Coin Change Total Number of Ways in DSA Rust - Time & Space 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?
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 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.
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=10 | About 50 operations |
| c=5, n=100 | About 500 operations |
| c=10, n=1000 | About 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.
Time Complexity: O(c * n)
This means the time needed grows proportionally to the number of coins times the amount we want to make.
[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.
Understanding this time complexity helps you explain how dynamic programming solutions scale and why nested loops matter in practice.
"What if we used recursion with memoization instead of this bottom-up approach? How would the time complexity change?"
Practice
Solution
Step 1: Understand the problem goal
The problem asks for counting all possible combinations of coins that sum up to the target amount.Step 2: Differentiate from similar problems
Unlike minimum coins problem, here we count ways, not minimize coins.Final Answer:
The total number of ways to make a target amount using given coin denominations. -> Option BQuick Check:
Count ways, not minimum coins [OK]
- Confusing total ways with minimum coins needed
- Thinking about maximum coin value instead of count
- Ignoring that order of coins does not matter
Solution
Step 1: Initialize vector size correctly
Vector size must be target + 1 to include amount 0 to target.Step 2: Set base case for amount 0
ways[0] = 1 means one way to make amount 0 (using no coins).Final Answer:
let mut ways = vec![0; target + 1]; ways[0] = 1; -> Option AQuick Check:
Vector size = target+1, ways[0] = 1 [OK]
- Using vector size target instead of target+1
- Setting ways[0] to 0 instead of 1
- Initializing all elements to 1 incorrectly
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]);Solution
Step 1: Calculate ways using coin 1
ways after coin 1: ways[0]=1, ways[1]=1, ways[2]=1, ways[3]=1Step 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]Final Answer:
2 -> Option CQuick Check:
Ways to make 3 with [1,2] = 2 [OK]
- Not updating ways correctly for each coin
- Confusing ways count with minimum coins
- Off-by-one errors in loops
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]);Solution
Step 1: Check inner loop range
Starting inner loop from 1 causes unnecessary checks; it should start from coin value to avoid redundant iterations.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.Final Answer:
The inner loop should start from coin, not 1. -> Option AQuick Check:
Inner loop start = coin [OK]
- Starting inner loop from 1 instead of coin
- Incorrect initialization of ways[0]
- Misunderstanding loop roles
Solution
Step 1: Initialize ways vector
ways = [1,0,0,0,0,0,0,0] for amounts 0 to 7.Step 2: Update ways with coin 1
ways after coin 1: ways[i] = 1 for all i from 0 to 7 (only using 1s).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]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]Final Answer:
4 -> Option DQuick Check:
Count ways with coins [1,3,5] for 7 = 4 [OK]
- Not updating ways correctly for each coin
- Forgetting to include ways[0] = 1
- Mixing up order of loops
Ready to prove it?
Pass the test and this pattern moves unknown
