Bird
Raised Fist0
DSA Javaprogramming~5 mins

Rod Cutting Problem in DSA Java - Time & Space Complexity

Choose your learning style10 modes available
LearnWhyDeepVisualTryChallengeProjectRecallTime
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 changes as the rod length grows.

How does the number of steps increase when the rod gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int cutRod(int[] prices, int n) {
    if (n == 0) return 0;
    int maxVal = Integer.MIN_VALUE;
    for (int i = 1; i <= n; i++) {
        maxVal = Math.max(maxVal, prices[i - 1] + cutRod(prices, n - i));
    }
    return maxVal;
}

This code tries all ways to cut the rod and picks the best price by checking every possible first cut.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: The for-loop tries all first cut lengths from 1 to n.
  • How many times: For each call with length n, it makes up to n recursive calls for smaller lengths.
How Execution Grows With Input

Each rod length n calls itself for all smaller lengths, causing many repeated calculations.

Input Size (n)Approx. Operations
10More than 100 calls
20More than 1,000,000 calls
30More than 1,000,000,000 calls

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

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 code runs in linear time because it just loops from 1 to n."

[OK] Correct: Each loop iteration calls the function again with smaller lengths, causing many repeated calls, so the total work grows exponentially.

Interview Connect

Understanding this exponential growth helps you appreciate why dynamic programming is used to speed up such problems by remembering past results.

Self-Check

"What if we store results of smaller rods in an array to avoid repeated calls? How would the time complexity change?"

Practice

(1/5)
1. What is the main goal of the Rod Cutting Problem in algorithms?
easy
A. To find the shortest rod length possible
B. To maximize the total profit by cutting the rod into pieces
C. To minimize the number of cuts made on the rod
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 optimization 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, 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 rod lengths in Java?
easy
A. int[] price = {1, 5, 8, 9};
B. int price[] = (1, 5, 8, 9);
C. int price = {1, 5, 8, 9};
D. int[] price = (1, 5, 8, 9);

Solution

  1. Step 1: Recall Java array declaration syntax

    In Java, arrays are declared with square brackets and initialized with curly braces.

  2. Step 2: Check each option

    int[] price = {1, 5, 8, 9}; uses correct syntax: int[] price = {1, 5, 8, 9};. Others use parentheses or wrong assignment.

  3. Final Answer:

    int[] price = {1, 5, 8, 9}; -> Option A

  4. Quick Check:

    Correct Java array syntax = int[] price = {1, 5, 8, 9}; [OK]
Hint: Use curly braces {} to initialize arrays in Java [OK]
Common Mistakes:
  • Using parentheses () instead of curly braces {}
  • Missing square brackets [] in declaration
  • Assigning array to int variable
3. What will be the output of the following Java code snippet for rod length 4?
int[] price = {1, 5, 8, 9};
int n = 4;
int maxVal = 0;
for (int i = 1; i <= n; i++) {
    maxVal = Math.max(maxVal, price[i-1] + price[n - i]);
}
System.out.println(maxVal);
medium
A. 13
B. 9
C. 10
D. 8

Solution

  1. Step 1: Trace the loop iterations and calculations

    For n=4, loop i=1 to 4:
    i=1: maxVal = max(0, price[0]+price[3]) = max(0,1+9)=10
    i=2: maxVal = max(10, price[1]+price[2])=max(10,5+8)=13
    i=3: maxVal = max(13, price[2]+price[1])=max(13,8+5)=13
    i=4: maxVal = max(13, price[3]+price[0])=max(13,9+1)=13

  2. Step 2: Identify the final maxVal value

    After loop, maxVal is 13, but code prints maxVal after loop ends. The code prints 13.

  3. Final Answer:

    13 -> Option A

  4. Quick Check:

    Max profit for length 4 = 13 [OK]
Hint: Check all splits and sum prices to find max [OK]
Common Mistakes:
  • Not considering all splits i=1 to n
  • Off-by-one errors in array indexing
  • Printing maxVal before loop ends
4. Identify the error in this Java method for rod cutting:
public int cutRod(int[] price, int n) {
    if (n <= 0) return 0;
    int maxVal = Integer.MIN_VALUE;
    for (int i = 1; i < n; i++) {
        maxVal = Math.max(maxVal, price[i-1] + cutRod(price, n - i));
    }
    return maxVal;
}
medium
A. Recursive call parameters are incorrect
B. Base case should return 1 instead of 0
C. Loop should run from i = 1 to i <= n
D. maxVal should be initialized to 0 instead of Integer.MIN_VALUE

Solution

  1. Step 1: Analyze the loop range

    The loop runs from i=1 to i<n, so it misses the case i=n which is needed to consider the whole rod as one piece.

  2. Step 2: Correct the loop condition

    The loop should run from i=1 to i<=n to include all possible cuts including the full length.

  3. Final Answer:

    Loop should run from i = 1 to i <= n -> Option C

  4. Quick Check:

    Include full length cut by looping to i <= n [OK]
Hint: Loop must include full rod length i = n [OK]
Common Mistakes:
  • Using i < n instead of i <= n
  • Incorrect base case return value
  • Wrong initialization of maxVal
5. Given a rod of length 5 and price array int[] price = {2, 5, 7, 8, 10};, what is the maximum profit using dynamic programming approach?
hard
A. 14
B. 10
C. 15
D. 12

Solution

  1. Step 1: Calculate max profit for each length up to 5

    Length 1: max = 2
    Length 2: max = max(5, 2+2=4) = 5
    Length 3: max = max(7, 5+2=7, 2+5=7) = 7
    Length 4: max = max(8, 7+2=9, 5+5=10, 2+7=9) = 10
    Length 5: max = max(10, 10+2=12, 7+5=12, 5+7=12, 2+10=12) = 12

  2. Step 2: Check if cutting into smaller pieces yields better profit

    Try cutting length 5 into 1+4, 2+3, 3+2, 4+1 etc. The best is 2+3 or 3+2 or 1+4 giving 12, but also check if full length price 10 is less than these.

  3. Step 3: Recalculate carefully

    Check all splits:
    1+4: 2 + 10 = 12
    2+3: 5 + 7 = 12
    3+2: 7 + 5 = 12
    4+1: 8 + 2 = 10
    But also check if cutting into 1+1+3 or 1+1+1+2 etc. can yield more.
    Try 1+1+3: 2+2+7=11
    Try 1+1+1+2: 2+2+2+5=11
    Try 1+1+1+1+1: 2*5=10
    So max is 12.

  4. Final Answer:

    12 -> Option D

  5. Quick Check:

    Max profit for length 5 = 12 [OK]
Hint: Check all splits and pick max sum for length 5 [OK]
Common Mistakes:
  • Choosing price of full length only
  • Not checking all split combinations
  • Confusing price array indices
Recommended

Ready to prove it?

Pass the test and this pattern moves unknown strong while earning +1 Grip toward your certificate.

10 questions~5 mineasy hard