Tabulation Bottom Up DP in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP table with zeros.

DSA Rust

let mut dp = vec![0; [1]];

Drag options to blanks, or click blank then click option'

A1

Bn - 1

Cn

Dn + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n instead of n + 1 causes index out of bounds errors.

Using n - 1 misses the last element.

The DP table size should be n + 1 to include the base case at index 0.

2fill in blank

medium

Complete the code to fill the DP table bottom-up for Fibonacci numbers.

DSA Rust

for i in 2..=[1] {
    dp[i] = dp[i - 1] + dp[i - 2];
}

Drag options to blanks, or click blank then click option'

An

Bn - 1

Cn + 1

Dn * 2

Attempts:

3 left

💡 Hint

Common Mistakes

Looping only to n - 1 misses the last value.

Looping beyond n + 1 causes index errors.

The loop should run up to n inclusive to compute dp[n].

3fill in blank

hard

Fix the error in the DP table update for the coin change problem.

DSA Rust

for coin in coins.iter() {
    for amount in [1]..=target {
        if amount >= *coin {
            dp[amount] += dp[amount - *coin];
        }
    }
}

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Atarget

B0

Ccoin

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Starting from 1 causes missing combinations for amount 0.

Starting from coin causes skipping smaller amounts.

The inner loop should start from 0 to include all amounts up to target.

4fill in blank

hard

Fill both blanks to create a DP table for the knapsack problem.

DSA Rust

let mut dp = vec![vec![0; [1] + 1]; [2] + 1];

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Acapacity

Bweights.len()

Cvalues.len()

Dcapacity + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Swapping rows and columns causes logic errors.

Using values.len() instead of weights.len() for rows is incorrect.

The DP table has rows for items (weights.len()) and columns for capacities (capacity + 1).

5fill in blank

hard

Fill all three blanks to complete the DP update for the knapsack problem.

DSA Rust

for i in 1..=[1] {
    for w in 0..=[2] {
        if weights[i - 1] <= w {
            dp[i][w] = dp[i - 1][w].max(dp[i - 1][w - [3]] + values[i - 1]);
        } else {
            dp[i][w] = dp[i - 1][w];
        }
    }
}

Drag options to blanks, or click blank then click option'

Aweights.len()

Bcapacity

Cweights[i - 1]

Dvalues.len()

Attempts:

3 left

💡 Hint

Common Mistakes

Using values.len() instead of weights.len() for items count.

Using capacity + 1 in loop range causes index errors.

Using values[i - 1] instead of weights[i - 1] in dp index.

Loop over all items and capacities; use weights[i - 1] to check and update dp.

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