# 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.  ```

To understand any DP problem clearly, we need to write first of all its recursive code and then go for optimization.

Recursion-Based Solution:

```Here for any value of n, we have 3 possibilities, for making the maximum segment count
if (n >= a) we can make 1 segment of length a + another possible segment from the length of n - a
if (n >= b) we can make 1 segment of length b + another possible segment from the length of n - b
if (n >= c) we can make 1 segment of length c + another possible segment from the length of n - c
so now we have to take the maximum possible segment above in 3 condition```

Below is an implementation for the same.

## 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)` `{` `    ``// Base case` `    ``if` `(n == 0) {` `        ``return` `0;` `    ``}`   `    ``int` `maxa = INT_MIN;` `    ``// Conditions`   `    ``// Making one segment of length a` `    ``if` `(n >= a) {` `        ``maxa = max(maxa,` `                   ``1 + maximumSegments(n - a, a, b, c));` `    ``}` `    ``// Making one segment of length b` `    ``if` `(n >= b) {` `        ``maxa = max(maxa,` `                   ``1 + maximumSegments(n - b, a, b, c));` `    ``}` `    ``// Making one segment of length c` `    ``if` `(n >= c) {` `        ``maxa = max(maxa,` `                   ``1 + maximumSegments(n - c, a, b, c));` `    ``}`   `    ``// Return maximum out of all possible segment` `    ``return` `maxa;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `n = 7, a = 5, b = 2, c = 5;`   `    ``// Function call` `    ``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`   `class` `GFG {`   `  ``static` `int` `INT_MIN = -``1000000000``;`   `  ``// Function to find the maximum number of segments` `  ``static` `int` `maximumSegments(``int` `n, ``int` `a, ``int` `b, ``int` `c) ` `  ``{`   `    ``// Base case` `    ``if` `(n == ``0``) {` `      ``return` `0``;` `    ``}`   `    ``int` `maxa = INT_MIN;` `    ``// Conditions`   `    ``// Making one segment of length a` `    ``if` `(n >= a) {` `      ``maxa = Math.max(maxa, ``1` `+ maximumSegments(n - a, a, b, c));` `    ``}` `    ``// Making one segment of length b` `    ``if` `(n >= b) {` `      ``maxa = Math.max(maxa, ``1` `+ maximumSegments(n - b, a, b, c));` `    ``}` `    ``// Making one segment of length c` `    ``if` `(n >= c) {` `      ``maxa = Math.max(maxa, ``1` `+ maximumSegments(n - c, a, b, c));` `    ``}`   `    ``// Return maximum out of all possible segment` `    ``return` `maxa;` `  ``}`   `  ``// Driver code` `  ``public` `static` `void` `main(String[] args) {` `    ``int` `n = ``7``, a = ``5``, b = ``2``, c = ``5``;`   `    ``// Function call` `    ``System.out.println(maximumSegments(n, a, b, c));` `  ``}` `}`   `// This code is contributed by ajaymakvana.`

## Python

 `# 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):` `    ``# Base case` `    ``if` `n ``=``=` `0``:` `        ``return` `0`   `    ``maxa ``=` `float``(``'-inf'``)` `    ``# Conditions`   `    ``# Making one segment of length a` `    ``if` `n >``=` `a:` `        ``maxa ``=` `max``(maxa,` `                   ``1` `+` `maximumSegments(n ``-` `a, a, b, c))` `    ``# Making one segment of length b` `    ``if` `n >``=` `b:` `        ``maxa ``=` `max``(maxa,` `                   ``1` `+` `maximumSegments(n ``-` `b, a, b, c))` `    ``# Making one segment of length c` `    ``if` `n >``=` `c:` `        ``maxa ``=` `max``(maxa,` `                   ``1` `+` `maximumSegments(n ``-` `c, a, b, c))`   `    ``# Return maximum out of all possible segment` `    ``return` `maxa`   `# Driver code` `if` `__name__ ``=``=` `'__main__'``:` `    ``n ``=` `7` `    ``a ``=` `5` `    ``b ``=` `2` `    ``c ``=` `5`   `    ``# Function call` `    ``print``(maximumSegments(n, a, b, c))`   ` ``# This code is contributed by divyansh2212`

## C#

 `using` `System;`   `namespace` `ConsoleApp {` `  ``class` `Program {` `    ``static` `void` `Main(``string``[] args)` `    ``{` `      ``int` `n = 7, a = 5, b = 2, c = 5;` `      ``Console.WriteLine(maximumSegments(n, a, b, c));` `    ``}`   `    ``// Function to find the maximum number of segments` `    ``static` `int` `maximumSegments(``int` `n, ``int` `a, ``int` `b, ``int` `c)` `    ``{` `      ``// Base case` `      ``if` `(n == 0) {` `        ``return` `0;` `      ``}`   `      ``int` `maxa = ``int``.MinValue;` `      ``// Conditions`   `      ``// Making one segment of length a` `      ``if` `(n >= a) {` `        ``maxa = Math.Max(` `          ``maxa, 1 + maximumSegments(n - a, a, b, c));` `      ``}` `      ``// Making one segment of length b` `      ``if` `(n >= b) {` `        ``maxa = Math.Max(` `          ``maxa, 1 + maximumSegments(n - b, a, b, c));` `      ``}` `      ``// Making one segment of length c` `      ``if` `(n >= c) {` `        ``maxa = Math.Max(` `          ``maxa, 1 + maximumSegments(n - c, a, b, c));` `      ``}`   `      ``// Return maximum out of all possible segments` `      ``return` `maxa;` `    ``}` `  ``}` `}`   `// This code is contributed by divyansh2212`

## Javascript

 `// Javascript 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) {` `    ``// Base case` `    ``if` `(n === 0) {` `        ``return` `0;` `    ``}`   `    ``let maxa = Number.MIN_SAFE_INTEGER;`   `    ``// Making one segment of length a` `    ``if` `(n >= a) {` `        ``maxa = Math.max(maxa,` `                   ``1 + maximumSegments(n - a, a, b, c));` `    ``}` `    ``// Making one segment of length b` `    ``if` `(n >= b) {` `        ``maxa = Math.max(maxa,` `                   ``1 + maximumSegments(n - b, a, b, c));` `    ``}` `    ``// Making one segment of length c` `    ``if` `(n >= c) {` `        ``maxa = Math.max(maxa,` `                   ``1 + maximumSegments(n - c, a, b, c));` `    ``}`   `    ``// Return maximum out of all possible segment` `    ``return` `maxa;` `}`   `// Driver code` `let n = 7, a = 5, b = 2, c = 5;`   `// Function call` `console.log(maximumSegments(n, a, b, c));`   `// This code is contributed by poojaagarwal2.`

Output

`2`

Time Complexity: O(3n)
Auxiliary Space : O(n)

Optimized 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] = 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

 ``

## Javascript

 ``

Output

`2`

Time Complexity: O(N), as we are using a loop to traverse N times.
Auxiliary Space: O(N), as we are using extra space for dp array.

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