Coin Change Minimum Coins in DSA Rust - Time & Space Complexity

Time Complexity: Coin Change Minimum Coins

O(amount x number_of_coins)

Understanding Time Complexity

We want to know how the time needed to find the minimum coins changes as the amount grows.

How does the program's work increase when the amount or coin types increase?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code finds the fewest coins needed to make a given amount using dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops: outer loop runs from 1 to amount, inner loop runs over all coin types.
• How many times: Outer loop runs amount times, inner loop runs number of coins times for each outer iteration.

How Execution Grows With Input

As the amount grows, the work grows by the amount times the number of coins.

Input Size (amount)	Approx. Operations
10	10 x number_of_coins
100	100 x number_of_coins
1000	1000 x number_of_coins

Pattern observation: The operations increase linearly with the amount and linearly with the number of coin types.

Final Time Complexity

Time Complexity: O(amount x number_of_coins)

This means the time needed grows proportionally to the amount and how many coin types you have.

Common Mistake

[X] Wrong: "The time depends only on the amount, not the number of coins."

[OK] Correct: Each coin type is checked for every amount value, so more coins mean more work.

Interview Connect

Understanding this helps you explain how dynamic programming improves efficiency and why nested loops matter in real problems.

Self-Check

"What if we used recursion with memoization instead of loops? How would the time complexity change?"

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