Why Tabulation Bottom Up DP in DSA Rust?

The Big Idea

Discover how building answers step-by-step can turn a slow guess into a fast solution!

The Scenario

Imagine you want to find the fastest way to climb stairs where each step depends on the previous ones. Doing this by trying every possible path manually is like guessing blindly.

The Problem

Trying every option manually takes too long and is confusing. You might repeat the same work many times, making it slow and error-prone.

The Solution

Tabulation builds a table from the bottom up, solving small parts first and using those answers to solve bigger parts. This way, you never repeat work and get the answer quickly.

Before vs After

✗ Before

fn climb_stairs(n: usize) -> usize {
    if n <= 1 { return 1; }
    climb_stairs(n - 1) + climb_stairs(n - 2)
}

✓ After

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

What It Enables

It lets you solve complex problems quickly by building answers step-by-step without repeating work.

Real Life Example

Planning the best route in a game where each move depends on previous moves, like jumping across stones in a river.

Key Takeaways

Manual guessing repeats work and is slow.

Tabulation uses a table to store answers from small to big.

This method saves time and avoids mistakes.

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