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.