Coin Change Total Number of Ways in DSA Python - Build from Scratch

Coin Change Total Number of Ways

📖 Scenario: You are helping a cashier who wants to find out how many different ways they can give change for a certain amount using given coin denominations.

🎯 Goal: Build a program that calculates the total number of ways to make change for a given amount using a list of coin denominations.

📋 What You'll Learn

Create a list of coin denominations

Create a variable for the target amount

Use a dynamic programming approach to calculate the total number of ways

Print the total number of ways to make change

💡 Why This Matters

🌍 Real World

Cashiers and vending machines use this logic to calculate how to give change in different ways.

💼 Career

Understanding dynamic programming and coin change problems is common in software engineering interviews and financial software development.

1

Create the list of coin denominations

Create a list called coins with these exact values: 1, 2, and 5.

DSA Python

# Create the list of coin denominations
# Your code here

Hint

Use square brackets to create a list and separate the numbers with commas.

2

Create the target amount variable

Create a variable called amount and set it to 5.

DSA Python

coins = [1, 2, 5]
# Create the target amount variable
# Your code here

Hint

Use a simple assignment to create the variable.

3

Calculate the total number of ways to make change

Create a list called dp with amount + 1 elements, all set to 0. Then set dp[0] to 1. Use a for loop with variable coin to iterate over coins. Inside it, use another for loop with variable x to iterate from coin to amount inclusive. Inside the inner loop, update dp[x] by adding dp[x - coin].

DSA Python

coins = [1, 2, 5]
amount = 5
# Create dp list and initialize
# Your code here

Hint

This is a classic dynamic programming approach. Initialize dp with zeros, set dp[0] to 1, then update dp for each coin.

4

Print the total number of ways to make change

Print the value of dp[amount].

DSA Python

coins = [1, 2, 5]
amount = 5
dp = [0] * (amount + 1)
dp[0] = 1
for coin in coins:
    for x in range(coin, amount + 1):
        dp[x] += dp[x - coin]
# Print the total number of ways
# Your code here

Hint

The output is the total number of ways to make change for 5 using coins 1, 2, and 5.

Practice

1. What does the "Coin Change Total Number of Ways" problem ask you to find?

easy

A. The total number of ways to make a target amount using given coin denominations.

B. The minimum number of coins needed to make a target amount.

C. The maximum value of coins that can be used to reach the target.

D. The total sum of all coin denominations.

Solution

1. Step 1: Understand the problem goal

   The problem asks for counting all possible ways to make a target amount using given coins, not minimizing or maximizing coins.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, this counts combinations, so answer is total ways to form amount.

3. Final Answer:

   The total number of ways to make a target amount using given coin denominations. -> Option A

4. Quick Check:

   Count ways, not minimum coins [OK]

Hint: Count combinations, not minimum coins [OK]

Common Mistakes:

• Confusing with minimum coin count problem
• Thinking about maximum coin value instead of ways
• Ignoring that order of coins does not matter

2. Which of the following is the correct Python syntax to initialize a list named dp with size 6 filled with zeros?

easy

A. dp = [6] * 0

B. dp = [0, 6]

C. dp = list(0, 6)

D. dp = [0] * 6

Solution

1. Step 1: Understand list initialization

   To create a list of size 6 filled with zeros, multiply a list containing zero by 6.

2. Step 2: Check each option

   dp = [0] * 6 uses correct syntax dp = [0] * 6. Others are invalid or create wrong lists.

3. Final Answer:

   dp = [0] * 6 -> Option D

4. Quick Check:

   List * size creates repeated elements [OK]

Hint: Use [0] * size to create zero-filled list [OK]

Common Mistakes:

• Using commas inside list incorrectly
• Trying to call list() with two arguments
• Multiplying list by zero instead of size

3. What is the output of the following code?

coins = [1, 2]
amount = 3
dp = [0] * (amount + 1)
dp[0] = 1
for coin in coins:
    for x in range(coin, amount + 1):
        dp[x] += dp[x - coin]
print(dp[amount])

medium

A. 1

B. 3

C. 2

D. 0

Solution

1. Step 1: Initialize dp and start loops

   dp = [1,0,0,0] initially. For coin=1, update dp for x=1 to 3 adding dp[x-1].

2. Step 2: Update dp for each coin

   After coin=1, dp = [1,1,1,1]. For coin=2, update dp[x] adding dp[x-2]: dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2.

3. Step 3: Final dp array and answer

   dp = [1,1,2,2], so dp[3] = 2 means 2 ways to make amount 3.

4. Final Answer:

   2 -> Option C

5. Quick Check:

   Ways to make 3 with [1,2] = 2 [OK]

Hint: Use dp array to accumulate ways for each coin [OK]

Common Mistakes:

• Mixing order of loops causing wrong counts
• Not initializing dp[0] = 1
• Confusing dp[x] update with assignment instead of addition

4. Find the error in this code snippet for coin change total ways:

coins = [1, 2, 3]
amount = 4
dp = [0] * (amount + 1)
for coin in coins:
    for x in range(coin, amount + 1):
        dp[x] += dp[x - coin]
print(dp[amount])

medium

A. The inner loop range is incorrect

B. dp array is not initialized with dp[0] = 1

C. The outer loop should be over amount, not coins

D. The print statement is outside the loop

Solution

1. Step 1: Check dp initialization

   dp[0] must be 1 to represent one way to make amount zero, but here dp[0] = 0.

2. Step 2: Understand impact of missing dp[0] = 1

   Without dp[0] = 1, dp[x] never updates correctly, so total ways remain zero.

3. Final Answer:

   dp array is not initialized with dp[0] = 1 -> Option B

4. Quick Check:

   dp[0] = 1 is essential [OK]

Hint: Always set dp[0] = 1 before loops [OK]

Common Mistakes:

• Forgetting dp[0] initialization
• Changing loop order incorrectly
• Misunderstanding dp array purpose

5. Given coins = [1, 3, 5] and amount = 7, how many total ways are there to make amount 7? Use the coin change total number of ways approach.

hard

A. 4

B. 5

C. 3

D. 6

Solution

1. Step 1: Initialize dp array and set dp[0] = 1

   dp = [1,0,0,0,0,0,0,0]

2. Step 2: Update dp for coin=1

   For x=1 to 7, dp[x] += dp[x-1], so dp = [1,1,1,1,1,1,1,1]

3. Step 3: Update dp for coin=3

   For x=3 to 7, dp[x] += dp[x-3]: dp[3]=1+1=2, dp[4]=1+1=2, dp[5]=1+1=2, dp[6]=1+2=3, dp[7]=1+2=3

4. Step 4: Update dp for coin=5

   For x=5 to 7, dp[x] += dp[x-5]: dp[5]=2+1=3, dp[6]=3+1=4, dp[7]=3+1=4

5. Final Answer:

   4 -> Option A

6. Quick Check:

   Count ways stepwise with dp array [OK]

Hint: Update dp for each coin sequentially [OK]

Common Mistakes:

• Not updating dp for all coins
• Mixing order of loops causing duplicates
• Forgetting dp[0] = 1 initialization