Coin Change Total Number of Ways in DSA Rust - Build from Scratch

Coin Change Total Number of Ways

📖 Scenario: Imagine you have different types of coins and you want to find out how many ways you can make a certain amount using these coins. This is like having a piggy bank with coins of different values and figuring out all the ways to pay a bill exactly.

🎯 Goal: You will write a Rust program that calculates the total number of ways to make a given amount using a list of coin values.

📋 What You'll Learn

Create a vector called coins with exact values 1, 2, 3

Create a variable called amount and set it to 4

Write a function called count_ways that takes amount and coins and returns the total number of ways to make the amount

Print the result of count_ways(amount, &coins)

💡 Why This Matters

🌍 Real World

This problem models how to count combinations in money transactions, budgeting, or resource allocation where different units can be combined in multiple ways.

💼 Career

Understanding dynamic programming and combinatorial counting is useful for software engineers working on finance apps, game development, and optimization problems.

1

Create the coin list

Create a vector called coins with the exact values 1, 2, 3.

DSA Rust

# Create a vector called coins with values 1, 2, 3
# Your code here

Hint

Use let coins = vec![1, 2, 3]; to create the vector.

2

Set the amount to make

Create a variable called amount and set it to 4.

DSA Rust

fn main() {
    let coins = vec![1, 2, 3];
    // Create a variable called amount and set it to 4
    // Your code here
}

Hint

Use let amount = 4; to set the amount.

3

Write the function to count ways

Write a function called count_ways that takes amount as usize and a reference to a vector coins of usize, and returns a usize representing the total number of ways to make the amount using the coins. Use dynamic programming with a vector dp where dp[i] stores the number of ways to make amount i. Initialize dp[0] to 1. For each coin, update dp[i] for i from coin to amount. Return dp[amount]. Add this function above main.

DSA Rust

// Write the count_ways function here

fn main() {
    let coins = vec![1, 2, 3];
    let amount = 4;
}

Hint

Use a vector dp to store ways, initialize dp[0] = 1, then update for each coin.

4

Print the total number of ways

Print the result of calling count_ways(amount, &coins) inside main using println!.

DSA Rust

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

fn main() {
    let coins = vec![1, 2, 3];
    let amount = 4;
    // Print the total number of ways to make amount using coins
    // Your code here
}

Hint

Use println!("{}", count_ways(amount, &coins)); to print the result.

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