Why Tabulation Bottom Up DP in DSA Go?

The Big Idea

Discover how building answers step-by-step can save you from endless repeated work!

The Scenario

Imagine you want to find the number of ways to climb stairs, but you try to count each path by hand, writing down every possible step combination.

The Problem

Counting all paths manually is slow and confusing. You might forget some paths or count some twice. It becomes impossible when the number of stairs grows.

The Solution

Tabulation Bottom Up DP builds the answer step by step from the smallest problem, storing results in a table. This way, you never repeat work and get the answer quickly.

Before vs After

✗ Before

func countWays(n int) int {
    if n == 0 || n == 1 {
        return 1
    }
    return countWays(n-1) + countWays(n-2)
}

✓ After

func countWays(n int) int {
    ways := make([]int, n+1)
    ways[0], ways[1] = 1, 1
    for i := 2; i <= n; i++ {
        ways[i] = ways[i-1] + ways[i-2]
    }
    return ways[n]
}

What It Enables

It enables solving complex problems efficiently by breaking them into smaller, manageable steps and remembering past results.

Real Life Example

Planning routes in a city where you want to know how many ways to reach a destination by taking one or two steps at a time without retracing steps.

Key Takeaways

Manual counting is slow and error-prone for big problems.

Tabulation builds solutions from the bottom up using a table.

This method saves time by avoiding repeated calculations.

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

D. Start solving from the largest problem and go downwards

Solution

1. Step 1: Understand tabulation approach

   Tabulation means filling a table starting from the smallest subproblems.

2. Step 2: Compare with other methods

   Unlike recursion, tabulation solves subproblems first and uses their answers to solve bigger problems.

3. Final Answer:

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

4. Quick Check:

   Tabulation = Bottom Up DP [OK]

Hint: Think bottom-up: solve small parts first, then bigger [OK]

Common Mistakes:

• Confusing tabulation with recursion
• Thinking tabulation starts from the largest problem
• Ignoring the use of a table to store results

2. Which of the following Go code snippets correctly initializes a DP table for tabulation with size n+1 and zero values?

easy

A. dp := []int{n+1}

B. dp := make([]int, n)

C. dp := new(int, n+1)

D. dp := make([]int, n+1)

Solution

1. Step 1: Understand Go slice initialization

   To create a slice of integers with length n+1 filled with zeros, use make([]int, n+1).

2. Step 2: Check other options

   dp := make([]int, n) creates size n, not n+1; dp := []int{n+1} is invalid syntax; dp := new(int, n+1) is invalid for slices.

3. Final Answer:

   dp := make([]int, n+1) -> Option D

4. Quick Check:

   make([]int, n+1) creates zero-filled slice [OK]

Hint: Use make([]int, size) for zero-filled slices [OK]

Common Mistakes:

• Using wrong slice size (n instead of n+1)
• Using new() which returns pointer, not slice
• Incorrect slice literal syntax

3. What is the output of the following Go code using bottom-up DP to compute Fibonacci numbers?

func fib(n int) int {
    dp := make([]int, n+1)
    dp[0], dp[1] = 0, 1
    for i := 2; i <= n; i++ {
        dp[i] = dp[i-1] + dp[i-2]
    }
    return dp[n]
}

fmt.Println(fib(5))

medium

A. 5

B. 8

C. 3

D. 13

Solution

1. Step 1: Trace DP table filling for n=5

   dp[0]=0, dp[1]=1; then dp[2]=1, dp[3]=2, dp[4]=3, dp[5]=5.

2. Step 2: Return dp[5]

   The function returns dp[5], which is 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Fibonacci(5) = 5 [OK]

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

Common Mistakes:

• Off-by-one errors in loop
• Wrong base cases initialization
• Confusing dp indices

4. Identify the bug in this bottom-up DP code for computing factorial in Go:

func factorial(n int) int {
    dp := make([]int, n+1)
    dp[0] = 1
    for i := 1; i < n; i++ {
        dp[i] = i * dp[i-1]
    }
    return dp[n]
}

medium

A. dp slice size should be n, not n+1

B. Loop should run until i <= n, not i < n

C. Return statement should be dp[n-1]

D. dp[0] should be 0, not 1

Solution

1. Step 1: Check loop range for factorial calculation

   Factorial needs to compute dp[n], so loop must include i = n (i <= n).

2. Step 2: Identify impact of loop range

   Current loop stops at i = n-1, so dp[n] remains zero, causing wrong return.

3. Final Answer:

   Loop should run until i <= n, not i < n -> Option B

4. Quick Check:

   Loop must cover all factorial steps [OK]

Hint: Factorial needs full range including n [OK]

Common Mistakes:

• Stopping loop before n
• Wrong base case initialization
• Returning wrong dp index

5. Given a list of coin denominations coins := []int{1, 2, 5} and an amount amount := 5, which bottom-up DP code snippet correctly computes the minimum number of coins needed to make the amount? Assume impossible cases return -1.

hard

A. dp := make([]int, amount+1); for i := range dp { dp[i] = amount+1 }; dp[0] = 0; for i := 1; i <= amount; i++ { for _, c := range coins { if c <= i && dp[i-c]+1 < dp[i] { dp[i] = dp[i-c] + 1 } } }; if dp[amount] > amount { return -1 } else { return dp[amount] }

B. dp := make([]int, amount); dp[0] = 0; for i := 1; i < amount; i++ { for _, c := range coins { if c <= i { dp[i] = dp[i-c] + 1 } } }; return dp[amount]

C. dp := make([]int, amount+1); dp[0] = 1; for i := 1; i <= amount; i++ { for _, c := range coins { if c <= i { dp[i] = dp[i-c] + 1 } } }; return dp[amount]

D. dp := make([]int, amount+1); for i := 0; i <= amount; i++ { dp[i] = 0 }; for i := 1; i <= amount; i++ { for _, c := range coins { if c <= i && dp[i-c] != 0 { dp[i] = dp[i-c] + 1 } } }; return dp[amount]

Solution

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

   Set dp size to amount+1, fill with amount+1 (impossible large), and dp[0]=0.

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

   For each amount i, check all coins c; if c <= i, update dp[i] = min(dp[i], dp[i-c]+1).

3. Step 3: Return result or -1 if impossible

   If dp[amount] > amount, return -1; else return dp[amount].

4. Final Answer:

   Option A code snippet -> Option A

5. Quick Check:

   Proper initialization and min update needed [OK]

Hint: Initialize dp with large number, update min coins [OK]

Common Mistakes:

• Wrong dp size or initialization
• Not checking if dp[i-c] is valid
• Incorrect base case dp[0]