Tabulation Bottom Up DP in DSA Rust - Build from Scratch

Tabulation Bottom Up DP: Fibonacci Sequence

📖 Scenario: You want to find the nth number in the Fibonacci sequence. This sequence starts with 0 and 1, and each next number is the sum of the two before it. Calculating this directly can be slow for big numbers, so we use a method called tabulation to build the answer step-by-step from the bottom up.

🎯 Goal: Build a Rust program that uses tabulation bottom up dynamic programming to find the nth Fibonacci number efficiently.

📋 What You'll Learn

Create a vector to store Fibonacci numbers up to n

Use a loop to fill the vector from the bottom up

Print the nth Fibonacci number

💡 Why This Matters

🌍 Real World

Calculating Fibonacci numbers efficiently is useful in computer graphics, financial models, and algorithm design where repeated calculations happen.

💼 Career

Understanding bottom up dynamic programming is essential for software engineers to optimize algorithms and solve complex problems efficiently.

1

Create the Fibonacci table vector

Create a vector called fib_table with size n + 1 filled with zeros. Also create a variable n and set it to 10.

DSA Rust

let n = 10;
// Your code here

Hint

Use vec![0; n + 1] to create a vector of zeros with length n + 1.

2

Initialize base cases in the table

Set fib_table[0] to 0 and fib_table[1] to 1 to start the Fibonacci sequence.

DSA Rust

let n = 10;
let mut fib_table = vec![0; n + 1];
// Your code here

Hint

Set the first two Fibonacci numbers manually before the loop.

3

Fill the table using a loop

Use a for loop with variable i from 2 to n inclusive. Inside the loop, set fib_table[i] to the sum of fib_table[i - 1] and fib_table[i - 2].

DSA Rust

let n = 10;
let mut fib_table = vec![0; n + 1];
fib_table[0] = 0;
fib_table[1] = 1;
// Your code here

Hint

Use 2..=n to include n in the loop range.

4

Print the nth Fibonacci number

Print the value of fib_table[n] using println!.

DSA Rust

let n = 10;
let mut fib_table = vec![0; n + 1];
fib_table[0] = 0;
fib_table[1] = 1;
for i in 2..=n {
    fib_table[i] = fib_table[i - 1] + fib_table[i - 2];
}
// Your code here

Hint

The 10th Fibonacci number is 55. Use println!("{}", fib_table[n]) to print it.

Practice

1. What is the main idea behind Tabulation Bottom Up DP in dynamic programming?

easy

A. Use random guesses to find the solution

B. Solve the problem by recursion without storing results

C. Start solving from the main problem and break it down recursively

D. Build a table from the smallest subproblems up to the main problem

Solution

1. Step 1: Understand bottom-up approach

   Tabulation means filling a table starting from the smallest subproblems, building up to the final answer.

2. Step 2: Compare with other methods

   Unlike recursion without memoization, bottom-up avoids repeated work by storing results in a table.

3. Final Answer:

   Build a table from the smallest subproblems up to the main problem -> Option D

4. Quick Check:

   Tabulation = Bottom-up DP [OK]

Hint: Think: solve small parts first, then bigger ones [OK]

Common Mistakes:

• Confusing bottom-up with top-down recursion
• Ignoring the use of a table to store results
• Thinking tabulation uses random guesses

2. Which of the following Rust code snippets correctly initializes a vector for tabulation with size 5 and all zeros?

easy

A. let mut dp = vec![0; 5];

B. let dp = vec![0;];

C. let mut dp = Vec::new(5);

D. let dp = vec!(5, 0);

Solution

1. Step 1: Check Rust vector initialization syntax

   To create a vector of size 5 filled with zeros, use vec![0; 5].

2. Step 2: Analyze options

   let mut dp = vec![0; 5]; uses correct syntax. let dp = vec![0;]; is missing size. let mut dp = Vec::new(5); is invalid because Vec::new() takes no size. let dp = vec!(5, 0); uses wrong macro syntax.

3. Final Answer:

   let mut dp = vec![0; 5]; -> Option A

4. Quick Check:

   vec![value; size] creates vector [OK]

Hint: Use vec![value; size] to initialize vectors [OK]

Common Mistakes:

• Using Vec::new() with size argument
• Incorrect macro syntax with vec!()
• Omitting size in vector initialization

3. What will be the output of the following Rust code using bottom-up DP to compute Fibonacci numbers?

fn fib(n: usize) -> usize {
    let mut dp = vec![0; n + 1];
    dp[1] = 1;
    for i in 2..=n {
        dp[i] = dp[i - 1] + dp[i - 2];
    }
    dp[n]
}

fn main() {
    println!("{}", fib(6));
}

medium

A. 13

B. 5

C. 8

D. 21

Solution

1. Step 1: Trace Fibonacci calculation for n=6

   dp array fills as: dp[0]=0, dp[1]=1, dp[2]=1, dp[3]=2, dp[4]=3, dp[5]=5, dp[6]=8.

2. Step 2: Output dp[6]

   The function returns dp[6] which is 8.

3. Final Answer:

   8 -> Option C

4. Quick Check:

   fib(6) = 8 [OK]

Hint: Remember Fibonacci sequence: 0,1,1,2,3,5,8... [OK]

Common Mistakes:

• Starting dp[1] as 0 instead of 1
• Off-by-one errors in loop range
• Confusing dp indices with Fibonacci numbers

4. Identify the error in this Rust bottom-up DP code for computing factorial and select the fix:

fn factorial(n: usize) -> usize {
    let mut dp = vec![1; n + 1];
    for i in 1..=n {
        dp[i] = dp[i - 1] * i;
    }
    dp[n]
}

fn main() {
    println!("{}", factorial(5));
}

medium

A. Initialize dp with zeros instead of ones

B. Change loop to start from 2 instead of 1

C. Use dp[i] = dp[i] * i instead of dp[i - 1] * i

D. Return dp[n-1] instead of dp[n]

Solution

1. Step 1: Analyze initialization and loop

   dp[0] = 1 is correct for factorial base case. Loop starting at 1 causes dp[1] = dp[0] * 1 = 1, which is correct but redundant.

2. Step 2: Fix loop start to avoid redundant calculation

   Starting loop at 2 avoids unnecessary multiplication and matches factorial definition better.

3. Final Answer:

   Change loop to start from 2 instead of 1 -> Option B

4. Quick Check:

   Factorial loop starts at 2 [OK]

Hint: Factorial base case at dp[0], loop from 2 [OK]

Common Mistakes:

• Starting loop at 1 causing redundant calculation
• Initializing dp with zeros breaking multiplication
• Returning wrong dp index

5. You want to find the minimum number of coins needed to make amount 7 using coins [1, 3, 4] with bottom-up DP in Rust. Which code snippet correctly implements this?

hard

A. let mut dp = vec![usize::MAX; 8]; dp[0] = 0; for i in 1..=7 { for &coin in &[1,3,4] { if i >= coin && dp[i - coin] != usize::MAX { dp[i] = dp[i].min(dp[i - coin] + 1); } } } dp[7]

B. let mut dp = vec![0; 8]; for i in 1..=7 { for &coin in &[1,3,4] { dp[i] += dp[i - coin]; } } dp[7]

C. let mut dp = vec![usize::MAX; 7]; dp[0] = 0; for i in 1..7 { for &coin in &[1,3,4] { if i > coin { dp[i] = dp[i - coin] + 1; } } } dp[6]

D. let mut dp = vec![0; 7]; dp[0] = 1; for i in 1..7 { for &coin in &[1,3,4] { if i >= coin { dp[i] = dp[i - coin] + 1; } } } dp[6]

Solution

1. Step 1: Initialize dp with max values and base case

   dp size must be 8 for amounts 0 to 7. dp[0] = 0 means zero coins needed for amount 0.

2. Step 2: Update dp with minimum coins for each amount

   For each amount i, check all coins. If coin fits and previous dp is valid, update dp[i] with minimum coins.

3. Step 3: Return dp[7]

   dp[7] holds minimum coins needed for amount 7.

4. Final Answer:

   let mut dp = vec![usize::MAX; 8]; dp[0] = 0; for i in 1..=7 { for &coin in &[1,3,4] { if i >= coin && dp[i - coin] != usize::MAX { dp[i] = dp[i].min(dp[i - coin] + 1); } } } dp[7] -> Option A

5. Quick Check:

   Minimum coins DP uses min and checks valid previous states [OK]

Hint: Use usize::MAX and min() to track minimum coins [OK]

Common Mistakes:

• Incorrect dp size causing index errors
• Not checking if previous dp value is valid
• Adding instead of taking minimum coins