Rod Cutting

Last Updated : 12 Aug, 2024

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Rod Cutting

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Arrays Dynamic Programming

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Medium

60.66%

1.4L

Given a rod of length n inches and an array of prices that includes prices of all pieces of size smaller than n. Determine the maximum value obtainable by cutting up the rod and selling the pieces. We are mainly given a price array such that the size of the price array is same as rod length and price[i] represents price of length i+1

Input : price[] = { 1 , 5, 8, 9, 10, 17 , 17 . 20}
Output : 22
Explanation : Rod length is 8 and the values of different pieces are given as the following,
length | 1 2 3 4 5 6 7 8
——————————
price | 1 5 8 9 10 17 17 20
The maximum obtainable value is 22 (by cutting in two pieces of lengths 2 and 6)

Input : price[] = { 3 , 5, 8, 9, 10, 17}
Output : 18
Explanation : Rod length is 7 And if the prices are as follows.
length | 1 2 3 4 5 6
————————
price | 3 4 6 7 9 10
The maximum obtainable value is 18 (by making 6 pieces each of length 1)

Input : price[] = { 3 }
Output : 3
Explanation: There is only 1 way to pick a piece of length 1

Table of Content

Naive Recursive Solution:

We generate all configurations of different pieces and find the highest-priced configuration. This solution is exponential in terms of time complexity. Let us see how this problem possesses both important properties of a Dynamic Programming (DP) Problem and can efficiently be solved using Dynamic Programming.

1) Optimal Substructure:

We can get the best price by making a cut at different positions and comparing the values obtained after a cut. We can recursively call the same function for a piece obtained after a cut.

Let cutRod(n) be the required (best possible price) value for a rod of length n. It can be written as follows.
cutRod(n) = max(price[i] + cutRod(n-i-1)) for all i in {0, 1 .. n-1}

C++

// A recursive solution for Rod cutting problem
#include <bits/stdc++.h>
using namespace std;

/* Returns the best obtainable price for a rod of length n
   and price[] as prices of different pieces */
int cutRod(int price[], int n)
{
    // if n is 0 we cannot cut the rod anymore.
    if (n ==0) {
       return 0;}
  
    int res = 0;
    for (int i=0; i<n; i++) 
       res = max(res, price[i] + cutRod(price, n-i-1));
  
    return res;
}

/* Driver program to test above functions */
int main()
{
    int arr[] = { 0, 5, 8, 9, 10, 17, 17, 20 };
    int size = sizeof(arr) / sizeof(arr[0]);
    cout << cutRod(arr, size);
    return 0;
}

C

#include <stdio.h>
#include <limits.h>

int cutRod(int price[], int idx, int n) {
  
    // Base case: if index is 0, then the only option
    // is to take multiple pieces of length 1
    if (idx == 0) return n * price[0];
  
    // Base case: if the rod length is 0, no
    // more cuts can be made
    if (n == 0) return 0;

    // Option 1: Do not cut the rod at this index
    int notCut = cutRod(price, idx - 1, n);
    int cut = INT_MIN;
    int rodLen = idx + 1;

    // Option 2: Cut the rod if the current 
    // length is within bounds
    if (rodLen <= n)
        cut = price[idx] + cutRod(price, idx, n - rodLen);

    // Return the maximum value between
    // cutting and not cutting
    return (notCut > cut) ? notCut : cut;
}

int main() {
    int price[] = {1, 5, 8, 9, 10, 17, 17, 20};
    int size = sizeof(price) / sizeof(price[0]);
    printf("%d\n", cutRod(price, size - 1, size));
    return 0;
}

Java

// Java recursive solution for Rod cutting problem
import java.io.*;
class GFG {

    /* Returns the best obtainable price for a rod of length
    n and price[] as prices of different pieces */
    static int cutRod(int price[], int index, int n)
    {
        // base case
      // if index ==0 that means we can get n number of pieces of unit length with price price[0]
        if (index == 0) {
            return n * price[0];
        }
      // if n, the size of the rod is 0 we cannot cut the rod anymore.
      if (n ==0) {
        return 0;
      }
      
        // At any index we have 2 options either
        // cut the rod of this length or not cut
        // it
        int notCut = cutRod(price, index - 1, n);
        int cut = Integer.MIN_VALUE;
        int rod_length = index + 1;
        int new_length = n - rod_length;
      
      // rod can be cut only if current rod length is than actual length of the rod
      
        if (rod_length <= n)
        //next cutting index can only be new length - 1
            cut = price[index]
                  + cutRod(price, new_length-1, new_length);

        return Math.max(notCut, cut);
    }

    /* Driver program to test above functions */
    public static void main(String args[])
    {
        int arr[] = { 1, 5, 8, 9, 10, 17, 17, 20};
        int size = arr.length;
        System.out.println("Maximum Obtainable Value is "
                           + cutRod(arr, size - 1, size));
    }
}

// This code is contributed by saikrishna padamatinti

Python

#  A recursive solution for Rod cutting problem

#  Returns the best obtainable price for a rod of length n
#   and price[] as prices of different pieces 
def cutRod(price, index, n):
    
    #  base case
    if index == 0:
        return n*price[0]
      
      #v if n is 0 we cannot cut the rod anymore.
    if (n ==0):
        return 0
    
    #   At any index we have 2 options either
    #   cut the rod of this length or not cut 
    #   it
    notCut = cutRod(price,index - 1,n)
    cut = float("-inf")
    rod_length = index + 1

    if (rod_length <= n):
        
        cut = price[index]+cutRod(price,index,n - rod_length)
  
    return max(notCut, cut)

#  Driver program to test above functions 
arr = [ 1, 5, 8, 9, 10, 17, 17, 20 ]
size = len(arr)
print("Maximum Obtainable Value is ",cutRod(arr, size - 1, size))

# This code is contributed by Vivek Maddeshiya

C#

// C# recursive solution for Rod cutting problem

using System;

public class GFG {
  static int max(int a, int b) {
    if (a > b) return a;
    return b;
  }

  static int MIN_VALUE = -1000000000;

  /* Returns the best obtainable price for a rod of 
  // length n and price[] as prices of different pieces */
  static int cutRod(int[] price, int index, int n) {
    // base case
    if (index == 0) {
      return n * price[0];
    }
    // if n is 0 we cannot cut the rod anymore.
      if (n ==0) {
        return 0;
      }
    
    // At any index we have 2 options either 
    // cut the rod of this length or not cut it
    int notCut = cutRod(price, index - 1, n);
    int cut = MIN_VALUE;
    int rod_length = index + 1;

    if (rod_length <= n)
      cut = price[index] + cutRod(price, index, n - rod_length);

    return max(notCut, cut);
  }

  // Driver program to test above functions
  public static void Main(string[] args) {
    int[] arr = {1, 5, 8, 9, 10, 17, 17, 20 };
    int size = arr.Length;
    Console.WriteLine("Maximum Obtainable Value is " + 
                      cutRod(arr, size - 1, size));
  }
}

// This code is contributed by ajaymakavana.

JavaScript

// A recursive solution for Rod cutting problem

/* Returns the best obtainable price for a rod of length n
   and price[] as prices of different pieces */
function cutRod(price,  index,  n)
{
    // base case
    if (index == 0) {
        return n * price[0];
    }
    // if n is 0 we cannot cut the rod anymore.
      if (n ==0) {
        return 0;
        }
    //At any index we have 2 options either
      //cut the rod of this length or not cut 
      //it
    let notCut = cutRod(price,index - 1,n);
    let cut = Number.MIN_VALUE;
    let rod_length = index + 1;

    if (rod_length <= n)
        cut = price[index]
               + cutRod(price,index,n - rod_length);
  
    return Math.max(notCut, cut);
}

/* Driver program to test above functions */

    let arr = [ 1, 5, 8, 9, 10, 17, 17, 20 ];
    let size = arr.length;
    console.log("Maximum Obtainable Value is "
         + cutRod(arr, size - 1, size));
   
   // This code is contributed by garg28harsh.

Output

Time Complexity: Exponential
Space Complexity: O(n) where n is the length of the price array.

2) Overlapping Subproblems

Considering the above implementation, the following is the recursion tree for a Rod of length 4.

cR() ---> cutRod() 
                            cR(4)
                  /        /           
                 /        /              
             cR(3)       cR(2)     cR(1)   cR(0)
            /  |         /         |
           /   |        /          |  
      cR(2) cR(1) cR(0) cR(1) cR(0) cR(0)
     /        |          |
    /         |          |   
  cR(1) cR(0) cR(0)      cR(0)
   /
 /
CR(0)

In the above partial recursion tree, cR(2) is solved twice. We can see that there are many subproblems that are solved again and again. Since the same subproblems are called again, this problem has the Overlapping Subproblems property. So the Rod Cutting problem has both properties (see this and this) of a dynamic programming problem. Like other typical Dynamic Programming(DP) problems, recomputations of the same subproblems can be avoided by either using memoization or tabulation.

Memoization (or Top-Down) Solution:

We use 1D memo array. The range of values change from 0 to n, we make an array of size n+1.
We initialize the memo array as -1 to indicate that all values are not computed initially.
Before we make a recursive call, we check if the value is already computed by checking for -1 in memo. After we compute values, we store them in memo.

C++

#include <bits/stdc++.h>
using namespace std;

/* The memo array is used to check if a solution
   is already computed or not */
int cutRodUtil(int price[], int n, vector<int>& memo) {
  
    // if n is 0, we cannot cut the rod anymore.
    if (n == 0) return 0;

    // If the result is already computed, return it.
    if (memo[n] != -1) return memo[n];

    int res = 0;
    for (int i = 0; i < n; i++) 
        res = max(res, price[i] + 
                cutRodUtil(price, n - i - 1, memo));

    // Store the computed result in memoization 
    // table and return it.
    return memo[n] = res;
}

int cutRod(int price[], int n) {
  
    // Initialize memoization table with
    // -1 to indicate uncomputed states.
    vector<int> memo(n + 1, -1);
  
    return cutRodUtil(price, n, memo);
}

/* Driver program to test above functions */
int main() {
    int arr[] = { 0, 5, 8, 9, 10, 17, 17, 20 };
    int size = sizeof(arr) / sizeof(arr[0]);
    cout << cutRod(arr, size);
    return 0;
}

Java

// A memoization solution for Rod cutting problem
import java.io.*;
import java.util.*;

/* Returns the best obtainable price for a rod of length n
and price[] as prices of different pieces */
class GFG {
  private static int cutRod(int price[], int index, int n,
                            int[][] dp)
  {
    
    // base case
    if (index == 0) {
      return n * price[0];
    }

    if (dp[index][n] != -1) {
      return dp[index][n];
    }

    // At any index we have 2 options either
    // cut the rod of this length or not cut
    // it
    int notCut = cutRod(price, index - 1, n, dp);
    int cut = Integer.MIN_VALUE;
    int rod_length = index + 1;

    if (rod_length <= n) {
      cut = price[index]
        + cutRod(price, index, n - rod_length,
                 dp);
    }

    return dp[index][n] = Math.max(cut, notCut);
  }

  /* Driver program to test above functions */
  public static void main(String[] args)
  {
    int arr[] = { 1, 5, 8, 9, 10, 17, 17, 20 };
    int size = arr.length;
    int dp[][] = new int[size][size + 1];
    for (int i = 0; i < size; i++) {
      Arrays.fill(dp[i], -1);
    }
    System.out.println(
      "Maximum Obtainable Value is "
      + cutRod(arr, size - 1, size, dp));
  }
}

// This code is contributed by Snigdha Patil

Python

# A memoization solution for Rod cutting problem

 # Returns the best obtainable price for 
  # a rod of length n and price[] as 
  # prices of different pieces
def cutRoad(price,index,n,dp):
  
    # base case
    if(index == 0):
        return n*price[0]
    if(dp[index][n] != -1):
        return dp[index][n]
      
    # At any index we have 2 options either 
    # cut the rod of this length or not cut it
    notCut = cutRoad(price,index-1,n,dp)
    cut = -5
    rod_length = index + 1
    if(rod_length <= n):
        cut = price[index] + cutRoad(price,index,n-rod_length,dp)
    dp[index][n] = max(notCut,cut)
    return dp[index][n]

# Driver program to test above functions
if __name__ == "__main__":
    arr = [1,5,8,9,10,17,17,20]
    size = len(arr)
    dp = []
    temp = []
    for i in range(0,size+1):
        temp.append(-1)
    for i in range(0,size):
        dp.append(temp)
    # print(dp)
    print("Maximum Obtainable Value is :",end=' ')
    print(cutRoad(arr,size-1,size,dp))

C#

// A memoization solution for Rod cutting problem
using System;

public class HelloWorld 
{
  
  // Returns the best obtainable price for 
  // a rod of length n and price[] as prices of different pieces
  private static int cutRod(int[] price, int index, 
                            int n, int[,] dp) 
  {

    // base case
    if (index == 0) {
      return n * price[0];
    }

    if (dp[index, n] != -1) {
      return dp[index, n];
    }

    // At any index we have 2 options either cut 
    // the rod of this length or not cut it
    int notCut = cutRod(price, index - 1, n, dp);
    int cut = Int32.MinValue;
    int rod_length = index + 1;

    if (rod_length <= n) {
      cut = price[index] + cutRod(price, index, n - rod_length, dp);
    }

    return dp[index, n] = Math.Max(cut, notCut);
  }

  //  Driver program to test above functions
  public static void Main(string[] args) {
    int[] arr = { 1, 5, 8, 9, 10, 17, 17, 20 };
    int size = arr.Length;
    int[,] dp = new int[size, size + 1];
    for (int i = 0; i < size; i++) {
      for (int j = 0; j < size + 1; j++) {
        dp[i, j] = -1;
      }
    }
    Console.WriteLine("Maximum Obtainable Value is " + 
                      cutRod(arr, size - 1, size, dp));
  }
}

// This code is contributed by ajaymakvana.

JavaScript

// A memoization solution for Rod cutting problem

// A utility function to get the maximum of two integers
function max(a, b) {
    return (a > b) ? a : b;
}

/* Returns the best obtainable price for a rod of length n
and price[] as prices of different pieces */
function cutRod(price, index, n, dp) {
  // base case
  if (index === 0) {
      return n * price[0];
  }
  if (dp[index][n] !== -1)
      return dp[index][n];

  // At any index we have 2 options either
  // cut the rod of this length or not cut
  // it
  let notCut = cutRod(price, index - 1, n, dp);
  let cut = Number.MIN_SAFE_INTEGER;
  let rod_length = index + 1;

  if (rod_length <= n)
      cut = price[index] + cutRod(price, index, n - rod_length, dp);

  return dp[index][n] = max(notCut, cut);
}

/* Driver program to test above functions */
function main() {
  let arr = [ 1, 5, 8, 9, 10, 17, 17, 20 ];
  let size = arr.length;
  let dp = new Array(size);
  for (let i = 0; i < size; i++) {
      dp[i] = new Array(size + 1).fill(-1);
  }
  console.log("Maximum Obtainable Value is " + cutRod(arr, size - 1, size, dp));
  return 0;
}

main();

Output

Time Complexity: O(n)
Auxiliary Space: O(n)

Dynamic Programming (Bottom Up) Solution:

We make a 1D array as there is one parameter that changes in recursion and size of this array is n+1 as the range goes from 0 to n.
We use a loop inside the main loop to fill the dp array in bottom up manner.

C++

#include <bits/stdc++.h>
using namespace std;

/* Returns the best obtainable price for a 
   rod of length n and price as prices of 
   different pieces */
int cutRod(vector<int>& price) {
  
    int n = price.size();
  
    // Create a dp array to store the maximum 
    // revenue obtainable for each length
    vector<int> dp(n + 1, 0);

    // Build the dp array in a bottom-up manner
    for (int i = 1; i <= n; i++) {
        int max_val = 0;
        for (int j = 0; j < i; j++) {
            max_val = max(max_val, price[j] + dp[i - j - 1]);
        }
        dp[i] = max_val;
    }
  
    return dp[n];
}

/* Driver program to test above functions */
int main() {
    vector<int> price = { 0, 5, 8, 9, 10, 17, 17, 20 };
    cout << cutRod(price);
    return 0;
}

C

// A Dynamic Programming solution for Rod cutting problem
#include<stdio.h>
#include<limits.h>

// A utility function to get the maximum of two integers
int max(int a, int b) { return (a > b)? a : b;}

/* Returns the best obtainable price for a rod of length n and
   price[] as prices of different pieces */
int cutRod(int price[], int n)
{
   int val[n+1];
   val[0] = 0;
   int i, j;

   // Build the table val[] in bottom up manner and return the last entry
   // from the table
   for (i = 1; i<=n; i++)
   {
       int max_val = INT_MIN;
       for (j = 0; j < i; j++)
         max_val = max(max_val, price[j] + val[i-j-1]);
       val[i] = max_val;
   }

   return val[n];
}

/* Driver program to test above functions */
int main()
{
    int arr[] = {1, 5, 8, 9, 10, 17, 17, 20};
    int size = sizeof(arr)/sizeof(arr[0]);
    printf("Maximum Obtainable Value is %d", cutRod(arr, size));
    getchar();
    return 0;
}

Java

// A Dynamic Programming solution for Rod cutting problem
import java.io.*;
class RodCutting
{
    /* Returns the best obtainable price for a rod of
       length n and price[] as prices of different pieces */
    static int cutRod(int price[],int n)
    {
        int val[] = new int[n+1];
        val[0] = 0;

        // Build the table val[] in bottom up manner and return
        // the last entry from the table
        for (int i = 1; i<=n; i++)
        {
            int max_val = Integer.MIN_VALUE;
            for (int j = 0; j < i; j++)
                max_val = Math.max(max_val, 
                                   price[j] + val[i-j-1]);
            val[i] = max_val;
        }

        return val[n];
    }

    /* Driver program to test above functions */
    public static void main(String args[])
    {
        int arr[] = new int[] {1, 5, 8, 9, 10, 17, 17, 20};
        int size = arr.length;
        System.out.println("Maximum Obtainable Value is " +
                            cutRod(arr, size));
    }
}
/* This code is contributed by Rajat Mishra */

Python

# A Dynamic Programming solution for Rod cutting problem
INT_MIN = -32767

# Returns the best obtainable price for a rod of length n and
# price[] as prices of different pieces
def cutRod(price, n):
    val = [0 for x in range(n+1)]
    val[0] = 0

    # Build the table val[] in bottom up manner and return
    # the last entry from the table
    for i in range(1, n+1):
        max_val = INT_MIN
        for j in range(i):
             max_val = max(max_val, price[j] + val[i-j-1])
        val[i] = max_val

    return val[n]

# Driver program to test above functions
arr = [1, 5, 8, 9, 10, 17, 17, 20]
size = len(arr)
print("Maximum Obtainable Value is " + str(cutRod(arr, size)))

# This code is contributed by Bhavya Jain

C#

// A Dynamic Programming solution 
// for Rod cutting problem
using System;
class GFG {

    /* Returns the best obtainable 
       price for a rod of length n
       and price[] as prices of 
       different pieces */
    static int cutRod(int []price,int n)
    {
        int []val = new int[n + 1];
        val[0] = 0;

        // Build the table val[] in
        // bottom up manner and return
        // the last entry from the table
        for (int i = 1; i<=n; i++)
        {
            int max_val = int.MinValue;
            for (int j = 0; j < i; j++)
                max_val = Math.Max(max_val, 
                          price[j] + val[i - j - 1]);
            val[i] = max_val;
        }

        return val[n];
    }
    
    // Driver Code
    public static void Main()
    {
        int []arr = new int[] {1, 5, 8, 9, 10, 17, 17, 20};
        int size = arr.Length;
        Console.WriteLine("Maximum Obtainable Value is " +
                                        cutRod(arr, size));
        
    }
}

// This code is contributed by Sam007

JavaScript

    // A Dynamic Programming solution
    // for Rod cutting problem
    
    /* Returns the best obtainable
       price for a rod of length n
       and price[] as prices of
       different pieces */
    function cutRod(price, n)
    {
        let val = new Array(n + 1);
        val[0] = 0;
 
        // Build the table val[] in
        // bottom up manner and return
        // the last entry from the table
        for (let i = 1; i<=n; i++)
        {
            let max_val = Number.MIN_VALUE;
            for (let j = 0; j < i; j++)
                max_val = Math.max(max_val, price[j] + val[i - j - 1]);
            val[i] = max_val;
        }
 
        return val[n];
    }
    
    let arr = [1, 5, 8, 9, 10, 17, 17, 20];
    let size = arr.length;
    console.log("Maximum Obtainable Value is " + cutRod(arr, size) + "n");

PHP

<?php
// A Dynamic Programming solution for
// Rod cutting problem

/* Returns the best obtainable price 
for a rod of length n and price[] as
prices of different pieces */
function cutRod( $price, $n)
{
    $val = array();
    $val[0] = 0;
    $i; $j;
    
    // Build the table val[] in bottom 
    // up manner and return the last
    // entry from the table
    for ($i = 1; $i <= $n; $i++)
    {
        $max_val = PHP_INT_MIN;
        
        for ($j = 0; $j < $i; $j++)
            $max_val = max($max_val, 
             $price[$j] + $val[$i-$j-1]);
        $val[$i] = $max_val;
    }
    
    return $val[$n];
}

// Driver program to test above functions
$arr = array(1, 5, 8, 9, 10, 17, 17, 20);
$size = count($arr);
echo "Maximum Obtainable Value is ", 
                      cutRod($arr, $size);
    
// This code is contributed by anuj_67.
?>

Output

Time Complexity : O(n^2)
Space Complexity: O(n)

Note : The above provided solution works better than the below solution as the below solution requires more extra space and computations.

Using the idea of Unbounded Knapsack:

This problem is very similar to the Unbounded Knapsack Problem, where there are multiple occurrences of the same item. Here we consider length of rod as capacity of knapsack. All lengths from 1 to n are considers are weights of items and prices as values of items.

Recursive Solution:

C++

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// Simple implementation of unbounded 
// knapsack problem.
int rodCutRec(int n, vector<int>& len,
              vector<int>& price, int idx) {
  
    // Base Case: if the rod length is 0 
    // or no pieces left to consider
    if (idx == 0) {
        return (n / len[0]) * price[0];
    }

    // If not taking the current length piece
    int notTake = rodCutRec(n, len, price, 
                                idx - 1);

    // If taking the current length piece
    int take = 0;
    if (len[idx] <= n) {
        take = price[idx] + rodCutRec(n-len[idx], 
                                len, price, idx);
    }

    // Return the maximum value between taking 
    // and not taking the current piece
    return max(take, notTake);
}

int rodCut(vector<int>& price) {
    int n = price.size();
  
    // Set up weights for unbounded
    // knapsack problem
    vector<int> len(n);
    for (int i = 0; i < n; ++i)
        len[i] = i + 1;  
 
    return rodCutRec(n, len, price, n - 1);
}

// Driver code
int main() {
    vector<int> price = {10, 30, 20};
    cout << rodCut(price) ;
    return 0;
}

C

// C program for above approach
#include <stdio.h>
#include <stdlib.h>

int max(int a, int b) 
{ 
  return (a > b) ? a : b; 
}

// Global Array for the
// purpose of memoization.
int t[9][9];

// A recursive program, using ,
// memoization, to implement the
// rod cutting problem(Top-Down).
int un_kp(int price[], int length[], 
                     int Max_len, int n)
{

    // The maximum price will be zero,
    // when either the length of the rod
    // is zero or price is zero.
    if (n == 0 || Max_len == 0) 
    {
        return 0;
    }

    // If the length of the rod is less
    // than the maximum length, Max_lene
    // will consider it.Now depending
    // upon the profit,
    // either Max_lene  we will take it
    // or discard it.
    if (length[n - 1] <= Max_len) 
    {
        t[n][Max_len]
            = max(price[n - 1]
                      + un_kp(price, length,
                              Max_len - length[n - 1], n),
                  un_kp(price, length, Max_len, n - 1));
    }

    // If the length of the rod is greater
    // than the permitted size, Max_len
    // we will  not consider it.
    else 
    {
        t[n][Max_len]
            = un_kp(price, length,
                             Max_len, n - 1);
    }

    // Max_lene Max_lenill return
    // the maximum value obtained,
    // Max_lenhich is present at the
    // nth roMax_len and Max_length column.
    return t[n][Max_len];
}

/* Driver program to test above functions */
int main()
{
    int price[] = { 1, 5, 8, 9, 10, 17, 17, 20 };
    int n = sizeof(price) / sizeof(price[0]);
    int length[n];
    for (int i = 0; i < n; i++) 
    {
        length[i] = i + 1;
    }
    int Max_len = n;

    // Function Call
    printf("Maximum obtained value  is %d \n",
           un_kp(price, length, n, Max_len));
}

Java

// Java program for above approach
import java.io.*;

class GFG {

    // Global Array for
    // the purpose of memoization.
    static int t[][] = new int[9][9];

    // A recursive program, using ,
    // memoization, to implement the
    // rod cutting problem(Top-Down).
    public static int un_kp(int price[], int length[],
                            int Max_len, int n)
    {

        // The maximum price will be zero,
        // when either the length of the rod
        // is zero or price is zero.
        if (n == 0 || Max_len == 0) {
            return 0;
        }

        // If the length of the rod is less
        // than the maximum length, Max_lene will
        // consider it.Now depending
        // upon the profit,
        // either Max_lene  we will take
        // it or discard it.
        if (length[n - 1] <= Max_len) {
            t[n][Max_len] = Math.max(
                price[n - 1]
                    + un_kp(price, length,
                            Max_len - length[n - 1], n),
                un_kp(price, length, Max_len, n - 1));
        }

        // If the length of the rod is
        // greater than the permitted size,
        // Max_len we will  not consider it.
        else {
            t[n][Max_len]
                = un_kp(price, length, Max_len, n - 1);
        }

        // Max_lene Max_lenill return the maximum
        // value obtained, Max_lenhich is present
        // at the nth roMax_len and Max_length column.
        return t[n][Max_len];
    }

    public static void main(String[] args)
    {

        int price[]
            = new int[] { 1, 5, 8, 9, 10, 17, 17, 20 };
        int n = price.length;
        int length[] = new int[n];
        for (int i = 0; i < n; i++) {
            length[i] = i + 1;
        }
        int Max_len = n;
        System.out.println(
            "Maximum obtained value is "
            + un_kp(price, length, n, Max_len));
    }
}

// This code is contributed by rajsanghavi9.

Python

# Python program for above approach

# Global Array for
# the purpose of memoization.
t = [[0 for i in range(9)] for j in range(9)]

# A recursive program, using ,
# memoization, to implement the
# rod cutting problem(Top-Down).
def un_kp(price, length, Max_len, n):

    # The maximum price will be zero,
    # when either the length of the rod
    # is zero or price is zero.
    if (n == 0 or Max_len == 0):
        return 0;
    

    # If the length of the rod is less
    # than the maximum length, Max_lene will
    # consider it.Now depending
    # upon the profit,
    # either Max_lene we will take
    # it or discard it.
    if (length[n - 1] <= Max_len):
        t[n][Max_len] = max(price[n - 1] + un_kp(price, length, Max_len - length[n - 1], n),
                un_kp(price, length, Max_len, n - 1));
    

    # If the length of the rod is
    # greater than the permitted size,
    # Max_len we will not consider it.
    else:
        t[n][Max_len] = un_kp(price, length, Max_len, n - 1);
    

    # Max_lene Max_lenill return the maximum
    # value obtained, Max_lenhich is present
    # at the nth roMax_len and Max_length column.
    return t[n][Max_len];


if __name__ == '__main__':

    price = [1, 5, 8, 9, 10, 17, 17, 20 ];
    n =len(price);
    length = [0]*n;
    for i in range(n):
        length[i] = i + 1;
    
    Max_len = n;
    print("Maximum obtained value is " ,un_kp(price, length, n, Max_len));

# This code is contributed by gauravrajput1

C#

// C# program for above approach
using System;
public class GFG {

  // Global Array for
  // the purpose of memoization.
  static int [,]t = new int[9,9];

  // A recursive program, using ,
  // memoization, to implement the
  // rod cutting problem(Top-Down).
  public static int un_kp(int []price, int []length, int Max_len, int n) {

    // The maximum price will be zero,
    // when either the length of the rod
    // is zero or price is zero.
    if (n == 0 || Max_len == 0) {
      return 0;
    }

    // If the length of the rod is less
    // than the maximum length, Max_lene will
    // consider it.Now depending
    // upon the profit,
    // either Max_lene we will take
    // it or discard it.
    if (length[n - 1] <= Max_len) {
      t[n,Max_len] = Math.Max(price[n - 1] + un_kp(price, length, Max_len - length[n - 1], n),
                              un_kp(price, length, Max_len, n - 1));
    }

    // If the length of the rod is
    // greater than the permitted size,
    // Max_len we will not consider it.
    else {
      t[n,Max_len] = un_kp(price, length, Max_len, n - 1);
    }

    // Max_lene Max_lenill return the maximum
    // value obtained, Max_lenhich is present
    // at the nth roMax_len and Max_length column.
    return t[n,Max_len];
  }

  // Driver code
  public static void Main(String[] args) {

    int []price = new int[] { 1, 5, 8, 9, 10, 17, 17, 20 };
    int n = price.Length;
    int []length = new int[n];
    for (int i = 0; i < n; i++) {
      length[i] = i + 1;
    }
    int Max_len = n;
    Console.WriteLine("Maximum obtained value is " + un_kp(price, length, n, Max_len));
  }
}

// This code is contributed by gauravrajput1.

JavaScript

// Javascript program for above approach

// Global Array for
// the purpose of memoization.
let t = new Array(9);
for (var i = 0; i < t.length; i++) {
    t[i] = new Array(2);
}

// A recursive program, using ,
// memoization, to implement the
// rod cutting problem(Top-Down).
function un_kp(price, length, Max_len, n)
{

    // The maximum price will be zero,
    // when either the length of the rod
    // is zero or price is zero.
    if (n == 0 || Max_len == 0) 
    {
        return 0;
    }

    // If the length of the rod is less
    // than the maximum length, Max_lene will
    // consider it.Now depending 
    // upon the profit,
    // either Max_lene  we will take 
    // it or discard it.
    if (length[n - 1] <= Max_len) 
    {
        t[n][Max_len]
            = Math.max(price[n - 1]
                      + un_kp(price, length,
                           Max_len - length[n - 1], n),
                  un_kp(price, length, Max_len, n - 1));
    }

    // If the length of the rod is
    // greater than the permitted size,
    // Max_len we will  not consider it.
    else 
    {
        t[n][Max_len]
            = un_kp(price, length, 
                              Max_len, n - 1);
    }

    // Max_lene Max_lenill return the maximum
    // value obtained, Max_lenhich is present
    // at the nth roMax_len and Max_length column.
    return t[n][Max_len];
}
    
    // Driver code
    
    let price = [ 1, 5, 8, 9, 10, 17, 17, 20 ];
    let n = price.length;
    let length = Array(n).fill(0);
    for (let i = 0; i < n; i++) {
        length[i] = i + 1;
    }
    let Max_len = n;

    // Function Call
    console.log("Maximum obtained value  is "
         + un_kp(price, length, n, Max_len));

Output

The above solution can be optimized either using memoization or dynamic programming.

Dynamic Programming Solution

We use a 2D array as there are two parameters that change during recursive calls.
We use 1 dimension for index and one for size of rod.

C++

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

// Function to return the maximum value that
// can be obtained from a rod of length n
// using dynamic programming
int rodCut(vector<int>& price) {
  
    // Set up for unbounded knapsack
    int n = price.size();
    vector<int> len(n);
    for (int i = 0; i < n; ++i)
        len[i] = i + 1;

    // Create a 2D dp array
    vector<vector<int>> dp(n, vector<int>(n + 1, 0));

    // Fill the dp array
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j <= n; ++j) {
          
            // If we are at the first row, we can 
            // only use the first length piece
            if (i == 0) {
                dp[i][j] = (j / len[0]) * price[0];
            } else {
              
                // Not taking the current length piece
                int notTake = dp[i - 1][j]; 
              
                // Taking the current length piece
                int take = 0;
                if (len[i] <= j) {
                    take = price[i] + dp[i][j - len[i]]; 
                }
              
                // Store the maximum value
                dp[i][j] = max(take, notTake); 
            }
        }
    }

    // The answer is in dp[n-1][n]
    return dp[n - 1][n];
}

// Driver code
int main() {
    vector<int> price = {10, 30, 20};
    cout << rodCut(price) << endl;
    return 0;
}

Java

// Java program for above approach
import java.io.*;

class GFG {

    public static int cutRod(int prices[], int n)
    {

        int mat[][] = new int[n + 1][n + 1];
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= n; j++) {
                if (i == 0 || j == 0) {
                    mat[i][j] = 0;
                }
                else {
                    if (i == 1) {
                        mat[i][j] = j * prices[i - 1];
                    }
                    else {
                        if (i > j) {
                            mat[i][j] = mat[i - 1][j];
                        }
                        else {
                            mat[i][j] = Math.max(
                                prices[i - 1]
                                    + mat[i][j - i],
                                mat[i - 1][j]);
                        }
                    }
                }
            }
        }

        return mat[n][n];
    }
    public static void main(String[] args)
    {

        int prices[]
            = new int[] { 1, 5, 8, 9, 10, 17, 17, 20 };
        int n = prices.length;

        System.out.println("Maximum obtained value is "
                           + cutRod(prices, n));
    }
}

Python

#  Python program for above approach


def cutRod(prices, n):
    mat = [[0 for i in range(n+1)]for j in range(n+1)]
    for i in range(1, n+1):
        for j in range(1, n+1):
            if i == 1:
                mat[i][j] = j*prices[i-1]
            else:
                if i > j:
                    mat[i][j] = mat[i-1][j]
                else:
                    mat[i][j] = max(prices[i-1]+mat[i][j-i], mat[i-1][j])
    return mat[n][n]


prices = [1, 5, 8, 9, 10, 17, 17, 20]
n = len(prices)

print("Maximum obtained value is ", cutRod(prices, n))

# This Code is Contributed By Vivek Maddeshiya

C#

// C# program for above approach
using System;
using System.Collections;
using System.Collections.Generic;

class GFG {

  public static int cutRod(int[] prices, int n)
  {

    int[, ] mat = new int[n + 1, n + 1];
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= n; j++) {
        if (i == 0 || j == 0) {
          mat[i, j] = 0;
        }
        else {
          if (i == 1) {
            mat[i, j] = j * prices[i - 1];
          }
          else {
            if (i > j) {
              mat[i, j] = mat[i - 1, j];
            }
            else {
              mat[i, j] = Math.Max(
                prices[i - 1]
                + mat[i, j - i],
                mat[i - 1, j]);
            }
          }
        }
      }
    }

    return mat[n, n];
  }
  public static void Main(string[] args)
  {

    int[] prices
      = new int[] { 1, 5, 8, 9, 10, 17, 17, 20 };
    int n = prices.Length;

    Console.WriteLine("Maximum obtained value is "
                      + cutRod(prices, n));
  }
}

// This code is contributed by Karandeep1234

JavaScript

// Javascript implementation of the above C++ code

function cutRod(prices, n) {
    // Initialize the matrix with all values as 0
    let mat = new Array(n + 1).fill(0).map(() => new Array(n + 1).fill(0));

    // Iterate over the matrix
    for (let i = 0; i <= n; i++) {
        for (let j = 0; j <= n; j++) {
            if (i === 0 || j === 0) {
                mat[i][j] = 0;
            } else {
                if (i === 1) {
                    mat[i][j] = j * prices[i - 1];
                } else {
                    if (i > j) {
                        mat[i][j] = mat[i - 1][j];
                    } else {
                        mat[i][j] = Math.max(prices[i - 1] + mat[i][j - i], mat[i - 1][j]);
                    }
                }
            }
        }
    }

    // Return the last element of the matrix
    return mat[n][n];
}

// Given prices array
let prices = [1, 5, 8, 9, 10, 17, 17, 20];
let n = prices.length;

console.log("Maximum obtained value is " + cutRod(prices, n));

Output

Time Complexity: O(n²)
Auxiliary Space: O(n²)

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