Coin Change Total Number of Ways in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Coin Change Ways Calculation

What is the output of this Rust code that calculates the total number of ways to make change for 5 using coins [1, 2, 3]?

DSA Rust

fn coin_change_ways(coins: &[usize], amount: usize) -> usize {
    let mut dp = vec![0; amount + 1];
    dp[0] = 1;
    for &coin in coins {
        for x in coin..=amount {
            dp[x] += dp[x - coin];
        }
    }
    dp[amount]
}

fn main() {
    let coins = vec![1, 2, 3];
    let amount = 5;
    println!("{}", coin_change_ways(&coins, amount));
}

A 6

B 3

C 5

D 4

💡 Hint

Think about how many combinations of coins sum to 5 using 1, 2, and 3.

The code uses dynamic programming to count ways to make amount 5 with coins 1, 2, and 3. The total ways are 6.

🧠 Conceptual

intermediate

Understanding DP Array Initialization

Why is the dp array initialized with dp[0] = 1 in the coin change ways algorithm?

A Because dp[0] is used to store the maximum coin value.

B Because dp[0] represents the first coin value.

C Because dp[0] stores the total number of coins available.

D Because there is exactly one way to make amount zero: use no coins.

💡 Hint

Think about how many ways to make zero amount exist.

dp[0] = 1 means there is one way to make amount zero: by choosing no coins.

🔧 Debug

advanced

Identify the Error in Coin Change Code

What error does this Rust code produce when compiling?

DSA Rust

fn coin_change_ways(coins: &[usize], amount: usize) -> usize {
    let mut dp = vec![0; amount + 1];
    dp[0] = 1
    for &coin in coins {
        for x in coin..=amount {
            dp[x] += dp[x - coin];
        }
    }
    dp[amount]
}

fn main() {
    let coins = vec![1, 2, 3];
    let amount = 5;
    println!("{}", coin_change_ways(&coins, amount));
}

A TypeError: cannot add usize to dp[x]

B SyntaxError: missing semicolon after dp[0] = 1

C RuntimeError: index out of bounds in dp array

D No error, code runs successfully

💡 Hint

Check the line dp[0] = 1 for syntax correctness.

The line dp[0] = 1 is missing a semicolon, causing a syntax error.

❓ Predict Output

advanced

Output with Different Coin Set

What is the output of this Rust code that calculates the total number of ways to make change for 10 using coins [2, 5, 3, 6]?

DSA Rust

fn coin_change_ways(coins: &[usize], amount: usize) -> usize {
    let mut dp = vec![0; amount + 1];
    dp[0] = 1;
    for &coin in coins {
        for x in coin..=amount {
            dp[x] += dp[x - coin];
        }
    }
    dp[amount]
}

fn main() {
    let coins = vec![2, 5, 3, 6];
    let amount = 10;
    println!("{}", coin_change_ways(&coins, amount));
}

A 5

B 4

C 6

D 7

💡 Hint

Count all combinations of coins that sum to 10.

The code counts 5 ways to make 10 using coins 2, 3, 5, and 6.

🧠 Conceptual

expert

Why Does Order of Coins Matter in DP?

In the coin change total number of ways algorithm, why does iterating over coins first and then amounts produce combinations without duplicates?

A Because iterating amounts first causes runtime errors.

B Because iterating coins first sorts the coins automatically.

C Because iterating coins first ensures each coin is only counted once per combination, avoiding permutations.

D Because iterating amounts first would count only the largest coin combinations.

💡 Hint

Think about how the order of loops affects counting unique combinations vs permutations.

Iterating coins first ensures combinations are counted without duplicates by adding ways coin by coin.

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