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

Consider a game where a player can score 3 or 5 or 10 points in a move. Given a total score n, find number of ways to reach the given score.

Examples:

Input: n = 20
Output: 4

There are following 4 ways to reach 20
(10, 10)
(5, 5, 10)
(5, 5, 5, 5)
(3, 3, 3, 3, 3, 5)

This problem is a variation of coin change problem

Method 1 : With the help of DP and CoinChange code

int count( int S[], int m, int n )
{
    int i, j, x, y;

    // We need n+1 rows as the table is consturcted in bottom up manner using 
    // the base case 0 value case (n = 0)
    int table[n+1][m];
    
    // Fill the enteries for 0 value case (n = 0)
    for (i=0; i<m; i++)
        table[0][i] = 1;

    // Fill rest of the table enteries in bottom up manner  
    for (i = 1; i < n+1; i++)
    {
        for (j = 0; j < m; j++)
        {
            // Count of solutions including S[j]
            x = (i-S[j] >= 0)? table[i - S[j]][j]: 0;

            // Count of solutions excluding S[j]
            y = (j >= 1)? table[i][j-1]: 0;

            // total count
            table[i][j] = x + y;
        }
    }
    return table[n][m-1];
}

Method 2 :   The idea is to create a table of size n+1 to store counts of all scores from 0 to n. For every possible move (3, 5 and 10), increment values in table.


#include <stdio.h>

int count(int n)
{
    int table[n+1], i;

    memset(table, 0, sizeof(table));

    table[0] = 1;

    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];
}



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