Python Program for Coin Change | DP-7

Write a Python program for a given integer array of coins[ ] of size N representing different types of denominations and an integer sum, the task is to find the number of ways to make a sum by using different denominations.

Examples:

Input: sum = 4, coins[] = {1,2,3},
Output: 4
Explanation: there are four solutions: {1, 1, 1, 1}, {1, 1, 2}, {2, 2}, {1, 3}.

Input: sum = 10, coins[] = {2, 5, 3, 6}
Output: 5
Explanation: There are five solutions: {2,2,2,2,2}, {2,2,3,3}, {2,2,6}, {2,3,5} and {5,5}.

Python Program for Coin Change using Recursion:

Recurrence Relation:

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

For each coin, there are 2 options.

• Include the current coin: Subtract the current coin’s denomination from the target sum and call the count function recursively with the updated sum and the same set of coins i.e., count(coins, n, sum – coins[n-1] )
• Exclude the current coin: Call the count function recursively with the same sum and the remaining coins. i.e., count(coins, n-1,sum ).

The final result will be the sum of both cases.

Base case:

• If the target sum (sum) is 0, there is only one way to make the sum, which is by not selecting any coin. So, count(0, coins, n) = 1.
• If the target sum (sum) is negative or no coins are left to consider (n == 0), then there are no ways to make the sum, so count(sum, coins, 0) = 0.
• Coin Change | DP-7

Below is the Implementation of the above approach:

Python3

# Recursive Python3 program for
# coin change problem.

# Returns the count of ways we can sum
# coins[0...n-1] coins to get sum "sum"


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


# Driver program to test above function
coins = [1, 2, 3]
n = len(coins)
print(count(coins, n, 5))

# This code is contributed by Smitha Dinesh Semwal

Output

5 

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

Python Program for Coin Change using Dynamic Programming (Memoization) :

The above recursive solution has Optimal Substructure and Overlapping Subproblems so Dynamic programming (Memoization) can be used to solve the problem. So 2D array can be used to store results of previously solved subproblems.

Step-by-step approach:

• Create a 2D dp array to store the results of previously solved subproblems.
• dp[i][j] will represent the number of distinct ways to make the sum j by using the first i coins.
• During the recursion call, if the same state is called more than once, then we can directly return the answer stored for that state instead of calculating again.

Below is the implementation of the above approach:

Python3

# Python program for the above approach

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


def count(coins, sum, n, dp):
# Base Case
    if (sum == 0):
        dp[n][sum] = 1
        return dp[n][sum]

    # If number of coins is 0 or sum is less than 0 then there is no way to make the sum.
    if (n == 0 or 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

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

    return dp[n][sum]


# Driver code
if __name__ == '__main__':
    tc = 1
    while (tc != 0):
        n = 3
        sum = 5
        coins = [1, 2, 3]
        dp = [[-1 for i in range(sum+1)] for j in range(n+1)]
        res = count(coins, sum, n, dp)
        print(res)
        tc -= 1

Output

5

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

Python Program for Coin Change using Dynamic Programming (Tabulation):

• 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 the above approach:

Python3

# Function to calculate the total distinct ways to make a sum using n coins of different denominations
def count(coins, n, target_sum):
    # 2D dp array where n is the number of coin denominations and target_sum is the target sum
    dp = [[0 for j in range(target_sum + 1)] for i in range(n + 1)]

    # 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 i in range(1, n + 1):
        for j in range(target_sum + 1):
            # 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][target_sum]

# Driver Code
if __name__ == "__main__":
    coins = [1, 2, 3]
    n = 3
    target_sum = 5
    print(count(coins, n, target_sum))

Output

5

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

Python Program for Coin Change using the Space Optimized 1D array:

In the above tabulation approach we are only using dp[i-1][j] and dp[i][j] etc, so we can do space optimization by only using a 1d dp array.

Step-by-step 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:

Python

# Dynamic Programming Python implementation of Coin
# Change problem


def count(coins, n, 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)
    # Initialize all table values as 0
    dp = [0 for k in range(sum+1)]

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

    # Pick all coins one by one and update the dp[] values
    # after the index greater than or equal to the value of the
    # picked coin
    for i in range(0, n):
        for j in range(coins[i], sum+1):
            dp[j] += dp[j-coins[i]]

    return dp[sum]


# Driver program to test above function
coins = [1, 2, 3]
n = len(coins)
sum = 5
x = count(coins, n, sum)
print(x)

# This code is contributed by Afzal Ansari

Output

5

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

Please refer complete article on Coin Change | DP-7 for more details!