Rod Cutting Problem in DSA C++ - Time & Space Complexity

Time Complexity: Rod Cutting Problem

O(2^n)

Understanding Time Complexity

We want to understand how the time needed to solve the rod cutting problem grows as the rod length increases.

Specifically, how does the number of steps change when we try all ways to cut the rod?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int rodCutting(int price[], int n) {
    if (n == 0) return 0;
    int max_val = INT_MIN;
    for (int i = 1; i <= n; i++) {
        max_val = std::max(max_val, price[i-1] + rodCutting(price, n - i));
    }
    return max_val;
}
    

This code tries all ways to cut a rod of length n to find the maximum price by recursively checking all smaller lengths.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop runs from 1 to n, and inside it calls the function recursively for smaller lengths.
• How many times: Each call makes up to n recursive calls, leading to many repeated calls for smaller lengths.

How Execution Grows With Input

Each rod length n calls itself for all smaller lengths, causing the number of calls to grow very fast.

Input Size (n)	Approx. Operations
10	Thousands of calls
100	More than billions of calls
1000	Extremely large, practically impossible to finish quickly

Pattern observation: The number of operations grows very fast, much faster than just doubling or squaring.

Final Time Complexity

Time Complexity: O(2^n)

This means the time needed roughly doubles with each increase in rod length, making it very slow for large n.

Common Mistake

[X] Wrong: "The recursive solution runs in linear time because it just loops from 1 to n."

[OK] Correct: Each loop iteration calls the function again, causing many repeated calls and exponential growth, not just a single loop.

Interview Connect

Understanding this problem helps you see why simple recursion can be slow and why techniques like memoization or dynamic programming are important to improve efficiency.

Self-Check

"What if we store results of smaller problems to avoid repeated calls? How would the time complexity change?"

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