CSES Solutions – Coin Combinations I

Consider a money system consisting of N coins. Each coin has a positive integer value. Your task is to calculate the number of distinct ways you can produce a money sum X using the available coins.

Examples:

Input: N = 3, X = 9, coins[] = {2, 3, 5}
Output: 8
Explanation: There are 8 number of ways to make sum = 9.

• {2, 2, 5} = 2 + 2 + 5 = 9
• {2, 5, 2} = 2 + 5 + 2 = 9
• {5, 2, 2} = 5 + 2 + 2 = 9
• {3, 3, 3} = 3 + 3 + 3 = 9
• {2, 2, 2, 3} = 2 + 2 + 2 + 3 = 9
• {2, 2, 3, 2} = 2 + 2 + 3 + 2 = 9
• {2, 3, 2, 2} = 2 + 3 + 2 + 2 = 9
• {3, 2, 2, 2} = 3 + 2 + 2 + 2 = 9

Input: N = 2, X = 3, coins[] = {1, 2}
Output: 3
Explanation: There are 3 ways to make sum = 3

• {1, 1, 1} = 1 + 1 + 1 = 3
• {1, 2} = 1 + 2 = 3
• {2, 1} = 2 + 1 = 3

Approach: To solve the problem, follow the below idea:

The problem can be solved using Dynamic Programming. We can maintain a dp[] array, such that dp[i] stores the number of distinct ways to produce sum = i. We can iterate i from 1 to X, and find the number of distinct ways to make sum = i. For any sum i, we assume that the last coin used was the jth coin where j will range from 0 to N – 1. For jth coin, the value will be coins[j], so the number of distinct ways to make sum = i, if the last coin used was the jth coin is equal to dp[i – coins[j]] + 1. Similarly, for every coin we can assume that this coin was the last coin used to make sum = i, and add the number of ways to the final answer. The formula for the ith sum will be: dp[i] = sum(dp[i – coins[j]]), for all j from 0 to N – 1.

Also, the above formula will be true only when coins[j] >= i.

Step-by-step algorithm:

• Maintain a dp[] array such that dp[i] stores the number of distinct ways to make sum = i.
• Initialize dp[0] = 1 as there is only 1 way to make sum = 0, that is to not take any coin.
• Iterate i from 1 to X and calculate number of ways to make sum = i.
• Iterate j over all possible coins assuming the jth coin to be last coin which was selected to make sum = i and add dp[i – coins[j]] to dp[i].
• After all the iterations, return the final answer as dp[X].

Below is the implementation of the algorithm:

C++

#include <bits/stdc++.h>
#define ll long long int
#define mod 1000000007
using namespace std;

// function to find the number of distinct ways to make sum
// = X
ll solve(ll N, ll X, vector<ll>& coins)
{
    // dp[] array such that dp[i] stores the number of
    // distinct ways to make sum = i
    ll dp[X + 1] = {};

    // There is only 1 way to make sum = 0, that is to not
    // select any coin
    dp[0] = 1;

    // Iterate over all possible sums from 1 to X
    for (int i = 1; i <= X; i++) {
        // Iterate over all the N coins
        for (int j = 0; j < N; j++) {
            // Check if it is possible to have jth coin as
            // the last coin to construct sum = i
            if (coins[j] > i)
                continue;
            dp[i] = (dp[i] + dp[i - coins[j]]) % mod;
        }
    }
    // Return the number of ways to make sum = X
    return dp[X];
}

int main()
{
    // Sample Input
    ll N = 3, X = 9;
    vector<ll> coins = { 2, 3, 5 };

    cout << solve(N, X, coins) << "\n";
}

Java

import java.util.*;

public class CoinChangeWays {
    static final int MOD = 1000000007;

    // Function to find the number of distinct ways to make sum = X
    static long solve(int N, int X, List<Integer> coins) {
        // dp[] array such that dp[i] stores the number of distinct ways
        // to make sum = i
        long[] dp = new long[X + 1];

        // There is only 1 way to make sum = 0, that is to not select any coin
        dp[0] = 1;

        // Iterate over all possible sums from 1 to X
        for (int i = 1; i <= X; i++) {
            // Iterate over all the N coins
            for (int j = 0; j < N; j++) {
                // Check if it is possible to have jth coin as
                // the last coin to construct sum = i
                if (coins.get(j) > i) {
                    continue;
                }
                dp[i] = (dp[i] + dp[i - coins.get(j)]) % MOD;
            }
        }

        // Return the number of ways to make sum = X
        return dp[X];
    }

    public static void main(String[] args) {
        // Sample Input
        int N = 3, X = 9;
        List<Integer> coins = Arrays.asList(2, 3, 5);

        System.out.println(solve(N, X, coins));
    }
}

// This code is contributed by shivamgupta0987654321

C#

using System;
using System.Collections.Generic;

public class CoinChangeWays
{
    static readonly int MOD = 1000000007;

    // Function to find the number of distinct ways to make sum = X
    static long Solve(int N, int X, List<int> coins)
    {
        // dp[] array such that dp[i] stores the number of distinct ways
        // to make sum = i
        long[] dp = new long[X + 1];

        // There is only 1 way to make sum = 0, that is to not select any coin
        dp[0] = 1;

        // Iterate over all possible sums from 1 to X
        for (int i = 1; i <= X; i++)
        {
            // Iterate over all the N coins
            for (int j = 0; j < N; j++)
            {
                // Check if it is possible to have jth coin as
                // the last coin to construct sum = i
                if (coins[j] > i)
                {
                    continue;
                }
                dp[i] = (dp[i] + dp[i - coins[j]]) % MOD;
            }
        }

        // Return the number of ways to make sum = X
        return dp[X];
    }

    public static void Main(string[] args)
    {
        // Sample Input
        int N = 3, X = 9;
        List<int> coins = new List<int> { 2, 3, 5 };

        Console.WriteLine(Solve(N, X, coins));
    }
}

JavaScript

// Function to find the number of distinct ways to make sum = X
function solve(N, X, coins) {
    // Initialize a dp array such that dp[i] stores the number of
    // distinct ways to make sum = i
    let dp = new Array(X + 1).fill(0);

    // There is only 1 way to make sum = 0, that is to not
    // select any coin
    dp[0] = 1;

    // Iterate over all possible sums from 1 to X
    for (let i = 1; i <= X; i++) {
        // Iterate over all the N coins
        for (let j = 0; j < N; j++) {
            // Check if it is possible to have jth coin as
            // the last coin to construct sum = i
            if (coins[j] > i)
                continue;
            dp[i] = (dp[i] + dp[i - coins[j]]) % 1000000007;
        }
    }
    // Return the number of ways to make sum = X
    return dp[X];
}

// Sample Input
let N = 3, X = 9;
let coins = [2, 3, 5];

console.log(solve(N, X, coins));

Python3

# Importing the required libraries
from typing import List

# Defining the mod constant
mod = 1000000007

def solve(N: int, X: int, coins: List[int]) -> int:
    # dp[] list such that dp[i] stores the number of
    # distinct ways to make sum = i
    dp = [0] * (X + 1)

    # There is only 1 way to make sum = 0, that is to not
    # select any coin
    dp[0] = 1

    # Iterate over all possible sums from 1 to X
    for i in range(1, X + 1):
        # Iterate over all the N coins
        for j in range(N):
            # Check if it is possible to have jth coin as
            # the last coin to construct sum = i
            if coins[j] > i:
                continue
            dp[i] = (dp[i] + dp[i - coins[j]]) % mod

    # Return the number of ways to make sum = X
    return dp[X]

def main():
    # Sample Input
    N = 3
    X = 9
    coins = [2, 3, 5]

    print(solve(N, X, coins))

# Calling the main function
if __name__ == "__main__":
    main()

Output

8

Time Complexity: O(N * X), where N is the size of array coins[] and X is the sum we make using the coins.
Auxiliary Space: O(X)