Given a stick of size N, find the number of ways in which it can be cut into K pieces such that length of every piece is greater than 0.
Examples :
Input : N = 5
K = 2
Output : 4
Input : N = 15
K = 5
Output : 1001
Solving this question is equivalent to solving the mathematics equation x1 + x2 + ….. + xK = N
We can solve this by using the bars and stars method in Combinatorics, from which we obtain the fact that the number of positive integral solutions to this equation is (N – 1)C(K – 1), where NCK is N! / ((N – K) ! * (K!)), where ! stands for factorial.
In C++ and Java, for large values of factorials, there might be overflow errors. In that case we can introduce a large prime number such as 107 + 7 to mod the answer. We can calculate nCr % p by using Lucas Theorem.
However, python can handle large values without overflow.
C++
// C++ program to calculate the number of ways// to divide a stick of length n into k pieces#include <bits/stdc++.h>using namespace std;// function to generate nCk or nChoosekunsigned long long nCr(unsigned long long n, unsigned long long r){ if (n < r) return 0; // Reduces to the form n! / n! if (r == 0) return 1; // nCr has been simplified to this form by // expanding numerator and denominator to // the form n(n - 1)(n - 2)...(n - r + 1) // ----------------------------- // (r!) // in the above equation, (n - r)! is cancelled // out in the numerator and denominator unsigned long long numerator = 1; for (int i = n; i > n - r; i--) numerator = (numerator * i); unsigned long long denominator = 1; for (int i = 1; i < r + 1; i++) denominator = (denominator * i); return (numerator / denominator);}// Returns number of ways to cut // a rod of length N into K pieces.unsigned long long countWays(unsigned long long N, unsigned long long K){ return nCr(N - 1, K - 1);}// Driver codeint main(){ unsigned long long N = 5; unsigned long long K = 2; cout << countWays(N, K); return 0;} |
Java
// Java program to find the number of ways in which// a stick of length n can be divided into K piecesimport java.io.*;import java.util.*;class GFG{ // function to generate nCk or nChoosek public static int nCr(int n, int r) { if (n < r) return 0; // Reduces to the form n! / n! if (r == 0) return 1; // nCr has been simplified to this form by // expanding numerator and denominator to // the form n(n - 1)(n - 2)...(n - r + 1) // ----------------------------- // (r!) // in the above equation, (n-r)! is cancelled // out in the numerator and denominator int numerator = 1; for (int i = n ; i > n - r ; i--) numerator = (numerator * i); int denominator = 1; for (int i = 1 ; i < r + 1 ; i++) denominator = (denominator * i); return (numerator / denominator); } // Returns number of ways to cut // a rod of length N into K pieces public static int countWays(int N, int K) { return nCr(N - 1, K - 1); } public static void main(String[] args) { int N = 5; int K = 2; System.out.println(countWays(N, K)); }} |
Python3
# Python program to find the number # of ways in which a stick of length # n can be divided into K pieces# function to generate nCk or nChoosekdef nCr(n, r): if (n < r): return 0 # reduces to the form n! / n! if (r == 0): return 1 # nCr has been simplified to this form by # expanding numerator and denominator to # the form n(n - 1)(n - 2)...(n - r + 1) # ----------------------------- # (r!) # in the above equation, (n-r)! is cancelled # out in the numerator and denominator numerator = 1 for i in range(n, n - r, -1): numerator = numerator * i denominator = 1 for i in range(1, r + 1): denominator = denominator * i return (numerator // denominator)# Returns number of ways to cut # a rod of length N into K pieces.def countWays(N, K) : return nCr(N - 1, K - 1);# Driver codeN = 5K = 2print(countWays(N, K)) |
C#
// C# program to find the number of // ways in which a stick of length n // can be divided into K piecesusing System;class GFG{ // function to generate nCk or nChoosek public static int nCr(int n, int r) { if (n < r) return 0; // Reduces to the form n! / n! if (r == 0) return 1; // nCr has been simplified to this form by // expanding numerator and denominator to // the form n(n - 1)(n - 2)...(n - r + 1) // ----------------------------- // (r!) // in the above equation, (n-r)! is cancelled // out in the numerator and denominator int numerator = 1; for (int i = n; i > n - r; i--) numerator = (numerator * i); int denominator = 1; for (int i = 1; i < r + 1; i++) denominator = (denominator * i); return (numerator / denominator); } // Returns number of ways to cut // a rod of length N into K pieces public static int countWays(int N, int K) { return nCr(N - 1, K - 1); } public static void Main() { int N = 5; int K = 2; Console.Write(countWays(N, K)); }}// This code is contributed by nitin mittal. |
PHP
<?php// PHP program to calculate the// number of ways to divide a // stick of length n into k pieces// function to generate nCk or nChoosekfunction nCr($n, $r){ if ($n < $r) return 0; // Reduces to the form n! / n! if ($r == 0) return 1; // nCr has been simplified to this form by // expanding numerator and denominator to // the form n(n - 1)(n - 2)...(n - r + 1) // ----------------------------- // (r!) // in the above equation, (n - r)! is cancelled // out in the numerator and denominator $numerator = 1; for ($i = $n; $i > $n - $r; $i--) $numerator = ($numerator * $i); $denominator = 1; for ($i = 1; $i < $r + 1; $i++) $denominator = ($denominator * $i); return (floor($numerator / $denominator));}// Returns number of ways to cut // a rod of length N into K pieces.function countWays($N, $K){ return nCr($N - 1, $K - 1);}// Driver code$N = 5;$K = 2;echo countWays($N, $K);return 0;// This code is contributed by nitin mittal.?> |
Javascript
<script>//Javascript Implementation// to calculate the number of ways// to divide a stick of length n into k pieces// function to generate nCk or nChoosekfunction nCr(n,r){ if (n < r) return 0; // Reduces to the form n! / n! if (r == 0) return 1; // nCr has been simplified to this form by // expanding numerator and denominator to // the form n(n - 1)(n - 2)...(n - r + 1) // ----------------------------- // (r!) // in the above equation, (n - r)! is cancelled // out in the numerator and denominator var numerator = 1; for (var i = n; i > n - r; i--) numerator = (numerator * i); var denominator = 1; for (var i = 1; i < r + 1; i++) denominator = (denominator * i); return Math.floor(numerator / denominator);}// Returns number of ways to cut // a rod of length N into K pieces.function countWays(N,K){ return nCr(N - 1, K - 1);}// Driver codevar N = 5;var K = 2;document.write(countWays(N, K));// This code is contributed by shubhamsingh10</script> |
Output
4
Time complexity: O(N)
Auxiliary space: O(1)
Exercise :
Extend the above problem with 0 length pieces allowed. Hint : The number of solutions can similarly be found by writing each xi as yi – 1, and we get an equation y1 + y2 + ….. + yK = N + K. The number of solutions to this equation is (N + K – 1)C(K – 1)
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