Related Articles

# Number of ways to cut a stick of length N into in even length at most K units long pieces

• Difficulty Level : Medium
• Last Updated : 01 Jun, 2021

Given a rod of length N units, the task is to find the number of ways to cut the rod into parts such that the length of each part is even and each part is at most K units.
Examples:

Input: N = 6, K = 4
Output:
Explanation:
Rod of length 6 units needs to be into parts having length at most 4 units. Hence cut the rod in three ways:
Way 1: 2 units + 2 units + 2 units
Way 2: 2 units + 4 units
Way 3: 4 units + 2 units
Input: N = 4, K = 2
Output:
Explanation:
Rod of length 4 units needs to be into parts having length at most 2 units. Hence cut the rod in 2 + 2 units.

Approach: The idea is to use dynamic programming where the optimal sub-structure is that the length of each part should be even. Count all the way to cut the rod by calling the function recursively for a piece obtained after a cut.
Below is the implementation of the above approach:

## C++14

 `// C++14 program for the above approach``#include ``using` `namespace` `std;` `// Recursive Function to count``// the total number of ways``int` `solve(``int` `n, ``int` `k, ``int` `mod, ``int` `dp[])``{``    ``// Base case if no-solution exist``    ``if` `(n < 0)``        ``return` `0;` `    ``// Condition if a solution exist``    ``if` `(n == 0)``        ``return` `1;` `    ``// Check if already calculated``    ``if` `(dp[n] != -1)``        ``return` `dp[n];` `    ``// Initialize counter``    ``int` `cnt = 0;``    ``for` `(``int` `i = 2; i <= k; i += 2) {``        ``// Recursive call``        ``cnt = (cnt % mod``               ``+ solve(n - i, k, mod, dp)``                     ``% mod)``              ``% mod;``    ``}` `    ``// Store the answer``    ``dp[n] = cnt;` `    ``// Return the answer``    ``return` `cnt;``}` `// Driver code``int` `main()``{` `    ``const` `int` `mod = 1e9 + 7;``    ``int` `n = 4, k = 2;``    ``int` `dp[n + 1];``    ``memset``(dp, -1, ``sizeof``(dp));``    ``int` `ans = solve(n, k, mod, dp);``    ``cout << ans << ``'\n'``;` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG{` `// Recursive Function to count``// the total number of ways``static` `int` `solve(``int` `n, ``int` `k, ``int` `mod, ``int` `dp[])``{``    ` `    ``// Base case if no-solution exist``    ``if` `(n < ``0``)``        ``return` `0``;` `    ``// Condition if a solution exist``    ``if` `(n == ``0``)``        ``return` `1``;` `    ``// Check if already calculated``    ``if` `(dp[n] != -``1``)``        ``return` `dp[n];` `    ``// Initialize counter``    ``int` `cnt = ``0``;``    ``for``(``int` `i = ``2``; i <= k; i += ``2``)``    ``{``        ` `        ``// Recursive call``        ``cnt = (cnt % mod + solve(n - i, k, mod,``               ``dp) % mod) % mod;``    ``}` `    ``// Store the answer``    ``dp[n] = cnt;` `    ``// Return the answer``    ``return` `cnt;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `mod = (``int``)(1e9 + ``7``);``    ``int` `n = ``4``, k = ``2``;``    ` `    ``int` `[]dp = ``new` `int``[n + ``1``];``    ``for``(``int` `i = ``0``; i < n + ``1``; i++)``        ``dp[i] = -``1``;``        ` `    ``int` `ans = solve(n, k, mod, dp);``    ` `    ``System.out.println(ans);``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program for the above approach` `# Recursive function to count``# the total number of ways``def` `solve(n, k, mod, dp):``    ` `    ``# Base case if no-solution exist``    ``if` `(n < ``0``):``        ``return` `0` `    ``# Condition if a solution exist``    ``if` `(n ``=``=` `0``):``        ``return` `1` `    ``# Check if already calculated``    ``if` `(dp[n] !``=` `-``1``):``        ``return` `dp[n]` `    ``# Initialize counter``    ``cnt ``=` `0``    ``for` `i ``in` `range``(``2``, k ``+` `1``, ``2``):``        ` `        ``# Recursive call``        ``cnt ``=` `((cnt ``%` `mod ``+``                ``solve(n ``-` `i, k, mod, dp) ``%``                                    ``mod) ``%` `mod)``                                   ` `    ``# Store the answer``    ``dp[n] ``=` `cnt` `    ``# Return the answer``    ``return` `int``(cnt)` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``mod ``=` `1e9` `+` `7``    ``n ``=` `4``    ``k ``=` `2``    ` `    ``dp ``=` `[``-``1``] ``*` `(n ``+` `1``)``    ``ans ``=` `solve(n, k, mod, dp)``    ` `    ``print``(ans)``    ` `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Recursive function to count``// the total number of ways``static` `int` `solve(``int` `n, ``int` `k, ``int` `mod, ``int` `[]dp)``{``    ` `    ``// Base case if no-solution exist``    ``if` `(n < 0)``        ``return` `0;` `    ``// Condition if a solution exist``    ``if` `(n == 0)``        ``return` `1;` `    ``// Check if already calculated``    ``if` `(dp[n] != -1)``        ``return` `dp[n];` `    ``// Initialize counter``    ``int` `cnt = 0;``    ``for``(``int` `i = 2; i <= k; i += 2)``    ``{``        ` `        ``// Recursive call``        ``cnt = (cnt % mod + solve(n - i, k, mod,``               ``dp) % mod) % mod;``    ``}` `    ``// Store the answer``    ``dp[n] = cnt;` `    ``// Return the answer``    ``return` `cnt;``}` `// Driver code``public` `static` `void` `Main(String[] args)``{``    ``int` `mod = (``int``)(1e9 + 7);``    ``int` `n = 4, k = 2;``    ` `    ``int` `[]dp = ``new` `int``[n + 1];``    ``for``(``int` `i = 0; i < n + 1; i++)``        ``dp[i] = -1;``        ` `    ``int` `ans = solve(n, k, mod, dp);``    ` `    ``Console.WriteLine(ans);``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``
Output:
`1`

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes arrow_drop_up