Coin Change Minimum Coins in DSA Rust - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the goal of the Coin Change Minimum Coins problem?

To find the smallest number of coins needed to make a given amount using available coin denominations.

beginner

In the Coin Change Minimum Coins problem, what does a dynamic programming table usually store?

It stores the minimum number of coins needed to make every amount from 0 up to the target amount.

intermediate

Why do we initialize the DP array with a large number (like amount + 1) in the Coin Change problem?

Because it represents an impossible high number of coins, so we can find the minimum by comparing and updating values.

intermediate

What does it mean if the DP value for the target amount remains the large initial number after processing?

It means the amount cannot be formed with the given coins, so no solution exists.

beginner

How does the Coin Change Minimum Coins problem relate to real life?

It is like trying to pay a bill with the fewest coins possible, saving time and effort.

What does the DP array represent in the Coin Change Minimum Coins problem?

A Sum of all coin values

B Maximum coins needed for each amount

C Number of coin types

D Minimum coins needed for each amount up to target

The DP array stores the minimum number of coins needed to make each amount from 0 to the target.

If you have coins [1, 3, 4] and want to make amount 6, what is the minimum number of coins needed?

A 2

B 3

C 4

D 1

Using coins 3 and 3, you can make 6 with 2 coins.

What initial value is commonly used to fill the DP array before computation?

A -1

B 0

C amount + 1

D 1

We use amount + 1 as a large number to represent impossible cases.

What does it mean if dp[amount] equals the initial large value after processing?

A Amount can be formed

B Amount cannot be formed

C Coins are unlimited

D Amount is zero

It means no combination of coins can form the amount.

Which approach is commonly used to solve the Coin Change Minimum Coins problem?

A Dynamic programming

B Greedy algorithm

C Divide and conquer

D Backtracking

Dynamic programming efficiently finds the minimum coins by building solutions for smaller amounts.

Explain how dynamic programming helps solve the Coin Change Minimum Coins problem.

Think about how you can reuse answers for smaller amounts to solve bigger ones.

Describe the steps to find the minimum number of coins needed for a given amount using coin denominations.

Start from zero and build up to the target amount.

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