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

Concept Flow - Coin Change Total Number of Ways

Start with amount=0

↓

Initialize ways array with ways[0

↓

For each coin

↓

For each amount from coin to target

↓

Update ways[amount

↓

Repeat for all coins

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Result: ways[target

We build up the number of ways to make each amount by adding ways contributed by each coin step by step.

Execution Sample

DSA Rust

let coins = [1usize, 2usize, 3usize];
let amount = 4usize;
let mut ways = vec![0usize; amount + 1];
ways[0] = 1;
for &coin in &coins {
    for x in coin..=amount {
        ways[x] += ways[x - coin];
    }
}

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

Execution Table

Step	Operation	Amount Index Updated	ways Array State	Explanation
0	Initialize ways[0]	0	[1, 0, 0, 0, 0]	Base case: 1 way to make amount 0 (no coins)
1	Use coin=1, update ways[1]	1	[1, 1, 0, 0, 0]	ways[1] += ways[0] → 0 + 1 = 1
2	Use coin=1, update ways[2]	2	[1, 1, 1, 0, 0]	ways[2] += ways[1] → 0 + 1 = 1
3	Use coin=1, update ways[3]	3	[1, 1, 1, 1, 0]	ways[3] += ways[2] → 0 + 1 = 1
4	Use coin=1, update ways[4]	4	[1, 1, 1, 1, 1]	ways[4] += ways[3] → 0 + 1 = 1
5	Use coin=2, update ways[2]	2	[1, 1, 2, 1, 1]	ways[2] += ways[0] → 1 + 1 = 2
6	Use coin=2, update ways[3]	3	[1, 1, 2, 2, 1]	ways[3] += ways[1] → 1 + 1 = 2
7	Use coin=2, update ways[4]	4	[1, 1, 2, 2, 3]	ways[4] += ways[2] → 1 + 2 = 3
8	Use coin=3, update ways[3]	3	[1, 1, 2, 3, 3]	ways[3] += ways[0] → 2 + 1 = 3
9	Use coin=3, update ways[4]	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 for each amount.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	After Step 5	After Step 6	After Step 7	After Step 8	After Step 9	After Step 10
ways	[0,0,0,0,0]	[1,0,0,0,0]	[1,1,0,0,0]	[1,1,1,0,0]	[1,1,1,1,0]	[1,1,1,1,1]	[1,1,2,1,1]	[1,1,2,2,1]	[1,1,2,2,3]	[1,1,2,3,3]	[1,1,2,3,4]
coin	N/A	1	1	1	1	2	2	2	3	3	3
amount index	N/A	1	2	3	4	2	3	4	3	4	N/A

Key Moments - 3 Insights

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

ways[0] = 1 means there is exactly one way to make amount 0: by choosing no coins. This base case allows the algorithm to build up counts for larger amounts (see step 0 in execution_table).

Why do we add ways[x - coin] to ways[x] instead of replacing it?

We add because each coin can contribute additional ways to make amount x. We accumulate all possible combinations from different coins (see steps 5-9 in execution_table).

Why do we loop coins first, then amounts?

Looping coins first ensures combinations are counted without duplicates in different orders. This order counts unique sets, not permutations (see the order of updates in execution_table).

Visual Quiz - 3 Questions

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

A 1

B 2

C 3

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 6

B Step 8

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 the coins array, how would ways[1] change after step 1?

A ways[1] would be 0

B ways[1] would be 1

C ways[1] would be 2

D ways[1] would be 3

💡 Hint

Without coin=1, no coin can make amount 1, so ways[1] stays 0 (see initial ways array).

Concept Snapshot

Coin Change Total Number of Ways:
- Use a ways array of size amount+1
- Initialize ways[0] = 1 (base case)
- For each coin, update ways[x] += ways[x - coin] for x >= coin
- Final answer is ways[amount]
- Counts combinations, not permutations

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 set to 1, meaning there is one way to make amount zero: by choosing no coins. Then, 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 counts instead of replacing, and why the coin loop is outer. The quiz tests understanding of specific steps and effects of removing coins. The snapshot summarizes the approach in a few lines.

Practice

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

easy

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

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

C. The maximum value of coins that can be used without exceeding 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 combinations of coins that sum up to the target amount.

2. Step 2: Differentiate from similar problems

   Unlike minimum coins problem, here we count ways, not minimize coins.

3. Final Answer:

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

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 about maximum coin value instead of count
• Ignoring that order of coins does not matter

2. Which of the following Rust code snippets correctly initializes a vector to store ways to make amounts up to `target` with 0 ways initially except for 0 amount which has 1 way?

easy

A. let mut ways = vec![0; target + 1]; ways[0] = 1;

B. let mut ways = vec![1; target]; ways[0] = 0;

C. let mut ways = vec![0; target]; ways[0] = 1;

D. let mut ways = vec![1; target + 1]; ways[0] = 0;

Solution

1. Step 1: Initialize vector size correctly

   Vector size must be target + 1 to include amount 0 to target.

2. Step 2: Set base case for amount 0

   ways[0] = 1 means one way to make amount 0 (using no coins).

3. Final Answer:

   let mut ways = vec![0; target + 1]; ways[0] = 1; -> Option A

4. Quick Check:

   Vector size = target+1, ways[0] = 1 [OK]

Hint: ways vector size = target+1, ways[0] = 1 [OK]

Common Mistakes:

• Using vector size target instead of target+1
• Setting ways[0] to 0 instead of 1
• Initializing all elements to 1 incorrectly

3. What is the output of the following Rust code snippet for coins = [1, 2] and target = 3?

let coins = vec![1usize, 2usize];
let target: usize = 3;
let mut ways = vec![0; target + 1];
ways[0] = 1;
for &coin in &coins {
    for amount in coin..=target {
        ways[amount] += ways[amount - coin];
    }
}
println!("{}", ways[target]);

medium

A. 4

B. 1

C. 2

D. 3

Solution

1. Step 1: Calculate ways using coin 1

   ways after coin 1: ways[0]=1, ways[1]=1, ways[2]=1, ways[3]=1

2. Step 2: Update ways using coin 2

   ways[2] += ways[0] => 1+1=2; ways[3] += ways[1] => 1+1=2; final ways = [1,1,2,2]

3. Final Answer:

   2 -> Option C

4. Quick Check:

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

Hint: Add ways for each coin incrementally [OK]

Common Mistakes:

• Not updating ways correctly for each coin
• Confusing ways count with minimum coins
• Off-by-one errors in loops

4. Identify the bug in this Rust code snippet for coin change ways calculation:

let coins = vec![1usize, 2usize, 3usize];
let target: usize = 4;
let mut ways = vec![0; target + 1];
ways[0] = 1;
for &coin in &coins {
    for amount in 1usize..=target {
        if amount >= coin {
            ways[amount] += ways[amount - coin];
        }
    }
}
println!("{}", ways[target]);

medium

A. The inner loop should start from coin, not 1.

B. ways[0] should be initialized to 0.

C. coins vector should be mutable.

D. The outer loop should iterate over amounts, not coins.

Solution

1. Step 1: Check inner loop range

   Starting inner loop from 1 causes unnecessary checks; it should start from coin value to avoid redundant iterations.

2. Step 2: Confirm initialization and loops

   ways[0] = 1 is correct; coins vector does not need to be mutable; outer loop over coins is correct.

3. Final Answer:

   The inner loop should start from coin, not 1. -> Option A

4. Quick Check:

   Inner loop start = coin [OK]

Hint: Inner loop starts at coin value to optimize [OK]

Common Mistakes:

• Starting inner loop from 1 instead of coin
• Incorrect initialization of ways[0]
• Misunderstanding loop roles

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

hard

A. 6

B. 5

C. 7

D. 4

Solution

1. Step 1: Initialize ways vector

   ways = [1,0,0,0,0,0,0,0] for amounts 0 to 7.

2. Step 2: Update ways with coin 1

   ways after coin 1: ways[i] = 1 for all i from 0 to 7 (only using 1s).

3. Step 3: Update ways with coin 3

   ways[3]+=ways[0]=1, ways[4]+=ways[1]=1, ways[5]+=ways[2]=1, ways[6]+=ways[3]=2, ways[7]+=ways[4]=2; ways now: [1,1,1,2,2,2,3,3]

4. Step 4: Update ways with coin 5

   ways[5] += ways[0] => 2 + 1 = 3; ways[6] += ways[1] => 3 + 1 = 4; ways[7] += ways[2] => 3 + 1 = 4; final ways: [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 incrementally for each coin [OK]

Common Mistakes:

• Not updating ways correctly for each coin
• Forgetting to include ways[0] = 1
• Mixing up order of loops