Coin Change Total Number of Ways in DSA Python

Mental Model

We want to find how many different ways we can make a total amount using given coin values, counting all combinations without caring about order.

Analogy: Imagine you have different types of candies and want to fill a jar with a certain number of candies. You want to know in how many ways you can fill the jar using any number of each candy type.

coins = [1, 2, 3]
amount = 4

Ways array (index = amount):
[1, 0, 0, 0, 0]
↑
Initially, 1 way to make 0 amount (use no coins).

Dry Run Walkthrough

Input: coins = [1, 2, 3], amount = 4

Goal: Find total number of ways to make amount 4 using coins 1, 2, and 3.

Step 1: Start with ways array: ways = [1, 0, 0, 0, 0] (1 way to make 0)

ways = [1, 0, 0, 0, 0]

Why: We know there is exactly 1 way to make amount 0: use no coins.

Step 2: Use coin 1: update ways for amounts 1 to 4 by adding ways[amount - 1]

ways = [1, 1, 1, 1, 1]

Why: For each amount, ways to make it includes ways to make amount minus 1 (coin 1).

Step 3: Use coin 2: update ways for amounts 2 to 4 by adding ways[amount - 2]

ways = [1, 1, 2, 2, 3]

Why: Add ways to make amount minus 2 (coin 2) to current ways.

Step 4: Use coin 3: update ways for amounts 3 to 4 by adding ways[amount - 3]

ways = [1, 1, 2, 3, 4]

Why: Add ways to make amount minus 3 (coin 3) to current ways.

Result:

ways = [1, 1, 2, 3, 4]
Total ways to make 4 = 4

Annotated Code

DSA Python

from typing import List

class CoinChange:
    def total_ways(self, coins: List[int], amount: int) -> int:
        # ways[i] will store number of ways to make amount i
        ways = [0] * (amount + 1)
        ways[0] = 1  # base case: 1 way to make amount 0

        for coin in coins:
            for x in range(coin, amount + 1):
                ways[x] += ways[x - coin]  # add ways to make (x - coin)

        return ways[amount]

if __name__ == '__main__':
    coins = [1, 2, 3]
    amount = 4
    cc = CoinChange()
    print(cc.total_ways(coins, amount))

ways = [0] * (amount + 1) ways[0] = 1

Initialize ways array with 0, set ways[0] = 1 because there is one way to make amount 0

for coin in coins:

Iterate over each coin to build up ways to make amounts

for x in range(coin, amount + 1):

For each amount from coin value to target, update ways

ways[x] += ways[x - coin]

Add ways to make amount (x - coin) to ways[x], accumulating combinations

return ways[amount]

Return total ways to make the target amount

Output

4

Complexity Analysis

Time: O(n * amount) because we iterate over each coin and for each coin we update ways for all amounts up to target

Space: O(amount) because we use a single array of size amount+1 to store ways counts

vs Alternative: Compared to recursive solutions without memoization which can be exponential, this dynamic programming approach is efficient and avoids repeated calculations.

Edge Cases

amount = 0

Returns 1 because there is exactly one way to make amount 0 (use no coins)

DSA Python

ways[0] = 1

coins = [] (no coins)

Returns 0 for any amount > 0 because no coins means no ways to make positive amount

DSA Python

for coin in coins:

coins contain coin larger than amount

Such coins are skipped because inner loop starts from coin value, so no update for amounts smaller than coin

DSA Python

for x in range(coin, amount + 1):

When to Use This Pattern

When asked to count the number of ways to make a total using given coin denominations, use the Coin Change Total Number of Ways pattern with dynamic programming to efficiently build solutions from smaller amounts.

Common Mistakes

Mistake: Counting permutations instead of combinations by changing the order of loops
Fix: Loop over coins first, then amounts, to count combinations without order sensitivity

Mistake: Not initializing ways[0] = 1, leading to zero ways for all amounts
Fix: Set ways[0] = 1 to represent the base case of making amount zero

Summary

Calculates how many different combinations of coins can make a target amount.

Use when you need to count combinations of coins to form a sum, ignoring order.

Build the solution bottom-up by adding ways to make smaller amounts using each coin.

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