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# Count number of ways to reach a given score in a game

• Difficulty Level : Medium
• Last Updated : 26 Sep, 2022

Consider a game where players can score 3, 5, or 10 points in a move. Given a total score of N, The task is to find the number of ways to reach the given score.

Examples:

Input: n = 20
Output: 4
Explanation: There are following 4 ways to reach 20: (10, 10), (5, 5, 10), (5, 5, 5, 5), (3, 3, 3, 3, 3, 5)

Input: n = 13
Output: 2
Explanation: There are following 2 ways to reach 13: (3, 5, 5), (3, 10)

Recommended Practice

The Given problem is a variation of coin change problem.

## Finding the Number of Ways By Using the Bottom Up Dynamic Programming:

Using the Bottom up Dynamic Programming Approach to find the Number of Ways to reach the Given score.

Follow the below steps to Implement the idea:

• Create an array table[] of size N+1 to store counts of all scores from 0 to N.
• For every possible move (3, 5, and 10), increment the number of ways to reach the current score x i.e. table[x] with ways in which those scores can be reached from where x is reachable i.e. (x – 3), (x – 5), (x – 10).
• Return table[N] .

Below is the Implementation of the above approach:

## C++

 `// A C++ program to count number of``// possible ways to a given score``// can be reached in a game where a``// move can earn 3 or 5 or 10``#include ``using` `namespace` `std;` `// Returns number of ways``// to reach score n``int` `count(``int` `n)``{``    ``// table[i] will store count``    ``// of solutions for value i.``    ``int` `table[n + 1], i;` `    ``// Initialize all table``    ``// values as 0``    ``for` `(``int` `j = 0; j < n + 1; j++)``        ``table[j] = 0;` `    ``// Base case (If given value is 0)``    ``table = 1;` `    ``// One by one consider given 3 moves``    ``// and update the table[] values after``    ``// the index greater than or equal to``    ``// the value of the picked move``    ``for` `(i = 3; i <= n; i++)``        ``table[i] += table[i - 3];` `    ``for` `(i = 5; i <= n; i++)``        ``table[i] += table[i - 5];` `    ``for` `(i = 10; i <= n; i++)``        ``table[i] += table[i - 10];` `    ``return` `table[n];``}` `// Driver Code``int` `main(``void``)``{``    ``int` `n = 20;``    ``cout << ``"Count for "` `<< n << ``" is "` `<< count(n) << endl;` `    ``n = 13;``    ``cout << ``"Count for "` `<< n << ``" is "` `<< count(n) << endl;``    ``return` `0;``}` `// This code is contributed``// by Shivi_Aggarwal`

## C

 `// A C program to count number of possible ways to a given``// score can be reached in a game where a move can earn 3 or``// 5 or 10``#include ` `// Returns number of ways to reach score n``int` `count(``int` `n)``{``    ``// table[i] will store count of solutions for``    ``// value i.``    ``int` `table[n + 1], i;` `    ``// Initialize all table values as 0``    ``for` `(``int` `i = 0; i < n + 1; i++) {``        ``table[i] = 0;``    ``}` `    ``// Base case (If given value is 0)``    ``table = 1;` `    ``// One by one consider given 3 moves and update the``    ``// table[] values after the index greater than or equal``    ``// to the value of the picked move``    ``for` `(i = 3; i <= n; i++)``        ``table[i] += table[i - 3];``    ``for` `(i = 5; i <= n; i++)``        ``table[i] += table[i - 5];``    ``for` `(i = 10; i <= n; i++)``        ``table[i] += table[i - 10];` `    ``return` `table[n];``}` `// Driver program``int` `main(``void``)``{``    ``int` `n = 20;``    ``printf``(``"Count for %d is %d\n"``, n, count(n));` `    ``n = 13;``    ``printf``(``"Count for %d is %d"``, n, count(n));``    ``return` `0;``}`

## Java

 `// Java program to count number of``// possible ways to a given score``// can be reached in a game where``// a move can earn 3 or 5 or 10``import` `java.util.Arrays;` `class` `GFG {``    ``// Returns number of ways to reach score n``    ``static` `int` `count(``int` `n)``    ``{``        ``// table[i] will store count of solutions for``        ``// value i.``        ``int` `table[] = ``new` `int``[n + ``1``], i;` `        ``// Initialize all table values as 0``        ``Arrays.fill(table, ``0``);` `        ``// Base case (If given value is 0)``        ``table[``0``] = ``1``;` `        ``// One by one consider given 3``        ``// moves and update the table[]``        ``// values after the index greater``        ``// than or equal to the value of``        ``// the picked move``        ``for` `(i = ``3``; i <= n; i++)``            ``table[i] += table[i - ``3``];``        ``for` `(i = ``5``; i <= n; i++)``            ``table[i] += table[i - ``5``];``        ``for` `(i = ``10``; i <= n; i++)``            ``table[i] += table[i - ``10``];` `        ``return` `table[n];``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``20``;``        ``System.out.println(``"Count for "` `+ n + ``" is "``                           ``+ count(n));` `        ``n = ``13``;``        ``System.out.println(``"Count for "` `+ n + ``" is "``                           ``+ count(n));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python program to count number of possible ways to a given``# score can be reached in a game where a move can earn 3 or``# 5 or 10.` `# Returns number of ways to reach score n.`  `def` `count(n):` `    ``# table[i] will store count of solutions for value i.``    ``# Initialize all table values as 0.``    ``table ``=` `[``0` `for` `i ``in` `range``(n``+``1``)]` `    ``# Base case (If given value is 0)``    ``table[``0``] ``=` `1` `    ``# One by one consider given 3 moves and update the``    ``# table[] values after the index greater than or equal``    ``# to the value of the picked move.``    ``for` `i ``in` `range``(``3``, n``+``1``):``        ``table[i] ``+``=` `table[i``-``3``]``    ``for` `i ``in` `range``(``5``, n``+``1``):``        ``table[i] ``+``=` `table[i``-``5``]``    ``for` `i ``in` `range``(``10``, n``+``1``):``        ``table[i] ``+``=` `table[i``-``10``]` `    ``return` `table[n]`  `# Driver Program``n ``=` `20``print``(``'Count for'``, n, ``'is'``, count(n))` `n ``=` `13``print``(``'Count for'``, n, ``'is'``, count(n))` `# This code is contributed by Soumen Ghosh`

## C#

 `// C# program to count number of``// possible ways to a given score``// can be reached in a game where``// a move can earn 3 or 5 or 10``using` `System;` `class` `GFG {` `    ``// Returns number of ways to reach``    ``// score n``    ``static` `int` `count(``int` `n)``    ``{` `        ``// table[i] will store count``        ``// of solutions for value i.``        ``int``[] table = ``new` `int``[n + 1];` `        ``// Initialize all table values``        ``// as 0``        ``for` `(``int` `j = 0; j < n + 1; j++)``            ``table[j] = 0;` `        ``// Base case (If given value is 0)``        ``table = 1;` `        ``// One by one consider given 3``        ``// moves and update the table[]``        ``// values after the index greater``        ``// than or equal to the value of``        ``// the picked move``        ``for` `(``int` `i = 3; i <= n; i++)``            ``table[i] += table[i - 3];``        ``for` `(``int` `i = 5; i <= n; i++)``            ``table[i] += table[i - 5];``        ``for` `(``int` `i = 10; i <= n; i++)``            ``table[i] += table[i - 10];` `        ``return` `table[n];``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 20;``        ``Console.WriteLine(``"Count for "` `+ n + ``" is "``                          ``+ count(n));` `        ``n = 13;``        ``Console.Write(``"Count for "` `+ n + ``" is "` `+ count(n));``    ``}``}` `// This code is contributed by nitin mittal.`

## PHP

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## Javascript

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Output

```Count for 20 is 4
Count for 13 is 2```

Time Complexity: O(N), Where N is the score to be obtained.
Auxiliary Space: O(N), Where N is the score to be obtained.

Exercise: How to count score when (10, 5, 5), (5, 5, 10) and (5, 10, 5) are considered as different sequences of moves. Similarly, (5, 3, 3), (3, 5, 3) and (3, 3, 5) are considered different.