Coin Change Minimum Coins in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp vector with a large number.

DSA Rust

let mut dp = vec![[1]; amount + 1];

Drag options to blanks, or click blank then click option'

A-1

B0

Cusize::MAX

Damount + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 0 will cause incorrect minimum calculations.

Using usize::MAX can cause overflow when adding 1.

We initialize dp with amount + 1 because it's larger than any possible minimum coins needed, acting as infinity.

2fill in blank

medium

Complete the code to set the base case for dp.

DSA Rust

dp[0] = [1];

Drag options to blanks, or click blank then click option'

A0

Busize::MAX

C1

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[0] to 1 or amount causes wrong results.

Using usize::MAX leads to overflow errors.

dp[0] = 0 means zero coins are needed to make amount zero.

3fill in blank

hard

Fix the error in the inner loop condition to avoid index out of bounds.

DSA Rust

for coin in coins.iter() {
    if coin <= &[1] {
        dp[i] = dp[i].min(dp[i - coin] + 1);
    }
}

Drag options to blanks, or click blank then click option'

Ai

Bdp[i]

Ccoin

Damount

Attempts:

3 left

💡 Hint

Common Mistakes

Comparing coin with amount instead of i causes logic errors.

Using coin or dp[i] in the condition causes type errors.

We compare coin with i (current amount) to ensure we don't access negative indices.

4fill in blank

hard

Fill both blanks to complete the outer loop and return statement.

DSA Rust

for [1] in 1..=[2] {
    for coin in coins.iter() {
        if coin <= &i {
            dp[i] = dp[i].min(dp[i - coin] + 1);
        }
    }
}

if dp[amount] > amount {
    -1
} else {
    dp[amount]
}

Drag options to blanks, or click blank then click option'

Ai

Bamount

Ccoin

Ddp

Attempts:

3 left

💡 Hint

Common Mistakes

Using coin as loop variable causes errors.

Looping beyond amount causes index errors.

The outer loop variable is i iterating from 1 to amount inclusive.

5fill in blank

hard

Fill all three blanks to complete the function signature, dp initialization, and return statement.

DSA Rust

fn coin_change([1]: Vec<i32>, [2]: i32) -> i32 {
    let amount = [2] as usize;
    let mut dp = vec![amount + 1; amount + 1];
    dp[0] = 0;

    for i in 1..=amount {
        for coin in [1].iter() {
            if (*coin as usize) <= i {
                dp[i] = dp[i].min(dp[i - *coin as usize] + 1);
            }
        }
    }

    if dp[amount] > amount {
        -1
    } else {
        dp[amount]
    }
}

Drag options to blanks, or click blank then click option'

Acoins

Bamount

Dnums

Attempts:

3 left

💡 Hint

Common Mistakes

Using inconsistent variable names causes errors.

Not casting amount to usize causes type errors.

The function takes coins vector and amount integer; dp uses amount; coins vector is iterated.

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