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

Concept Flow - Coin Change Total Number of Ways

Start with empty ways array

↓

Set ways[0

↓

For each coin

↓

For each amount from coin to target

↓

Update ways[amount

↓

Repeat for all coins

↓

Result is ways[target

We build an array where each position shows how many ways to make that amount. We start with 1 way to make 0. For each coin, we add ways to make amounts including that coin.

Execution Sample

DSA C

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

Calculate how many ways to make amount 4 using coins 1, 2, and 3.

Execution Table

Step	Operation	Amount Updated	ways Array State	Explanation
0	Initialize ways	-	[1, 0, 0, 0, 0]	ways[0] = 1 because 1 way to make 0 amount
1	Use coin 1	1	[1, 1, 0, 0, 0]	ways[1] += ways[0] → 0 + 1 = 1
2	Use coin 1	2	[1, 1, 1, 0, 0]	ways[2] += ways[1] → 0 + 1 = 1
3	Use coin 1	3	[1, 1, 1, 1, 0]	ways[3] += ways[2] → 0 + 1 = 1
4	Use coin 1	4	[1, 1, 1, 1, 1]	ways[4] += ways[3] → 0 + 1 = 1
5	Use coin 2	2	[1, 1, 2, 1, 1]	ways[2] += ways[0] → 1 + 1 = 2
6	Use coin 2	3	[1, 1, 2, 2, 1]	ways[3] += ways[1] → 1 + 1 = 2
7	Use coin 2	4	[1, 1, 2, 2, 3]	ways[4] += ways[2] → 1 + 2 = 3
8	Use coin 3	3	[1, 1, 2, 3, 3]	ways[3] += ways[0] → 2 + 1 = 3
9	Use coin 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 ways to make amount 4

💡 All coins processed, final ways array shows total ways to make target amount

Variable Tracker

Variable	Start	After coin 1	After coin 2	After coin 3	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	-	1 to 4	2 to 4	3 to 4	-

Key Moments - 3 Insights

Why do we start ways[0] with 1?

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 other amounts, as shown in execution_table step 0.

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

Adding ways[amount - coin] means we count all ways to make the smaller amount before adding the current coin. This accumulates all combinations including the current coin, as seen in steps 1-9.

Why do we process coins one by one instead of all at once?

Processing coins one by one ensures we count unique combinations without duplicates. Each coin updates ways array cumulatively, as shown in variable_tracker and execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 5, what is ways[2] after using coin 2?

A 2

B 1

C 0

D 3

💡 Hint

Check the ways array state column at step 5 in execution_table

At which step does ways[4] first become greater than 1?

A Step 4

B Step 9

C Step 7

D Step 3

💡 Hint

Look at ways[4] values in execution_table rows for steps 4 to 9

If we remove coin 1 from coins, what happens to ways[1] after processing all coins?

A It stays 1

B It becomes 0

C It becomes 2

D It becomes 3

💡 Hint

Check variable_tracker ways array after coin 1 and think about how ways[1] is updated

Concept Snapshot

Coin Change Total Number of Ways:
- Use a ways array of size target+1
- Initialize ways[0] = 1 (base case)
- For each coin, update ways[amount] += ways[amount - coin]
- Result is ways[target]
- Counts combinations, order of coins does not matter

Full Transcript

This visualization shows how to find the total number of ways to make a target amount using given coin denominations. 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 by adding the number of ways to make the current amount minus the coin's value. This accumulates all possible combinations. 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 coins are processed one by one. The quiz tests understanding of array updates and the effect of removing coins. The snapshot summarizes the approach simply.

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 combine coins to reach a target amount.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, this counts combinations, not minimum coins.

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 total ways with minimum coins needed
• Thinking it finds maximum coin value
• Assuming it sums coin denominations

2. Which of the following is the correct way to declare the DP array for the Coin Change Total Number of Ways problem in C?

easy

A. int dp[target];

B. int dp[target+1];

C. int dp[target+1][coins];

D. int dp[coins+1];

Solution

1. Step 1: Understand DP array size

   We need to store ways for amounts from 0 to target, so size is target+1.

2. Step 2: Choose correct declaration

   One-dimensional array of size target+1 is used to store ways for each amount.

3. Final Answer:

   int dp[target+1]; -> Option B

4. Quick Check:

   Array size = target+1 [OK]

Hint: DP array size must cover 0 to target inclusive [OK]

Common Mistakes:

• Using dp[target] misses amount 0
• Using 2D array unnecessarily
• Using dp[coins+1] which is unrelated

3. What is the output of the following C code snippet for coins = {1, 2} and target = 3?

int coins[] = {1, 2};
int target = 3;
int dp[4] = {1, 0, 0, 0};
for (int i = 0; i < 2; i++) {
    for (int j = coins[i]; j <= target; j++) {
        dp[j] += dp[j - coins[i]];
    }
}
printf("%d", dp[target]);

medium

A. 1

B. 3

C. 2

D. 4

Solution

1. Step 1: Initialize dp array

   dp = [1,0,0,0] means 1 way to make 0 amount.

2. Step 2: Process coin 1

   For j=1 to 3: dp[1]+=dp[0]=1, dp[2]+=dp[1]=1, dp[3]+=dp[2]=1; dp=[1,1,1,1]

3. Step 3: Process coin 2

   For j=2 to 3: dp[2]+=dp[0]=1+1=2, dp[3]+=dp[1]=1+1=2; dp=[1,1,2,2]

4. Final Answer:

   2 -> Option C

5. Quick Check:

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

Hint: Add ways from smaller amounts using each coin [OK]

Common Mistakes:

• Not initializing dp[0] to 1
• Updating dp array incorrectly
• Mixing order of loops

4. Identify the error in this code snippet for calculating total ways to make change:

int coins[] = {1, 2, 3};
int target = 4;
int dp[5] = {0};
dp[0] = 1;
for (int i = 0; i <= 3; i++) {
    for (int j = coins[i]; j <= target; j++) {
        dp[j] += dp[j - coins[i]];
    }
}
printf("%d", dp[target]);

medium

A. Loop i should run from 0 to 2, not 0 to 3.

B. dp array size is too small.

C. dp[0] should be initialized to 0.

D. Inner loop should start from 0, not coins[i].

Solution

1. Step 1: Check loop bounds for coins array

   coins has 3 elements indexed 0 to 2, but loop runs i=0 to 3 (4 iterations) causing out-of-bounds access.

2. Step 2: Correct loop range

   Change loop to i=0 to 2 to avoid accessing coins[3] which is invalid.

3. Final Answer:

   Loop i should run from 0 to 2, not 0 to 3. -> Option A

4. Quick Check:

   Array index out-of-bounds error [OK]

Hint: Loop over coins must match array length exactly [OK]

Common Mistakes:

• Using wrong loop bounds causing crashes
• Incorrect dp initialization
• Starting inner loop incorrectly

5. Given coins = {1, 3, 5} and target = 7, how many total ways are there to make change? Use the bottom-up DP approach.

hard

A. 6

B. 5

C. 7

D. 4

Solution

1. Step 1: Initialize dp array

   dp[0]=1, dp[1..7]=0 initially.

2. Step 2: Process coin 1

   Update dp for amounts 1 to 7 by adding dp[j-1]: dp=[1,1,1,1,1,1,1,1]

3. Step 3: Process coin 3

   Update dp for j=3 to 7: dp[3]+=dp[0]=2, dp[4]+=dp[1]=2, dp[5]+=dp[2]=2, dp[6]+=dp[3]=3, dp[7]+=dp[4]=3; dp=[1,1,1,2,2,2,3,3]

4. Step 4: Process coin 5

   Update dp for j=5 to 7: dp[5]+=dp[0]=3, dp[6]+=dp[1]=4, dp[7]+=dp[2]=4; dp=[1,1,1,2,2,3,4,4]

5. Final Answer:

   4 -> Option D

6. Quick Check:

   Count ways with coins {1,3,5} for 7 = 4 [OK]

Hint: Add ways for each coin incrementally [OK]

Common Mistakes:

• Not updating dp correctly for each coin
• Forgetting dp[0] = 1 initialization
• Mixing order of coin processing