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DSA Rustprogramming~5 mins

Coin Change Total Number of Ways in DSA Rust - Time & Space Complexity

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Time Complexity: Coin Change Total Number of Ways
O(c * n)
Understanding Time Complexity

We want to understand how the time needed to find the total ways to make change grows as the amount or number of coins increases.

How does the work increase when we have more coins or a bigger amount?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn count_ways(coins: &[usize], amount: usize) -> usize {
    let mut dp = vec![0; amount + 1];
    dp[0] = 1;
    for &coin in coins {
        for x in coin..=amount {
            dp[x] += dp[x - coin];
        }
    }
    dp[amount]
}

This code calculates how many ways we can make the given amount using the provided coin denominations.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

  • Primary operation: Nested loops where the outer loop goes through each coin, and the inner loop goes through amounts from the coin value up to the target amount.
  • How many times: Outer loop runs once per coin (let's say c coins), inner loop runs up to amount (n) times for each coin.
How Execution Grows With Input

As we add more coins or increase the amount, the number of operations grows roughly by multiplying these two factors.

Input Size (coins c, amount n)Approx. Operations
c=5, n=10About 50 operations
c=5, n=100About 500 operations
c=10, n=1000About 10,000 operations

Pattern observation: Doubling the number of coins or the amount roughly doubles the work, so the total work grows like the product of coins and amount.

Final Time Complexity

Time Complexity: O(c * n)

This means the time needed grows proportionally to the number of coins times the amount we want to make.

Common Mistake

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

[OK] Correct: Each coin adds a full pass through the amounts, so more coins mean more work, not just the amount.

Interview Connect

Understanding this time complexity helps you explain how dynamic programming solutions scale and why nested loops matter in practice.

Self-Check

"What if we used recursion with memoization instead of this bottom-up approach? How would the time complexity change?"

Practice

(1/5)
1. What does the "Coin Change Total Number of Ways" problem ask you to find?
easy
A. The minimum number of coins needed to make a target amount.
B. The total number of ways to make a target amount using given coin denominations.
C. The maximum value of coins that can be used without exceeding the target.
D. The total sum of all coin denominations.

Solution

  1. Step 1: Understand the problem goal

    The problem asks for counting all possible combinations of coins that sum up to the target amount.
  2. Step 2: Differentiate from similar problems

    Unlike minimum coins problem, here we count ways, not minimize coins.
  3. Final Answer:

    The total number of ways to make a target amount using given coin denominations. -> Option B
  4. Quick Check:

    Count ways, not minimum coins [OK]
Hint: Count combinations, not minimum coins [OK]
Common Mistakes:
  • Confusing total ways with minimum coins needed
  • Thinking about maximum coin value instead of count
  • Ignoring that order of coins does not matter
2. Which of the following Rust code snippets correctly initializes a vector to store ways to make amounts up to `target` with 0 ways initially except for 0 amount which has 1 way?
easy
A. let mut ways = vec![0; target + 1]; ways[0] = 1;
B. let mut ways = vec![1; target]; ways[0] = 0;
C. let mut ways = vec![0; target]; ways[0] = 1;
D. let mut ways = vec![1; target + 1]; ways[0] = 0;

Solution

  1. Step 1: Initialize vector size correctly

    Vector size must be target + 1 to include amount 0 to target.
  2. Step 2: Set base case for amount 0

    ways[0] = 1 means one way to make amount 0 (using no coins).
  3. Final Answer:

    let mut ways = vec![0; target + 1]; ways[0] = 1; -> Option A
  4. Quick Check:

    Vector size = target+1, ways[0] = 1 [OK]
Hint: ways vector size = target+1, ways[0] = 1 [OK]
Common Mistakes:
  • Using vector size target instead of target+1
  • Setting ways[0] to 0 instead of 1
  • Initializing all elements to 1 incorrectly
3. What is the output of the following Rust code snippet for coins = [1, 2] and target = 3?
let coins = vec![1usize, 2usize];
let target: usize = 3;
let mut ways = vec![0; target + 1];
ways[0] = 1;
for &coin in &coins {
    for amount in coin..=target {
        ways[amount] += ways[amount - coin];
    }
}
println!("{}", ways[target]);
medium
A. 4
B. 1
C. 2
D. 3

Solution

  1. Step 1: Calculate ways using coin 1

    ways after coin 1: ways[0]=1, ways[1]=1, ways[2]=1, ways[3]=1
  2. Step 2: Update ways using coin 2

    ways[2] += ways[0] => 1+1=2; ways[3] += ways[1] => 1+1=2; final ways = [1,1,2,2]
  3. Final Answer:

    2 -> Option C
  4. Quick Check:

    Ways to make 3 with [1,2] = 2 [OK]
Hint: Add ways for each coin incrementally [OK]
Common Mistakes:
  • Not updating ways correctly for each coin
  • Confusing ways count with minimum coins
  • Off-by-one errors in loops
4. Identify the bug in this Rust code snippet for coin change ways calculation:
let coins = vec![1usize, 2usize, 3usize];
let target: usize = 4;
let mut ways = vec![0; target + 1];
ways[0] = 1;
for &coin in &coins {
    for amount in 1usize..=target {
        if amount >= coin {
            ways[amount] += ways[amount - coin];
        }
    }
}
println!("{}", ways[target]);
medium
A. The inner loop should start from coin, not 1.
B. ways[0] should be initialized to 0.
C. coins vector should be mutable.
D. The outer loop should iterate over amounts, not coins.

Solution

  1. Step 1: Check inner loop range

    Starting inner loop from 1 causes unnecessary checks; it should start from coin value to avoid redundant iterations.
  2. Step 2: Confirm initialization and loops

    ways[0] = 1 is correct; coins vector does not need to be mutable; outer loop over coins is correct.
  3. Final Answer:

    The inner loop should start from coin, not 1. -> Option A
  4. Quick Check:

    Inner loop start = coin [OK]
Hint: Inner loop starts at coin value to optimize [OK]
Common Mistakes:
  • Starting inner loop from 1 instead of coin
  • Incorrect initialization of ways[0]
  • Misunderstanding loop roles
5. Given coins = [1, 3, 5] and target = 7, how many total ways are there to make the target amount? Use the coin change total number of ways approach.
hard
A. 6
B. 5
C. 7
D. 4

Solution

  1. Step 1: Initialize ways vector

    ways = [1,0,0,0,0,0,0,0] for amounts 0 to 7.
  2. Step 2: Update ways with coin 1

    ways after coin 1: ways[i] = 1 for all i from 0 to 7 (only using 1s).
  3. Step 3: Update ways with coin 3

    ways[3]+=ways[0]=1, ways[4]+=ways[1]=1, ways[5]+=ways[2]=1, ways[6]+=ways[3]=2, ways[7]+=ways[4]=2; ways now: [1,1,1,2,2,2,3,3]
  4. Step 4: Update ways with coin 5

    ways[5] += ways[0] => 2 + 1 = 3; ways[6] += ways[1] => 3 + 1 = 4; ways[7] += ways[2] => 3 + 1 = 4; final ways: [1,1,1,2,2,3,4,4]
  5. Final Answer:

    4 -> Option D
  6. Quick Check:

    Count ways with coins [1,3,5] for 7 = 4 [OK]
Hint: Add ways incrementally for each coin [OK]
Common Mistakes:
  • Not updating ways correctly for each coin
  • Forgetting to include ways[0] = 1
  • Mixing up order of loops
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