Maximum number of segments of lengths a, b and c

Given a positive integer N, find the maximum number of segments of lengths a, b and c that can be formed from N .

Examples :

Input : N = 7, a = 5, b, = 2, c = 5 
Output : 2 
N can be divided into 2 segments of lengths
2 and 5. For the second example,

Input : N = 17, a = 2, b = 1, c = 3 
Output : 17 
N can be divided into 17 segments of 1 or 8 
segments of 2 and 1 segment of 1. But 17 segments
of 1 is greater than 9 segments of 2 and 1.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach : The approach used is Dynamic Programming. The base dp₀ will be 0 as initially it has no segments. After that, iterate from 1 to n, and for each of the 3 states i.e, dp_i+a, dp_i+b and dp_i+c, store the maximum value obtained by either using or not using the a, b or c segment.
The 3 states to deal with are :

dp_i+a=max(dp_i+1, dp_i+a);
dp_i+b=max(dp_i+1, dp_i+b);
dp_i+c=max(dp_i+1, dp_i+c);

Below is the implementation of above idea :

C++

// C++ implementation to divide N into  
// maximum number of segments  
// of length a, b and c 
#include <bits/stdc++.h> 
using namespace std; 
  
// function to find the maximum 
// number of segments 
int maximumSegments(int n, int a,  
                    int b, int c) 
{ 
    // stores the maximum number of  
    // segments each index can have 
    int dp[n + 1]; 
      
    // initialize with -1 
    memset(dp, -1, sizeof(dp)); 
  
    // 0th index will have 0 segments 
    // base case 
    dp[0] = 0;  
  
    // traverse for all possible  
    // segments till n 
    for (int i = 0; i < n; i++)  
    { 
        if (dp[i] != -1) { 
              
            // conditions 
        if(i + a <= n )    //avoid buffer overflow 
                dp[i + a] = max(dp[i] + 1,  
                            dp[i + a]); 
                              
        if(i + b <= n ) //avoid buffer overflow 
                dp[i + b] = max(dp[i] + 1,  
                            dp[i + b]); 
                              
        if(i + c <= n )    //avoid buffer overflow 
                dp[i + c] = max(dp[i] + 1,  
                            dp[i + c]); 
        } 
    } 
    return dp[n]; 
} 
  
// Driver code 
int main() 
{ 
    int n = 7, a = 5, b = 2, c = 5; 
    cout << maximumSegments(n, a, b, c);  
    return 0; 
} 

Java

// Java implementation to divide N into 
// maximum number of segments 
// of length a, b and c 
import java.util.*; 
  
class GFG  
{ 
      
    // function to find the maximum 
    // number of segments 
    static int maximumSegments(int n, int a,  
                            int b, int c) 
    { 
        // stores the maximum number of 
        // segments each index can have 
        int dp[] = new int[n + 10]; 
  
        // initialize with -1 
        Arrays.fill(dp, -1); 
  
        // 0th index will have 0 segments 
        // base case 
        dp[0] = 0;  
  
        // traverse for all possible  
        // segments till n 
        for (int i = 0; i < n; i++)  
        { 
            if (dp[i] != -1)  
            { 
  
                // conditions 
                if(i + a <= n )    //avoid buffer overflow 
                dp[i + a] = Math.max(dp[i] + 1,  
                                    dp[i + a]); 
                                      
                if(i + b <= n )    //avoid buffer overflow 
                dp[i + b] = Math.max(dp[i] + 1,      
                                    dp[i + b]); 
                                      
                if(i + c <= n )    //avoid buffer overflow 
                dp[i + c] = Math.max(dp[i] + 1,  
                                    dp[i + c]); 
            } 
        } 
        return dp[n]; 
    } 
  
    // Driver code 
    public static void main(String arg[]) 
    { 
        int n = 7, a = 5, b = 2, c = 5; 
        System.out.print(maximumSegments(n, a, b, c)); 
    } 
} 
  
// This code is contributed by Anant Agarwal. 

Python3

# Python implementation  
# to divide N into maximum  
# number of segments of  
# length a, b and c 
  
# function to find  
# the maximum number  
# of segments 
def maximumSegments(n, a, b, c) : 
  
    # stores the maximum  
    # number of segments  
    # each index can have 
    dp = [-1] * (n + 10) 
  
    # 0th index will have  
    # 0 segments base case 
    dp[0] = 0
  
    # traverse for all possible 
    # segments till n 
    for i in range(0, n) : 
      
        if (dp[i] != -1) : 
          
            # conditions 
            if(i + a <= n ): # avoid buffer overflow     
                dp[i + a] = max(dp[i] + 1,  
                            dp[i + a]) 
                              
            if(i + b <= n ): # avoid buffer overflow     
                dp[i + b] = max(dp[i] + 1,  
                            dp[i + b]) 
                              
            if(i + c <= n ): # avoid buffer overflow     
                dp[i + c] = max(dp[i] + 1,  
                            dp[i + c]) 
  
    return dp[n] 
  
# Driver code 
n = 7
a = 5
b = 2
c = 5
print (maximumSegments(n, a,  
                    b, c)) 
  
# This code is contributed by  
# Manish Shaw(manishshaw1) 

C#

// C# implementation to divide N into 
// maximum number of segments 
// of length a, b and c 
using System; 
  
class GFG  
{ 
      
    // function to find the maximum 
    // number of segments 
    static int maximumSegments(int n, int a,  
                            int b, int c) 
    { 
        // stores the maximum number of 
        // segments each index can have 
        int []dp = new int[n + 10]; 
  
        // initialize with -1 
        for(int i = 0; i < n + 10; i++) 
        dp[i]= -1; 
          
  
        // 0th index will have 0 segments 
        // base case 
        dp[0] = 0;  
  
        // traverse for all possible 
        // segments till n 
        for (int i = 0; i < n; i++)  
        { 
            if (dp[i] != -1)  
            { 
  
                // conditions 
                if(i + a <= n )    // avoid buffer overflow 
                dp[i + a] = Math.Max(dp[i] + 1,  
                                    dp[i + a]); 
                                      
                if(i + b <= n )    // avoid buffer overflow 
                dp[i + b] = Math.Max(dp[i] + 1,  
                                    dp[i + b]); 
                                      
                if(i + c <= n )    // avoid buffer overflow 
                dp[i + c] = Math.Max(dp[i] + 1,  
                                    dp[i + c]); 
            } 
        } 
        return dp[n]; 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int n = 7, a = 5, b = 2, c = 5; 
        Console.Write(maximumSegments(n, a, b, c)); 
    } 
} 
  
// This code is contributed by nitin mittal 

PHP

<?php 
// PHP implementation to divide  
// N into maximum number of  
// segments of length a, b and c 
  
// function to find the maximum 
// number of segments 
function maximumSegments($n, $a,  
                        $b, $c) 
{ 
    // stores the maximum  
    // number of segments  
    // each index can have 
    $dp = array(); 
  
    // initialize with -1 
    for($i = 0; $i < $n + 10; $i++) 
        $dp[$i]= -1; 
      
  
    // 0th index will have  
    // 0 segments base case 
    $dp[0] = 0;  
  
    // traverse for all possible 
    // segments till n 
    for ($i = 0; $i < $n; $i++)  
    { 
        if ($dp[$i] != -1)  
        { 
            // conditions 
            if($i + $a <= $n )    // avoid buffer overflow 
            $dp[$i + $a] = max($dp[$i] + 1,  
                            $dp[$i + $a]); 
                              
            if($i + $b <= $n )    // avoid buffer overflow 
            $dp[$i + $b] = max($dp[$i] + 1,  
                            $dp[$i + $b]); 
                              
            if($i + $c <= $n )    // avoid buffer overflow 
            $dp[$i + $c] = max($dp[$i] + 1,  
                            $dp[$i + $c]); 
        } 
    } 
    return $dp[$n]; 
} 
  
// Driver code 
$n = 7; $a = 5;  
$b = 2; $c = 5; 
echo (maximumSegments($n, $a,  
                    $b, $c)); 
  
// This code is contributed by  
// Manish Shaw(manishshaw1) 
?> 

Output :

Time complexity : O(n)

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Suggest a Topic

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

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