Coin Change Minimum Coins in DSA Rust

Mental Model

We want to find the smallest number of coins that add up to a target amount by trying all coin options step by step.

Analogy: Imagine you have different sized boxes and want to fill a big box exactly using the fewest small boxes possible. You try each box size and see which combination uses the least boxes.

amount: 0 1 2 3 4 5
coins: [1, 2, 5]
min_coins: [0, ∞, ∞, ∞, ∞, ∞]
↑ start with 0 amount needing 0 coins

Dry Run Walkthrough

Input: coins: [1, 2, 5], amount: 5

Goal: Find the minimum number of coins needed to make amount 5

Step 1: Set min_coins[0] = 0 because 0 coins are needed for amount 0

amount: 0 1 2 3 4 5
min_coins: [0, ∞, ∞, ∞, ∞, ∞]

Why: Base case: no coins needed to make zero amount

Step 2: Check coin 1 for amounts 1 to 5, update min_coins if smaller

amount: 0 1 2 3 4 5
min_coins: [0, 1, 2, 3, 4, 5]

Why: Using coin 1 repeatedly, min coins equals amount

Step 3: Check coin 2 for amounts 2 to 5, update min_coins if smaller

amount: 0 1 2 3 4 5
min_coins: [0, 1, 1, 2, 2, 3]

Why: Using coin 2 reduces coins needed for even amounts

Step 4: Check coin 5 for amount 5, update min_coins if smaller

amount: 0 1 2 3 4 5
min_coins: [0, 1, 1, 2, 2, 1]

Why: Using coin 5 directly makes amount 5 with 1 coin

Result:

amount: 0 1 2 3 4 5
min_coins: [0, 1, 1, 2, 2, 1]
Answer: 1 coin needed for amount 5

Annotated Code

DSA Rust

fn coin_change(coins: &[i32], amount: i32) -> i32 {
    let mut min_coins = vec![amount + 1; (amount + 1) as usize];
    min_coins[0] = 0; // base case: 0 coins for amount 0

    for a in 1..=amount {
        for &coin in coins {
            if coin <= a {
                let prev = min_coins[(a - coin) as usize];
                if prev != amount + 1 && prev + 1 < min_coins[a as usize] {
                    min_coins[a as usize] = prev + 1; // update with fewer coins
                }
            }
        }
    }

    if min_coins[amount as usize] > amount {
        -1 // no solution
    } else {
        min_coins[amount as usize]
    }
}

fn main() {
    let coins = vec![1, 2, 5];
    let amount = 5;
    let result = coin_change(&coins, amount);
    println!("Minimum coins needed: {}", result);
}

min_coins[0] = 0; // base case: 0 coins for amount 0

initialize base case for zero amount

for a in 1..=amount {

iterate through all amounts from 1 to target

for &coin in coins {

try each coin for current amount

if coin <= a {

only consider coin if it does not exceed current amount

if prev + 1 < min_coins[a as usize] {

update min_coins if using this coin reduces count

min_coins[a as usize] = prev + 1; // update with fewer coins

store new minimum coin count for amount a

if min_coins[amount as usize] > amount {

check if solution exists or return -1

Output

Minimum coins needed: 1

Complexity Analysis

Time: O(n * amount) because we check each coin for every amount up to target

Space: O(amount) because we store minimum coins needed for each amount

vs Alternative: Compared to recursive brute force which is exponential, this dynamic programming approach is efficient and avoids repeated work

Edge Cases

amount = 0

Returns 0 immediately since no coins are needed

DSA Rust

min_coins[0] = 0; // base case: 0 coins for amount 0

coins array empty

Returns -1 because no coins available to make any amount

DSA Rust

if min_coins[amount as usize] > amount { -1 }

amount cannot be formed by given coins

Returns -1 indicating no solution

DSA Rust

if min_coins[amount as usize] > amount { -1 }

When to Use This Pattern

When asked to find the minimum number of items to reach a target sum using given options, use the coin change minimum coins pattern with dynamic programming to build solutions bottom-up.

Common Mistakes

Mistake: Not initializing min_coins[0] to 0, causing incorrect base case
Fix: Set min_coins[0] = 0 before starting loops

Mistake: Not checking if coin value is less than or equal to current amount before using it
Fix: Add condition if coin <= amount before updating min_coins

Mistake: Returning min_coins[amount] without checking if it was updated, leading to wrong answers
Fix: Check if min_coins[amount] > amount and return -1 if true

Summary

Finds the fewest coins needed to make a target amount using given coin denominations.

Use when you want to minimize the number of items to reach a sum from fixed options.

Build the solution from smallest amounts up, updating minimum coins needed step by step.

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