Rod Cutting Problem in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to return the maximum price for a rod of length n.

DSA C

int cutRod(int price[], int n) {
    if (n <= 0) return 0;
    int max_val = [1];
    for (int i = 1; i <= n; i++) {
        int val = price[i - 1] + cutRod(price, n - i);
        if (val > max_val) max_val = val;
    }
    return max_val;
}

Drag options to blanks, or click blank then click option'

A1

B-1

C0

DINT_MIN

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing max_val to 0 causes incorrect results when all prices are negative.

Using 1 or -1 does not correctly represent the smallest integer.

We initialize max_val with INT_MIN to ensure any computed value will be larger, allowing correct maximum calculation.

2fill in blank

medium

Complete the code to implement the bottom-up dynamic programming approach for rod cutting.

DSA C

int cutRodDP(int price[], int n) {
    int val[n+1];
    val[0] = 0;
    for (int i = 1; i <= n; i++) {
        int max_val = [1];
        for (int j = 1; j <= i; j++) {
            max_val = max(max_val, price[j-1] + val[i-j]);
        }
        val[i] = max_val;
    }
    return val[n];
}

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A0

B-1

CINT_MIN

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using INT_MIN causes unnecessary complexity in bottom-up approach.

Starting max_val at -1 or 1 can cause incorrect maximum calculations.

In bottom-up DP, max_val starts at 0 because prices are non-negative and we want to find the maximum.

3fill in blank

hard

Fix the error in the recursive rod cutting function to avoid infinite recursion.

DSA C

int cutRod(int price[], int n) {
    if (n == 0) return 0;
    int max_val = INT_MIN;
    for (int i = 1; i <= n; i++) {
        int val = price[i - 1] + cutRod(price, [1]);
        if (val > max_val) max_val = val;
    }
    return max_val;
}

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An - i

Bn + i

Ci

Dn

Attempts:

3 left

💡 Hint

Common Mistakes

Using n + i causes infinite recursion.

Using n or i directly does not reduce problem size correctly.

We reduce the rod length by i to avoid infinite recursion and correctly compute subproblems.

4fill in blank

hard

Fill both blanks to create a dictionary comprehension that maps rod lengths to their maximum prices for lengths greater than 0.

DSA C

int maxPrices[[1]];
for (int i = 1; i <= [2]; i++) {
    maxPrices[i-1] = cutRodDP(price, i);
}

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An

Bn+1

Cn-1

Dn*2

Attempts:

3 left

💡 Hint

Common Mistakes

Using n causes out-of-bound errors for length n.

Using n-1 misses the last rod length.

We use n+1 to include length n in the loop and array size to store max prices for all lengths up to n.

5fill in blank

hard

Fill all three blanks to complete the memoized rod cutting function.

DSA C

int memoCutRod(int price[], int n, int memo[]) {
    if (n == 0) return 0;
    if (memo[[1]] >= 0) return memo[[2]];
    int max_val = INT_MIN;
    for (int i = 1; i <= n; i++) {
        int val = price[i - 1] + memoCutRod(price, n - i, memo);
        if (val > max_val) max_val = val;
    }
    memo[[3]] = max_val;
    return max_val;
}

Drag options to blanks, or click blank then click option'

An

Bn-1

C0

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n causes out-of-bound errors.

Using 0 or 1 does not correspond to current rod length.

We use n-1 as index because array indices start at 0 for rod length n.

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