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

Concept Flow - Coin Change Total Number of Ways

Start with empty ways array

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Set ways[0

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For each coin

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For each amount from coin to target

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Update ways[amount

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Repeat for all coins

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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.