j=1 to j

2. Step 2: Correct the loop condition

   Changing inner loop to j <= i ensures all possible cuts including full length are considered.

3. Final Answer:

   Inner loop should run till j <= i, not j < i -> Option D

4. Quick Check:

   Include full length cut by using j <= i [OK]

Hint: Inner loop must include full length cut (j <= i) [OK]

Common Mistakes:

• Using j < i misses full rod cut
• Incorrect dp array size causes errors
• Initializing max_val to 0 is correct here

5. Given a rod of length 5 and prices price = {0, 2, 5, 7, 8, 10}, what is the maximum profit and the cuts to achieve it?
Use the rod cutting logic to find the best cuts and profit.

hard

A. Profit = 12, Cuts = [2, 3]

B. Profit = 11, Cuts = [1, 4]

C. Profit = 10, Cuts = [5]

D. Profit = 13, Cuts = [1, 1, 3]

Solution

1. Step 1: Calculate maximum profit using dp

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

2. Step 2: Determine cuts for maximum profit

   Profit 12 achieved by cutting rod into lengths 2 and 3 (price[2]+price[3]=5+7=12).

3. Final Answer:

   Profit = 12, Cuts = [2, 3] -> Option A

4. Quick Check:

   Max profit 12 with cuts 2 and 3 [OK]

Hint: Check all cut pairs for max sum profit [OK]

Common Mistakes:

• Choosing no cuts or single cut only
• Not checking all combinations for max profit
• Confusing profit values with cut lengths