Tabulation Bottom Up DP in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Bottom-Up DP Fibonacci Implementation

What is the output of this Rust code that uses bottom-up tabulation to compute Fibonacci numbers?

DSA Rust

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

fn main() {
    println!("{}", fibonacci(7));
}

A 34

B 21

C 8

D 13

💡 Hint

Remember that Fibonacci sequence starts with 0, 1, 1, 2, 3, 5, 8, 13, 21...

The code computes Fibonacci numbers using bottom-up DP. fibonacci(7) returns dp[7], which is the 7th Fibonacci number (0-indexed): 0,1,1,2,3,5,8,13. So output is 13.

❓ Predict Output

intermediate

Output of Bottom-Up DP for Coin Change Ways

What is the output of this Rust code that calculates the number of ways to make change for amount 5 using coins [1, 2, 5] with bottom-up DP?

DSA Rust

fn coin_change_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]
}

fn main() {
    println!("{}", coin_change_ways(&[1, 2, 5], 5));
}

A 4

B 3

C 2

D 5

💡 Hint

Count all combinations of coins that sum to 5.

The code counts ways to make 5 using coins 1,2,5. The ways are: 5; 2+2+1; 2+1+1+1; 1+1+1+1+1. Total 4 ways.

🔧 Debug

advanced

Identify the Error in Bottom-Up DP for Longest Increasing Subsequence

What is the output of this Rust code that attempts to find the length of the longest increasing subsequence using bottom-up DP? Note the bug in the update logic.

DSA Rust

fn length_of_lis(nums: &[i32]) -> usize {
    let n = nums.len();
    let mut dp = vec![1; n];
    for i in 1..n {
        for j in 0..i {
            if nums[i] > nums[j] {
                dp[i] = dp[i].max(dp[j] + 1);
            }
        }
    }
    *dp.iter().max().unwrap()
}

fn main() {
    let nums = vec![10,9,2,5,3,7,101,18];
    println!("{}", length_of_lis(&nums));
}

A Outputs 5

B Outputs 1

C Outputs 4

D Outputs 3

💡 Hint

Check how dp[i] is updated inside the inner loop. It lacks a max comparison.

The code outputs 4, which is correct for this input (LIS length is 4, e.g., 2<3<7<18). However, it has a bug: dp[i] is set to dp[j]+1 without comparing to current dp[i] (should be dp[i] = dp[i].max(dp[j] + 1)). It takes the dp from the last qualifying j, not the maximum. This fails on other inputs, e.g., [10,20,1,30] outputs 2 instead of 3.

🧠 Conceptual

advanced

Understanding Time Complexity of Bottom-Up DP

What is the time complexity of the bottom-up DP solution for the 0/1 Knapsack problem with n items and capacity W?

A O(n + W)

B O(n * W)

C O(n^2 * W)

D O(2^n)

💡 Hint

Consider the nested loops over items and capacity.

The bottom-up DP uses a 2D table of size n by W, filling each cell once, so time complexity is O(n * W).

🚀 Application

expert

Final DP Table State for Edit Distance Problem

Given the strings s1 = "kitten" and s2 = "sitting", what is the value in the bottom-right cell of the DP table after running the bottom-up edit distance algorithm?

A 3

B 4

C 5

D 2

💡 Hint

Edit distance counts minimum insertions, deletions, and substitutions to convert s1 to s2.

The edit distance between "kitten" and "sitting" is 3: substitute 'k'->'s', substitute 'e'->'i', insert 'g'.

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