Coin Change Total Number of Ways in DSA Python - Execution Trace

Concept Flow - Coin Change Total Number of Ways

Start with ways array initialized to 0 except ways[0

↓

For each coin in coins

→For each amount from coin to target

↓

Update ways[amount

↓

Repeat for all coins

↓

Return ways[target

We start with an array that counts ways to make amounts. For each coin, we update ways for amounts that can include that coin. Finally, ways[target] gives total ways.

Execution Sample

DSA Python

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

This code calculates how many ways to make amount 4 using coins 1, 2, and 3.

Execution Table

Step	Operation	Coin	Amount	ways Array State	Explanation
0	Initialize ways	-	-	[1, 0, 0, 0, 0]	ways[0] = 1 means 1 way to make amount 0 (no coins)
1	Use coin 1	1	1	[1, 1, 0, 0, 0]	ways[1] += ways[0] → 0 + 1 = 1
2	Use coin 1	1	2	[1, 1, 1, 0, 0]	ways[2] += ways[1] → 0 + 1 = 1
3	Use coin 1	1	3	[1, 1, 1, 1, 0]	ways[3] += ways[2] → 0 + 1 = 1
4	Use coin 1	1	4	[1, 1, 1, 1, 1]	ways[4] += ways[3] → 0 + 1 = 1
5	Use coin 2	2	2	[1, 1, 2, 1, 1]	ways[2] += ways[0] → 1 + 1 = 2
6	Use coin 2	2	3	[1, 1, 2, 2, 1]	ways[3] += ways[1] → 1 + 1 = 2
7	Use coin 2	2	4	[1, 1, 2, 2, 3]	ways[4] += ways[2] → 1 + 2 = 3
8	Use coin 3	3	3	[1, 1, 2, 3, 3]	ways[3] += ways[0] → 2 + 1 = 3
9	Use coin 3	3	4	[1, 1, 2, 3, 4]	ways[4] += ways[1] → 3 + 1 = 4
10	End	-	-	[1, 1, 2, 3, 4]	All coins processed, ways[4] = 4 total ways

💡 All coins processed and amounts updated, final ways array shows total ways to make target.

Variable Tracker

Variable	Start	After Step 4	After Step 7	After Step 9	Final
ways	[1, 0, 0, 0, 0]	[1, 1, 1, 1, 1]	[1, 1, 2, 2, 3]	[1, 1, 2, 3, 4]	[1, 1, 2, 3, 4]
coin	-	1	2	3	3
amount	-	4	4	4	-

Key Moments - 3 Insights

Why do we start ways[0] with 1 instead of 0?

ways[0] = 1 means there is exactly one way to make amount 0: by choosing no coins. This is the base case needed to build up counts for larger amounts, as seen in execution_table step 0.

Why do we add ways[amount - coin] to ways[amount] inside the loop?

Adding ways[amount - coin] means we count all ways to make the smaller amount (amount - coin) and then add the current coin to reach amount. This accumulates all combinations including the current coin, shown in steps 1-9.

Why do we loop coins first, then amounts, not the other way?

Looping coins first ensures combinations are counted without duplicates. Each coin adds ways to amounts in order, preventing counting the same combination multiple times. This order is reflected in the execution_table rows.

Visual Quiz - 3 Questions

Look at the execution_table at step 7, what is the value of ways[4]?

A 1

B 3

C 2

D 4

💡 Hint

Check the 'ways Array State' column at step 7 in execution_table.

At which step does ways[3] first become 3?

A Step 8

B Step 4

C Step 7

D Step 9

💡 Hint

Look at the 'ways Array State' column for ways[3] in execution_table rows.

If we remove coin 1 from coins, how would ways[4] change after all steps?

A It would become 3

B It would stay 4

C It would become 1

D It would become 0

💡 Hint

Without coin 1, ways to make amounts rely on coins 2 and 3 only; check how ways[4] updates in execution_table steps for coin 2 and 3.

Concept Snapshot

Coin Change Total Number of Ways:
- Use a ways array of size target+1, ways[0]=1
- For each coin, update ways[amount] += ways[amount - coin]
- Loop coins outer, amounts inner to avoid duplicates
- ways[target] gives total combinations
- Dynamic programming builds solutions bottom-up

Full Transcript

This visualization shows how to find the total number of ways to make a target amount using given coins. We start with an array 'ways' where ways[0] is 1, meaning one way to make zero amount. For each coin, we update the ways array for all amounts from that coin's value up to the target. We add ways[amount - coin] to ways[amount], counting all ways that include the current coin. The execution table traces each update step by step, showing how the ways array changes. Key moments clarify why ways[0] starts at 1, why we add ways[amount - coin], and why the coin loop is outer. The quiz tests understanding of the array state at specific steps and the effect of removing a coin. The snapshot summarizes the approach as a dynamic programming solution building up from smaller amounts to the target.

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