Rod Cutting Problem in DSA C - Execution Trace

Concept Flow - Rod Cutting Problem

Start with full rod length N

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Try all cut lengths i from 1 to N

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Calculate price for cut length i + best price for remaining length N-i

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Store max price for length N

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Repeat for all lengths from 1 to N

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Result: max price for full rod length

We try all possible first cuts, combine their prices with the best prices for the leftover rod, and store the maximum price for each rod length.

Execution Sample

DSA C

int rodCut(int length, int prices[]) {
  int dp[length+1];
  dp[0] = 0;
  for (int i=1; i<=length; i++) {
    int max_val = 0;
    for (int j=1; j<=i; j++) {
      max_val = (max_val > prices[j-1] + dp[i-j] ? max_val : prices[j-1] + dp[i-j]);
    }
    dp[i] = max_val;
  }
  return dp[length];
}

This code finds the maximum price for a rod of given length by trying all cuts and using dynamic programming to store best prices.

Execution Table

Step	Operation	Current Length i	Cut Length j	Calculation	Max Price for i	dp Array State	Visual State
1	Initialize dp[0]	0	-	-	0	[0]	[]
2	Calculate max price for length 1	1	1	prices[0] + dp[0] = 1 + 0 = 1	1	[0,1]	[1]
3	Calculate max price for length 2	2	1	prices[0] + dp[1] = 1 + 1 = 2	2	[0,1,2]	[2]
4	Calculate max price for length 2	2	2	prices[1] + dp[0] = 5 + 0 = 5	5	[0,1,5]	[5]
5	Calculate max price for length 3	3	1	prices[0] + dp[2] = 1 + 5 = 6	6	[0,1,5,6]	[6]
6	Calculate max price for length 3	3	2	prices[1] + dp[1] = 5 + 1 = 6	6	[0,1,5,6]	[6]
7	Calculate max price for length 3	3	3	prices[2] + dp[0] = 8 + 0 = 8	8	[0,1,5,8]	[8]
8	Calculate max price for length 4	4	1	prices[0] + dp[3] = 1 + 8 = 9	9	[0,1,5,8,9]	[9]
9	Calculate max price for length 4	4	2	prices[1] + dp[2] = 5 + 5 = 10	10	[0,1,5,8,10]	[10]
10	Calculate max price for length 4	4	3	prices[2] + dp[1] = 8 + 1 = 9	10	[0,1,5,8,10]	[10]
11	Calculate max price for length 4	4	4	prices[3] + dp[0] = 9 + 0 = 9	10	[0,1,5,8,10]	[10]
12	Calculate max price for length 5	5	1	prices[0] + dp[4] = 1 + 10 = 11	11	[0,1,5,8,10,11]	[11]
13	Calculate max price for length 5	5	2	prices[1] + dp[3] = 5 + 8 = 13	13	[0,1,5,8,10,13]	[13]
14	Calculate max price for length 5	5	3	prices[2] + dp[2] = 8 + 5 = 13	13	[0,1,5,8,10,13]	[13]
15	Calculate max price for length 5	5	4	prices[3] + dp[1] = 9 + 1 = 10	13	[0,1,5,8,10,13]	[13]
16	Calculate max price for length 5	5	5	prices[4] + dp[0] = 10 + 0 = 10	13	[0,1,5,8,10,13]	[13]
17	Return max price for full length	5	-	-	13	[0,1,5,8,10,13]	[13]

💡 All lengths from 1 to 5 processed, dp array fully computed, maximum price for rod length 5 is 13

Variable Tracker

Variable	Start	After Step 2	After Step 4	After Step 7	After Step 11	After Step 16	Final
dp	[0]	[0,1]	[0,1,5]	[0,1,5,8]	[0,1,5,8,10]	[0,1,5,8,10,13]	[0,1,5,8,10,13]
max_val	0	1	5	8	10	13	13
i (current length)	0	1	2	3	4	5	5
j (cut length)	-	1	2	3	4	5	-

Key Moments - 3 Insights

Why do we check all cut lengths j from 1 to i for each rod length i?

Because the best price for length i might come from cutting the rod into different pieces, not just one piece. The execution_table rows 3-16 show how each cut length j is tried to find the maximum price.

Why do we add prices[j-1] + dp[i-j] in the calculation?

prices[j-1] is the price for the first cut of length j, and dp[i-j] is the best price for the remaining rod length i-j. Adding them gives the total price for that cut combination, as shown in the Calculation column in execution_table.

Why does dp[0] start at 0?

dp[0] represents the price for a rod of length 0, which is zero because no rod means no price. This base case allows building up solutions for larger lengths, as seen in Step 1 of execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at Step 7, what is the max price for rod length 3?

A 8

B 5

C 6

D 9

💡 Hint

Check the Max Price for i column at Step 7 in execution_table

At which step does the dp array first contain the value 10?

A Step 11

B Step 9

C Step 10

D Step 8

💡 Hint

Look at the dp Array State column in execution_table rows around Step 9 and Step 11

If prices[1] changed from 5 to 7, how would max price for rod length 2 change at Step 4?

A It would remain 5

B It would become 2

C It would become 7

D It would become 12

💡 Hint

Look at Calculation in Step 4: prices[1] + dp[0], changing prices[1] affects max price for length 2

Concept Snapshot

Rod Cutting Problem:
- Goal: Maximize price by cutting rod into pieces
- Use dp array where dp[i] = max price for length i
- For each length i, try all cuts j (1 to i)
- dp[i] = max(prices[j-1] + dp[i-j])
- dp[0] = 0 base case
- Result: dp[length] is max price

Full Transcript

The Rod Cutting Problem finds the best way to cut a rod to get the highest price. We use a dp array where each position stores the best price for that rod length. Starting from length 0 with price 0, we build up to the full rod length. For each length i, we try all possible first cuts j and calculate the total price as prices[j-1] plus the best price for the leftover rod dp[i-j]. We keep the maximum of these values in dp[i]. After filling dp for all lengths, dp[length] gives the maximum price. The execution table shows each step, the calculations, and how dp array changes. Key points include trying all cuts, adding prices and dp values, and starting dp[0] at zero. The visual quiz tests understanding of dp values at specific steps and effects of price changes.