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)
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 <iostream>using namespace std;// Returns number of ways// to reach score nint 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[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 Codeint 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 <stdio.h>// Returns number of ways to reach score nint 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[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 programint 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 10import 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 Programn = 20print('Count for', n, 'is', count(n))n = 13print('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 10using 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[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 (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
<?php// PHP 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 function counts($n) { // table[i] will store count // of solutions for value i. // Initialize all table // values as 0 for($j = 0; $j < $n + 1; $j++) $table[$j] = 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 $n = 20; echo "Count for ";echo($n);echo (" is "); echo counts($n); $n = 13;echo ("\n") ;echo "Count for ";echo($n);echo (" is " ); echo counts($n); // This code is contributed // by Shivi_Aggarwal ?> |
Javascript
<script>// A JavaScript 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 nfunction count(n){ // table[i] will store count // of solutions for value i. let table = new Array(n + 1), i; // Initialize all table // values as 0 for(let j = 0; j < n + 1; j++) table[j] = 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 let n = 20; document.write("Count for " + n + " is " + count(n) + "<br>"); n = 13; document.write("Count for "+ n + " is " + count(n) + "<br>");// This code is contributed by Surbhi Tyagi.</script> |
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.
Another Approach:
- Define a function countWays that takes an integer parameter n representing the total score to be obtained and returns an integer representing the number of ways to reach the given score.
- Create a vector ways of size n + 1 to store the counts of all scores from 0 to n. Initialize all the values of the vector to 0.
- Set the base case: If the given value n is 0, set ways[0] to 1, as there is one way to obtain 0 score, i.e., by not making any moves.
- Define an integer array moves of size 3, which contains all the possible moves, i.e., 3, 5, and 10.
- Using nested loops, iterate over all the possible moves and scores that can be obtained using those moves. Starting from the moves[i] score to the total score n, update the ways[j] value with the sum of ways[j – moves[i]] and the current value of ways[j]. This is because the number of ways to obtain the score j using the move moves[i] is equal to the sum of the number of ways to obtain the score j – moves[i] using the same move and the current number of ways to obtain the score j using all the possible moves.
- Return the value of ways[n], which represents the number of ways to obtain the given score n using the given moves.
- Define the main function that calls the countWays function for two different values of n, i.e., 20 and 13, and print the number of ways obtained using the given moves for each value of 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 <iostream>#include <vector>using namespace std;// Returns number of ways// to reach score nint countWays(int n){// ways[i] will store count// of solutions for value i.vector<int> ways(n + 1, 0); // Base case (If given value is 0)ways[0] = 1;// Consider all possible movesint moves[] = {3, 5, 10};for (int i = 0; i < 3; i++) { for (int j = moves[i]; j <= n; j++) { ways[j] += ways[j - moves[i]]; }}return ways[n]; }// Driver Codeint main(void){int n = 20;cout << "Count for " << n << " is " << countWays(n) << endl; n = 13;cout << "Count for " << n << " is " << countWays(n) << endl;return 0;} |
Java
// A 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 10import java.util.*;public class ScoreCount { // Returns number of ways // to reach score n static int countWays(int n) { // ways[i] will store count // of solutions for value i. int[] ways = new int[n + 1]; // Base case (If given value is 0) ways[0] = 1; // Consider all possible moves int[] moves = { 3, 5, 10 }; for (int i = 0; i < 3; i++) { for (int j = moves[i]; j <= n; j++) { ways[j] += ways[j - moves[i]]; } } return ways[n]; } // Driver Code public static void main(String[] args) { int n = 20; System.out.println("Count for " + n + " is " + countWays(n)); n = 13; System.out.println("Count for " + n + " is " + countWays(n)); }}// This code is contributed by sarojmcy2e |
Python3
# 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 10def countWays(n: int) -> int: # ways[i] will store count of solutions for value i ways = [0] * (n + 1) # Base case (If given value is 0) ways[0] = 1 # Consider all possible moves moves = [3, 5, 10] for i in range(3): for j in range(moves[i], n+1): ways[j] += ways[j - moves[i]] return ways[n]# Driver Coden = 20print(f"Count for {n} is {countWays(n)}")n = 13print(f"Count for {n} is {countWays(n)}") |
C#
using System;using System.Collections.Generic;public class GFG{ // Returns number of ways // to reach score n public static int countWays(int n) { // ways[i] will store count // of solutions for value i. List<int> ways = new List<int>(); for (int i = 0; i <= n; i++) ways.Add(0); // Base case (If given value is 0) ways[0] = 1; // Consider all possible moves int[] moves = { 3, 5, 10 }; for (int i = 0; i < 3; i++) { for (int j = moves[i]; j <= n; j++) { ways[j] += ways[j - moves[i]]; } } return ways[n]; } // Driver Code public static void Main(string[] args) { int n = 20; Console.WriteLine("Count for {0} is {1}", n, countWays(n)); n = 13; Console.WriteLine("Count for {0} is {1}", n, countWays(n)); }} |
Javascript
// A JavaScript 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 10function countWays(n) { // ways[i] will store count of solutions for value i const ways = new Array(n + 1).fill(0); // Base case (If given value is 0) ways[0] = 1; // Consider all possible moves const moves = [3, 5, 10]; for (let i = 0; i < 3; i++) { for (let j = moves[i]; j <= n; j++) { ways[j] += ways[j - moves[i]]; } } return ways[n];}// Driver Codelet n = 20;console.log(`Count for ${n} is ${countWays(n)}`);n = 13;console.log(`Count for ${n} is ${countWays(n)}`); |
Output
Count for 20 is 4 Count for 13 is 2
Time Complexity: O(N)
Auxiliary Space: O(N)
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.
This article is contributed by Rajeev Arora. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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