Coin Change Minimum Coins in DSA Rust - Build from Scratch

Coin Change Minimum Coins

📖 Scenario: You are helping a cashier who needs to give change using the fewest coins possible. The cashier has coins of certain fixed values.We want to find the minimum number of coins needed to make a specific amount.

🎯 Goal: Build a Rust program that calculates the minimum number of coins needed to make a given amount using a list of coin values.

📋 What You'll Learn

Create a vector called coins with the exact values 1, 3, and 5

Create a variable called amount and set it to 11

Write a function called min_coins that takes coins: &Vec<i32> and amount: i32 and returns i32

Use a vector called dp inside min_coins to store minimum coins needed for each amount from 0 to amount

Print the minimum number of coins needed for the given amount

💡 Why This Matters

🌍 Real World

Cashiers and vending machines need to give change using the fewest coins to save time and reduce coin usage.

💼 Career

Understanding dynamic programming and optimization problems is useful for software engineers working on financial software, embedded systems, and algorithm design.

1

Create the coin values vector

Create a vector called coins with the exact values 1, 3, and 5.

DSA Rust

# Create a vector called coins with values 1, 3, and 5
# Your code here

Hint

Use let coins = vec![1, 3, 5]; to create the vector.

2

Set the amount to change

Create a variable called amount and set it to 11 inside the main function, below the coins vector.

DSA Rust

fn main() {
    let coins = vec![1, 3, 5];
    // Create a variable called amount and set it to 11
    // Your code here
}

Hint

Use let amount = 11; to create the variable.

3

Write the min_coins function

Write a function called min_coins that takes coins: &Vec<i32> and amount: i32 and returns i32. Inside, create a vector called dp of size amount + 1 filled with amount + 1. Set dp[0] to 0. Use a for loop with variable i from 1 to amount inclusive. Inside, use a for loop with variable coin iterating over coins. If coin <= i, update dp[i] to the minimum of dp[i] and dp[i - coin] + 1. Return dp[amount] if it is not greater than amount, else return -1.

DSA Rust

fn main() {
    let coins = vec![1, 3, 5];
    let amount = 11;
}

// Write the min_coins function below
// Your code here

Hint

Use dynamic programming with a vector dp to store minimum coins for each amount.

4

Print the minimum coins needed

Call the min_coins function with &coins and amount inside main. Print the result using println!("{}", result); where result is the returned value.

DSA Rust

fn main() {
    let coins = vec![1, 3, 5];
    let amount = 11;

    fn min_coins(coins: &Vec, amount: i32) -> i32 {
        let mut dp = vec![amount + 1; (amount + 1) as usize];
        dp[0] = 0;
        for i in 1..=amount {
            for &coin in coins.iter() {
                if coin  amount {
            -1
        } else {
            dp[amount as usize]
        }
    }

    // Call min_coins and print the result
    // Your code here
}

Hint

Call min_coins(&coins, amount) and print the result.

Practice

1. What does the "Coin Change Minimum Coins" problem ask you to find?

easy

A. The smallest number of coins needed to make a given amount

B. The total number of different coins available

C. The largest coin value in the list

D. The sum of all coin values

Solution

1. Step 1: Understand the problem goal

   The problem asks for the minimum number of coins needed to make a specific amount using given coin denominations.

2. Step 2: Identify the correct interpretation

   Among the options, only the smallest number of coins needed matches the problem's goal.

3. Final Answer:

   The smallest number of coins needed to make a given amount -> Option A

4. Quick Check:

   Minimum coins = smallest count [OK]

Hint: Focus on 'minimum coins' needed for the amount [OK]

Common Mistakes:

• Confusing total coin count with minimum coins
• Thinking about coin values instead of counts
• Ignoring the target amount

2. Which Rust code snippet correctly initializes a vector of coin denominations for the coin change problem?

easy

A. let coins = {1, 2, 5};

B. let coins = [1, 2, 5];

C. let coins = vec![1, 2, 5];

D. let coins = (1, 2, 5);

Solution

1. Step 1: Identify Rust vector initialization syntax

   In Rust, vectors are created using vec![elements], which allows dynamic sizing.

2. Step 2: Compare options

   Only let coins = vec![1, 2, 5]; uses the vec! macro correctly. let coins = [1, 2, 5]; creates an array with fixed size. let coins = (1, 2, 5); creates a tuple. let coins = {1, 2, 5}; has invalid syntax.

3. Final Answer:

   let coins = vec![1, 2, 5]; -> Option C

4. Quick Check:

   vec! macro creates vector [OK]

Hint: Use vec![...] to create dynamic coin lists [OK]

Common Mistakes:

• Using tuple or array instead of vector
• Using curly braces which are invalid here
• Forgetting the vec! macro

3. What is the output of the following Rust code snippet for minimum coins to make amount 11 with coins [1, 2, 5]?

fn min_coins(coins: &[i32], amount: i32) -> i32 {
    let mut dp = vec![amount + 1; (amount + 1) as usize];
    dp[0] = 0;
    for i in 1..=amount {
        for &coin in coins {
            if coin <= i {
                dp[i as usize] = dp[i as usize].min(dp[(i - coin) as usize] + 1);
            }
        }
    }
    if dp[amount as usize] > amount { -1 } else { dp[amount as usize] }
}

fn main() {
    let coins = vec![1, 2, 5];
    println!("{}", min_coins(&coins, 11));
}

medium

A. 5

B. 3

C. 4

D. -1

Solution

1. Step 1: Understand the dynamic programming array dp

   dp[i] stores minimum coins needed for amount i. Initialized with amount+1 (impossible high value), dp[0]=0.

2. Step 2: Trace dp updates for amount 11

   Using coins 1,2,5, minimum coins for 11 is 3 (5+5+1). The code finds dp[11] = 3.

3. Final Answer:

   3 -> Option B

4. Quick Check:

   11 = 5+5+1 uses 3 coins [OK]

Hint: Check dp array for minimum coins at target amount [OK]

Common Mistakes:

• Confusing dp index with coin values
• Not initializing dp[0] to zero
• Returning amount instead of -1 when no solution

4. Identify the bug in this Rust function for minimum coins:

fn min_coins(coins: &[i32], amount: i32) -> i32 {
    let mut dp = vec![0; (amount + 1) as usize];
    for i in 1..=amount {
        dp[i as usize] = i;
        for &coin in coins {
            if coin <= i {
                dp[i as usize] = dp[i as usize].min(dp[(i - coin) as usize] + 1);
            }
        }
    }
    dp[amount as usize]
}

medium

A. Loop range should start from 0, not 1

B. Coins vector should be mutable

C. Missing return statement

D. dp vector initialized with zeros instead of large values

Solution

1. Step 1: Check dp initialization

   dp is initialized with zeros, which causes min comparisons to fail because 0 is less than any valid coin count.

2. Step 2: Correct initialization

   dp should be initialized with a large value (like amount+1) to represent impossible states, allowing min to work correctly.

3. Final Answer:

   dp vector initialized with zeros instead of large values -> Option D

4. Quick Check:

   dp init affects min logic [OK]

Hint: Initialize dp with large values, not zeros [OK]

Common Mistakes:

• Starting loop at 0 instead of 1
• Forgetting to return dp[amount]
• Making coins mutable unnecessarily

5. Given coin denominations [1, 3, 4] and amount 6, what is the minimum number of coins needed? Use the coin change minimum coins approach.

hard

A. 2

B. 3

C. 4

D. -1

Solution

1. Step 1: Analyze possible combinations for amount 6

   Using coins 1,3,4, possible sums: 3+3=6 (2 coins), 4+1+1=6 (3 coins), 1+1+1+1+1+1=6 (6 coins).

2. Step 2: Find minimum coins

   The minimum coins needed is 2 (two 3-value coins).

3. Final Answer:

   2 -> Option A

4. Quick Check:

   6 = 3+3 uses 2 coins [OK]

Hint: Try largest coins first to minimize count [OK]

Common Mistakes:

• Choosing 4+1+1 instead of 3+3
• Assuming no solution exists
• Counting coins incorrectly