Maximize the number of segments of length p, q and r

Given a rod of length L, the task is to cut the rod in such a way that the total number of segments of length p, q and r is maximized. The segments can only be of length p, q, and r.

Examples:

Input: l = 11, p = 2, q = 3, r = 5
Output: 5
Segments of 2, 2, 2, 2 and 3

Input: l = 7, p = 2, q = 5, r = 5
Output: 2
Segments of 2 and 5

Approach: As the solution for maximum number of cuts that can be made in a given length depends on the maximum number of cuts previously made in shorter lengths, this question could be solved by the approach of Dynamic Programming. Suppose we are given a length ‘l’. For finding the maximum number of cuts that can be made in length ‘l’, find number of cuts made in shorter previous length ‘l-p’,’l-q’,’l-r’ lengths respectively. The required answer would be the max(l-p,l-q,l-r)+1 as one more cut should be needed after this to cut length ‘l’. So for solving this problem for given length, find maximum number of cuts that can be made in lengths ranging from ‘1’ to ‘l’.
Example:



l = 11, p = 2, q = 3, r = 5
Analysing lengths from 1 to 11:

  1. Not possible to cut->0
  2. Possible cut is of lengths 2->1 (2)
  3. Possible cut is of lengths 3->1 (3)
  4. Possible cuts are of lengths max(arr[4-2],arr[4-3])+1->2 (2,2)
  5. Possible cuts are of lengths max(arr[5-2],arr[5-3])+1->2 (2,3)
  6. Possible cuts are of lengths max(arr[6-2],arr[6-3],arr[6-5])+1->3 (2,2,2)
  7. Possible cuts are of lengths max(arr[7-2],arr[7-3],arr[7-5])+1->3 (2,3,2) or (2,2,3)
  8. Possible cuts are of lengths max(arr[8-2],arr[8-3],arr[8-5])+1->4 (2,2,2,2)
  9. Possible cuts are of lengths max(arr[9-2],arr[9-3],arr[9-5])+1->4 (2,3,2,2) or (2,2,3,2) or (2,2,2,3)
  10. Possible cuts are of lengths max(arr[10-2],arr[10-3],arr[10-5])+1->5 (2,2,2,2,2)
  11. Possible cuts are of lengths max(arr[11-2],arr[11-3],arr[11-5])+1->5 (2,3,2,2,2) or (2,2,3,2,2) or (2,2,2,3,2) or (2,2,2,2,3)

Algorithm:

  1. Initialise an array DP[]={-1} and DP[0]=0.
  2. Run a loop from ‘1’ to ‘l’
  3. If DP[i]=-1 means it’s not possible to divide it using giving segments p,q,r so continue;
  4. DP[i+p]=max(DP[i+p],DP[i]+1)
  5. DP[i+q]=max(DP[i+q],DP[i]+1)
  6. DP[i+r]=max(DP[i+r],DP[i]+1)
  7. print DP[l]

Pseudo Code:

DP[l+1]={-1}
DP[0]=0
for(i from 0 to l)
  if(DP[i]==-1)
  continue
  DP[i+p]=max(DP[i+p],DP[i]+1)
  DP[i+q]=max(DP[i+q],DP[i]+1)
  DP[i+r]=max(DP[i+r],DP[i]+1)

print(DP[l])

Implementation:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to maximize the number
// of segments of length p, q and r
#include <bits/stdc++.h>
using namespace std;
  
// Function that returns the maximum number
// of segments possible
int findMaximum(int l, int p, int q, int r)
{
  
    // Array to store the cut at each length
    int dp[l + 1];
  
    // All values with -1
    memset(dp, -1, sizeof(dp));
  
    // if length of rod is 0 then total cuts will be 0
    // so, initialize the dp[0] with 0
    dp[0] = 0;
  
    for (int i = 0; i <= l; i++) {
  
        // if certain length is not possible
        if (dp[i] == -1)
            continue;
  
        // if a segment of p is possible
        if (i + p <= l)
            dp[i + p] = max(dp[i + p], dp[i] + 1);
  
        // if a segment of q is possible
        if (i + q <= l)
            dp[i + q] = max(dp[i + q], dp[i] + 1);
  
        // if a segment of r is possible
        if (i + r <= l)
            dp[i + r] = max(dp[i + r], dp[i] + 1);
    }
  
    // return value corresponding to length l
    return dp[l];
}
  
// Driver Code
int main()
{
    int l = 11, p = 2, q = 3, r = 5;
  
    // Calling Function
    int ans = findMaximum(l, p, q, r);
    cout << ans;
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to maximize 
// the number of segments 
// of length p, q and r
import java.io.*;
  
class GFG 
{
      
// Function that returns
// the maximum number
// of segments possible
static int findMaximum(int l, int p, 
                       int q, int r)
{
  
    // Array to store the
    // cut at each length
    int dp[] = new int[l + 1];
  
    // All values with -1
    for(int i = 0; i < l + 1; i++)
    dp[i] = -1;
      
  
    // if length of rod is 0 
    // then total cuts will 
    // be 0 so, initialize 
    // the dp[0] with 0
    dp[0] = 0;
  
    for (int i = 0; i <= l; i++) 
    {
  
        // if certain length 
        // is not possible
        if (dp[i] == -1)
            continue;
  
        // if a segment of
        // p is possible
        if (i + p <= l)
            dp[i + p] = Math.max(dp[i + p], 
                                 dp[i] + 1);
  
        // if a segment of 
        // q is possible
        if (i + q <= l)
            dp[i + q] = Math.max(dp[i + q],
                                 dp[i] + 1);
  
        // if a segment of
        // r is possible
        if (i + r <= l)
            dp[i + r] = Math.max(dp[i + r], 
                                 dp[i] + 1);
    }
  
    // return value corresponding 
    // to length l
    return dp[l];
}
  
// Driver Code
public static void main (String[] args)
{
    int l = 11, p = 2
        q = 3, r = 5;
  
    // Calling Function
    int ans = findMaximum(l, p, q, r);
    System.out.println( ans);
}
}
  
// This code is contributed
// by anuj_67.

chevron_right


Python 3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python 3 program to 
# maximize the number
# of segments of length 
# p, q and r
  
# Function that returns 
# the maximum number
# of segments possible
def findMaximum( l, p, q, r):
  
    # Array to store the cut 
    # at each length
    # All values with -1
    dp = [-1]*(l + 1
  
    # if length of rod is 0 then 
    # total cuts will be 0
    # so, initialize the dp[0] with 0
    dp[0] = 0
  
    for i in range( l+1) :
  
        # if certain length is not 
        # possible
        if (dp[i] == -1):
            continue
  
        # if a segment of p is possible
        if (i + p <= l):
            dp[i + p] = (max(dp[i + p], 
                        dp[i] + 1))
  
        # if a segment of q is possible
        if (i + q <= l):
            dp[i + q] = (max(dp[i + q], 
                        dp[i] + 1))
  
        # if a segment of r is possible
        if (i + r <= l):
            dp[i + r] = (max(dp[i + r],
                        dp[i] + 1))
  
    # return value corresponding 
    # to length l
    return dp[l]
  
# Driver Code
if __name__ == "__main__":
    l = 11
    p = 2
    q = 3
    r = 5
  
    # Calling Function
    ans = findMaximum(l, p, q, r)
    print(ans)
  
# This code is contributed by 
# ChitraNayal

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to maximize 
// the number of segments 
// of length p, q and r
using System;
  
class GFG 
{
      
// Function that returns
// the maximum number
// of segments possible
static int findMaximum(int l, int p, 
                       int q, int r)
{
  
    // Array to store the
    // cut at each length
    int []dp = new int[l + 1];
  
    // All values with -1
    for(int i = 0; i < l + 1; i++)
    dp[i] = -1;
      
  
    // if length of rod is 0 
    // then total cuts will 
    // be 0 so, initialize 
    // the dp[0] with 0
    dp[0] = 0;
  
    for (int i = 0; i <= l; i++) 
    {
  
        // if certain length 
        // is not possible
        if (dp[i] == -1)
            continue;
  
        // if a segment of
        // p is possible
        if (i + p <= l)
            dp[i + p] = Math.Max(dp[i + p], 
                                 dp[i] + 1);
  
        // if a segment of 
        // q is possible
        if (i + q <= l)
            dp[i + q] = Math.Max(dp[i + q],
                                 dp[i] + 1);
  
        // if a segment of
        // r is possible
        if (i + r <= l)
            dp[i + r] = Math.Max(dp[i + r], 
                                 dp[i] + 1);
    }
  
    // return value corresponding 
    // to length l
    return dp[l];
}
  
// Driver Code
public static void Main ()
{
    int l = 11, p = 2, 
        q = 3, r = 5;
  
    // Calling Function
    int ans = findMaximum(l, p, q, r);
    Console.WriteLine(ans);
}
}
  
// This code is contributed
// by anuj_67.

chevron_right


Output:

5

Complexity Analysis:

  • Time Complexity: O(N).
    Use of a single for loop till length ‘N’.
  • Auxiliary Space: O(N).
    Use of an array ‘DP’ to keep track of segments.

competitive-programming-img




My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : vt_m, chitranayal, bidibaaz123