Coin Change Problem in C++

Last Updated : 07 Aug, 2024

In this article, we will learn how to count all combinations of coins to make a given value sum using the C++ programming language. The Coin Change Problem involves finding the number of ways to make change for a given amount using a set of coin denominations.

Example:

Consider an integer array coins[] representing different coin denominations and an integer amount, determine the number of combinations of coins that sum up to the given amount.

Input:
coins = [1, 2, 5]
amount = 5

Output:
4

Explanation:
The different combinations are:
1) 1 + 1 + 1 + 1 + 1
2) 1 + 1 + 1 + 2
3) 1 + 2 + 2
4) 5

How to Solve Coin Change Problem in C++?

Method 1: Count All Combinations Using Recursion
Method 2: Count All Combinations Using Dynamic Programming (Memoization)
Method 3: Count All Combinations Using Dynamic Programming (Tabulation)
Method 4: Count All Combinations Using Dynamic Programming (Space Optimized)

1. Count All Combinations Using Recursion

In the recursive approach, we will generate all possible combinations and count those that sum to the given amount using recursion.

Recurrence Relation:

count(coins,n,sum) = count(coins,n,sum-count[n-1]) + count(coins,n-1,sum)

Coin-Change-With-Recursion — Coin Change Using Recursion

Approach:

Base Condition:
If the amount is 0, there is one way to make change (using no coins), so return 1.
If the amount is negative or there are no coins left, return 0 as no valid combination can be formed.
For each coin, recursively count the combinations by
including the coin (reduce the amount by the coin's value) and
excluding the coin (move to the next coin).

Below is the implementation of the above approach:

C++

// C++ program to solve coin change problem using recursion
#include <iostream>
#include <vector>
using namespace std;

// Returns the count of ways we can
// sum coins[0...n-1] coins to get sum "sum"
int count(int coins[], int n, int sum)
{

    // If sum is 0 then there is 1 solution
    // (do not include any coin)
    if (sum == 0)
        return 1;

    // If sum is less than 0 then no
    // solution exists
    if (sum < 0)
        return 0;

    // If there are no coins and sum
    // is greater than 0, then no
    // solution exist
    if (n <= 0)
        return 0;

    // count is sum of solutions (i)
    // including coins[n-1] (ii) excluding coins[n-1]
    return count(coins, n, sum - coins[n - 1]) + count(coins, n - 1, sum);
}

int main()
{
    int i, j;
    int coins[] = {1, 2, 3};
    int n = sizeof(coins) / sizeof(coins[0]);
    int sum = 5;

    cout << " " << count(coins, n, sum);

    return 0;
}

Output

Time Complexity: O(2^sum)
Auxiliary Space: O(sum)

2. Count All Combinations Using Dynamic Programming (Memoization)

We can further optimize the recursive solution by storing the results of previously solved subproblems in a 2D array to avoid redundant calculations.

Approach:

Initialize Memoization Table by creating a 2D array dp of size (n+1) x (amount+1) and initialize all values to -1.
For each subproblem:
if the result is already computed, return it from the table.
Otherwise, compute the result recursively, store it in the memoization table, and return the stored value.

Below is the implementation of the above approach:

C++

// C++ program to solve coin change problem  using Dynamic Programming (Memoization

#include <iostream>
#include <vector>
using namespace std;

// Recursive function to count the numeber of distinct ways
// to make the sum by using n coins

int count(vector<int> &coins, int n, int sum, vector<vector<int>> &dp)
{
    // Base Case
    if (sum == 0)
        return dp[n][sum] = 1;

    // If number of coins is 0 or sum is less than 0 then
    // there is no way to make the sum.
    if (n == 0 || sum < 0)
        return 0;

    // If the subproblem is previously calculated then
    // simply return the result
    if (dp[n][sum] != -1)
        return dp[n][sum];

    // Two options for the current coin
    return dp[n][sum] = count(coins, n, sum - coins[n - 1], dp) + count(coins, n - 1, sum, dp);
}
int32_t main()
{
    int tc = 1;
    // cin >> tc;
    while (tc--)
    {
        int n, sum;
        n = 3, sum = 5;
        vector<int> coins = {1, 2, 3};
        // 2d dp array to store previously calculated
        // results
        vector<vector<int>> dp(n + 1, vector<int>(sum + 1, -1));
        int res = count(coins, n, sum, dp);
        cout << res << endl;
    }
}

Output

Time Complexity: O(N*sum), where N is the number of coins and sum is the target sum.
Auxiliary Space: O(N*sum)

3. Count All Combinations Using Dynamic Programming (Tabulation)

We can use the bottom-up approach by building the solution iteratively using a table to store intermediate results.

Approach:

Create a 2D dp array with rows and columns equal to the number of coin denominations and target sum.
dp[0][0] will be set to 1 which represents the base case where the target sum is 0, and there is only one way to make the change by not selecting any coin.
Iterate through the rows of the dp array (i from 1 to n), representing the current coin being considered.
The inner loop iterates over the target sums (j from 0 to sum).
Add the number of ways to make change without using the current coin, i.e., dp[i][j] += dp[i-1][j].
Add the number of ways to make change using the current coin, i.e., dp[i][j] += dp[i][j-coins[i-1]].
dp[n][sum] will contain the total number of ways to make change for the given target sum using the available coin denominations.

Below is the implementation of above approach:

C++

// C++ program to solve coin change problem  using Dynamic Programming (Tabulation)

#include <iostream>
#include <vector>

using namespace std;

// Returns total distinct ways to make sum using n coins of
// different denominations
int count(vector<int> &coins, int n, int sum)
{
    // 2d dp array where n is the number of coin
    // denominations and sum is the target sum
    vector<vector<int>> dp(n + 1, vector<int>(sum + 1, 0));

    // Represents the base case where the target sum is 0,
    // and there is only one way to make change: by not
    // selecting any coin
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 0; j <= sum; j++)
        {

            // Add the number of ways to make change without
            // using the current coin,
            dp[i][j] += dp[i - 1][j];

            if ((j - coins[i - 1]) >= 0)
            {

                // Add the number of ways to make change
                // using the current coin
                dp[i][j] += dp[i][j - coins[i - 1]];
            }
        }
    }
    return dp[n][sum];
}

// Driver Code
int main()
{
    vector<int> coins{1, 2, 3};
    int n = 3;
    int sum = 5;
    cout << count(coins, n, sum);
    return 0;
}

Output

Time complexity : O(N*sum)
Auxiliary Space : O(N*sum)

4. Count All Combinations Using Dynamic Programming (Space Optimized)

In the above tabulation approach we are only using dp[i-1][j] and dp[i][j] , so we can further optimize the space complexity of the tabulation method by using only a 1D array.

Approach:

Create a 1D dp array, dp[i] represents the number of ways to make the sum i using the given coin denominations.
The outer loop iterates over the coins, and the inner loop iterates over the target sums. For each dp[j], it calculates the number of ways to make change using the current coin denomination and the previous results stored in dp.
dp[sum] contains the total number of ways to make change for the given target sum using the available coin denominations. This approach optimizes space by using a 1D array instead of a 2D DP table.

Below is the implementation of the above approach:

C++

// C++ program to solve coin change problem using Dynamic Programming (Space Optimized)
#include <cstring> // Include the cstring header for memset
#include <iostream>
#include <vector>

using namespace std;

int count(const vector<int> &coins, int n, int sum)
{
    // dp[i] will be storing the number of solutions for value i.
    // We need sum+1 rows as the dp is constructed in bottom up manner using the base case (sum = 0)
    vector<int> dp(sum + 1, 0);

    // Base case (If given value is 0)
    dp[0] = 1;

    // Pick all coins one by one and update the table values after
    // the index greater than or equal to the value of the picked coin
    for (int i = 0; i < n; i++)
    {
        for (int j = coins[i]; j <= sum; j++)
        {
            dp[j] += dp[j - coins[i]];
        }
    }
    return dp[sum];
}

int main()
{
    vector<int> coins = {1, 2, 3};
    int n = coins.size();
    int sum = 5;
    cout << count(coins, n, sum) << endl;
    return 0;
}

Output

Time complexity : O(N*sum)
Auxiliary Space : O(sum)

daswanta_kumar_routhu

Improve

Article Tags :

C++
CPP-DSA

Coin Change Problem in C++

1. Count All Combinations Using Recursion

Approach:

2. Count All Combinations Using Dynamic Programming (Memoization)

Approach:

3. Count All Combinations Using Dynamic Programming (Tabulation)

Approach:

4. Count All Combinations Using Dynamic Programming (Space Optimized)

Approach:

Explore

C++ Basics

Core Concepts

OOP in C++

Standard Template Library(STL)

Practice & Problems

Thank You!

What kind of Experience do you want to share?