# 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, dpi+a, dpi+b and dpi+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 :

dpi+a=max(dpi+1, dpi+a);
dpi+b=max(dpi+1, dpi+b);
dpi+c=max(dpi+1, dpi+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 ` `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;  ` ` `  `    ``// 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;  ` ` `  `        ``// 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

 ` `

Output :

`2`

Time complexity : O(n)

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