# Maximise number of cuts in a rod if it can be cut only in given 3 sizes

Given a rod of length

N

meters, and the rod can be cut in only 3 sizes

A

,

B

and

C

. The task is to maximizes the number of cuts in rod. If it is impossible to make cut then print

-1

.

Examples:

Input: N = 17, A = 10, B = 11, C = 3 Output: 3 Explanation: The maximum cut can be obtain after making 2 cut of length 3 and one cut of length 11. Input: N = 10, A = 9, B = 7, C = 11 Output: -1 Explanation: It is impossible to make any cut so output will be -1.

Naive Approach:

• Let us assume x, y, and z numbers of rods of sizes A, B, and C respectively are cut. And this can be written as a linear equation: x*A + y*B + z*C = N
• Now, simply iterate over all possible value of x and y and compute z using (N – x*A + y*B) / c.
• If x*A + y*B + z*C = N, then it is one of the possible answers.
• Finally compute the maximum value of x + y + z.

Time Complexity:

O(N

2

)

Auxiliary Space:

O(1)

Efficient Approach:

The problem can be solve using Dynamic Programming.

1. Create a dp[] array of size N and initialise all value to INT_MIN.
2. Set dp[0] = 0, as it will be base case for our approach.
3. Iterate from 1 to N and check if it is possible to make a cut of any of possible length i.e A, B and C, and update dp[i] to minimum of all.
4. dp[i] = min (dp[i], 1 + subresult) where, subresult = min(dp[i – A], min(dp[i – B], dp[i – C]))

1. Here is the implementation of the above approach:
2. ## C++

 `// A Dynamic Programming solution for ` `// Maximum Rod cutting problem `   `#include ` `using` `namespace` `std; `   `// function that eturns the maximum ` `// number of rods that can be ` `// made from the rod of length N ` `int` `cuttingRod(``int` `arr[], ``int` `N) ` `{ ` `    ``int` `dp[N + 1]; `   `    ``// Initializing the number of rods we ` `    ``// can make from length 0 ` `    ``dp[0] = 0; `   `    ``// Iterating over lengths that can ` `    ``// be formed ` `    ``for` `(``int` `i = 1; i <= N; i++) { `   `        ``// Initializing the possible ` `        ``// cuts as infinite ` `        ``dp[i] = INT_MIN; `   `        ``// Cutting the desired lengths ` `        ``for` `(``int` `j = 0; j < 3; j++) { `   `            ``// Checking whether the length of ` `            ``// rod becomes 0 or if after cutting ` `            ``// the rod, it becomes useless ` `            ``if` `((i - arr[j]) >= 0 ` `                ``&& dp[i - arr[j]] != INT_MIN) { `   `                ``// Choosing the maximum ` `                ``// possible desired ` `                ``// length cuts to be made ` `                ``dp[i] = max(dp[i - arr[j]] + 1, ` `                            ``dp[i]); ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `dp[N]; ` `} `   `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 17; ` `    ``int` `arr[] = { 10, 11, 3 }; ` `    ``cout << cuttingRod(arr, N); ` `    ``return` `0; ` `} `

## Java

 `// A Dynamic Programming solution for` `// Maximum Rod cutting problem` `class` `GFG{`   `// Function that eturns the maximum` `// number of rods that can be` `// made from the rod of length N` `static` `int` `cuttingRod(``int` `arr[], ``int` `N)` `{` `    ``int` `[]dp = ``new` `int``[N + ``1``];`   `    ``// Initializing the number of rods we` `    ``// can make from length 0` `    ``dp[``0``] = ``0``;`   `    ``// Iterating over lengths that can` `    ``// be formed` `    ``for``(``int` `i = ``1``; i <= N; i++)` `    ``{` `        `  `        ``// Initializing the possible` `        ``// cuts as infinite` `        ``dp[i] = Integer.MIN_VALUE;`   `        ``// Cutting the desired lengths` `        ``for``(``int` `j = ``0``; j < ``3``; j++)` `        ``{` `            `  `            ``// Checking whether the length of ` `            ``// rod becomes 0 or if after cutting ` `            ``// the rod, it becomes useless ` `            ``if` `((i - arr[j]) >= ``0` `&& ` `              ``dp[i - arr[j]] != Integer.MIN_VALUE) ` `            ``{` `                `  `                ``// Choosing the maximum` `                ``// possible desired` `                ``// length cuts to be made` `                ``dp[i] = Math.max(dp[i - arr[j]] + ``1``,` `                                 ``dp[i]);` `            ``}` `        ``}` `    ``}` `    ``return` `dp[N];` `}`   `// Driver code` `public` `static` `void` `main(String[] args)` `{` `    ``int` `N = ``17``;` `    ``int` `arr[] = { ``10``, ``11``, ``3` `};` `    `  `    ``System.out.print(cuttingRod(arr, N));` `}` `}`   `// This code is contributed by Princi Singh`

## Python3

 `# A Dynamic Programming solution for` `# Maximum Rod cutting problem` `import` `sys`   `# Function that returns the maximum` `# number of rods that can be` `# made from the rod of length N` `def` `cuttingRod(arr, N):`   `    ``dp ``=` `(N ``+` `1``) ``*` `[``0``]`   `    ``# Initializing the number of rods we` `    ``# can make from length 0` `    ``dp[``0``] ``=` `0`   `    ``# Iterating over lengths that can` `    ``# be formed` `    ``for` `i ``in` `range` `(``1``, N ``+` `1``):`   `        ``# Initializing the possible` `        ``# cuts as infinite` `        ``dp[i] ``=` `-``sys.maxsize ``-` `1`   `        ``# Cutting the desired lengths` `        ``for` `j ``in` `range``(``3``):`   `            ``# Checking whether the length of` `            ``# rod becomes 0 or if after cutting` `            ``# the rod, it becomes useless` `            ``if` `((i ``-` `arr[j]) >``=` `0` `and` `              ``dp[i ``-` `arr[j]] !``=` `-``sys.maxsize``-``1``):`   `                ``# Choosing the maximum` `                ``# possible desired` `                ``# length cuts to be made` `                ``dp[i] ``=` `max``(dp[i ``-` `arr[j]] ``+` `1``,` `                            ``dp[i])` `    ``return` `dp[N]`   `# Driver code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``N ``=` `17` `    ``arr ``=` `[ ``10``, ``11``, ``3` `]` `    `  `    ``print``(cuttingRod(arr, N))`   `# This code is contributed by chitranayal`

## C#

 `// A Dynamic Programming solution for` `// Maximum Rod cutting problem` `using` `System;`   `class` `GFG{`   `// Function that eturns the maximum` `// number of rods that can be` `// made from the rod of length N` `static` `int` `cuttingRod(``int``[] arr, ``int` `N)` `{` `    ``int` `[]dp = ``new` `int``[N + 1];`   `    ``// Initializing the number of rods we` `    ``// can make from length 0` `    ``dp[0] = 0;`   `    ``// Iterating over lengths that can` `    ``// be formed` `    ``for``(``int` `i = 1; i <= N; i++)` `    ``{` `        `  `        ``// Initializing the possible` `        ``// cuts as infinite` `        ``dp[i] = Int32.MinValue;`   `        ``// Cutting the desired lengths` `        ``for``(``int` `j = 0; j < 3; j++)` `        ``{` `            `  `            ``// Checking whether the length of ` `            ``// rod becomes 0 or if after cutting ` `            ``// the rod, it becomes useless ` `            ``if` `((i - arr[j]) >= 0 && ` `              ``dp[i - arr[j]] != Int32.MinValue) ` `            ``{` `                `  `                ``// Choosing the maximum` `                ``// possible desired` `                ``// length cuts to be made` `                ``dp[i] = Math.Max(dp[i - arr[j]] + 1,` `                                 ``dp[i]);` `            ``}` `        ``}` `    ``}` `    ``return` `dp[N];` `}`   `// Driver code` `public` `static` `void` `Main()` `{` `    ``int` `N = 17;` `    ``int``[] arr = { 10, 11, 3 };` `    `  `    ``Console.Write(cuttingRod(arr, N));` `}` `}`   `// This code is contributed by code_hunt`

## Javascript

 `// A Dynamic Programming solution for ` `// Maximum Rod cutting problem `   `// function that eturns the maximum ` `// number of rods that can be ` `// made from the rod of length N ` `const cuttingRod = (arr, N) => {` `    ``const dp = ``new` `Array(N + 1);` `    `  `    ``// Initializing the number of rods we ` `    ``// can make from length 0 ` `    ``dp[0] = 0;` `    `  `    ``// Iterating over lengths that can ` `    ``// be formed ` `    ``for` `(let i = 1; i <= N; i++) {` `        `  `        ``// Initializing the possible ` `        ``// cuts as infinite ` `        ``dp[i] = Number.MIN_SAFE_INTEGER;` `        `  `         ``// Cutting the desired lengths ` `        ``for` `(let j = 0; j < 3; j++) {` `            `  `             ``// Checking whether the length of ` `            ``// rod becomes 0 or if after cutting ` `            ``// the rod, it becomes useless ` `            ``if` `((i - arr[j]) >= 0 && dp[i - arr[j]] !== Number.MIN_SAFE_INTEGER) {` `                `  `                ``// Choosing the maximum ` `                ``// possible desired ` `                ``// length cuts to be made ` `                ``dp[i] = Math.max(dp[i - arr[j]] + 1, dp[i]);` `            ``}` `        ``}` `    ``}` `    `  `    ``return` `dp[N];` `}`   `// Driver code ` `const N = 17;` `const arr = [10, 11, 3];` `console.log(cuttingRod(arr, N));`

3. Output

```3

```
4. Time Complexity:
5. O (N)
6. Auxiliary Space:
7. O (N)

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