dp[i] represents the maximum revenue obtainable from a rod of length i. For each length, we try all possible first cuts and take the maximum.

1. Create dp array of size n + 1, initialized to 0.
2. For each length i from 1 to n:
   • For each cut length j from 1 to i:
     • dp[i] = max(dp[i], price[j-1] + dp[i-j]).
3. Return dp[n].

• Time Complexity: O(N^2).
• Space Complexity: O(N).

def rod_cutting(price, n):
    dp = [0] * (n + 1)
    
    for i in range(1, n + 1):
        for j in range(1, i + 1):
            dp[i] = max(dp[i], price[j - 1] + dp[i - j])
    
    return dp[n]