Tabulation Bottom Up DP in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP table with zeros.

DSA C++

int dp[n+1] = {0};
for (int i = 0; i <= n; i++) {
    dp[i] = [1];
}

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

A0

B1

C-1

Di

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 instead of 0 causes incorrect base values.

Using -1 can cause wrong results in summations.

We initialize each element of the dp array to 0 to start the bottom-up calculation.

2fill in blank

medium

Complete the code to fill the DP table for the Fibonacci sequence.

DSA C++

dp[0] = 0;
dp[1] = 1;
for (int i = 2; i <= n; i++) {
    dp[i] = dp[i-1] [1] dp[i-2];
}

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

A/

B*

C-

D+

Attempts:

3 left

💡 Hint

Common Mistakes

Using multiplication instead of addition.

Using subtraction causes wrong sequence.

Fibonacci numbers are the sum of the two previous numbers.

3fill in blank

hard

Fix the error in the code to correctly compute the minimum coin count for amount 'amt'.

DSA C++

for (int i = 1; i <= amt; i++) {
    dp[i] = INT_MAX;
    for (int coin : coins) {
        if (i - coin >= 0 && dp[i - coin] != INT_MAX) {
            dp[i] = std::min(dp[i], dp[i - coin] [1] 1);
        }
    }
}

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

A*

B+

C/

D-

Attempts:

3 left

💡 Hint

Common Mistakes

Using subtraction causes wrong coin count.

Using multiplication or division is incorrect here.

We add 1 to the previous minimum to count the current coin.

4fill in blank

hard

Fill both blanks to complete the code for counting ways to make change.

DSA C++

dp[0] = 1;
for (int coin : coins) {
    for (int i = [1]; i <= amount; i++) {
        dp[i] = dp[i] [2] dp[i - coin];
    }
}

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

Acoin

B+=

C-=

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Starting loop from 0 causes invalid indices.

Using '-=' instead of '+=' causes wrong counts.

We start from coin value and add ways from dp[i - coin].

5fill in blank

hard

Complete the code to complete the code for Longest Common Subsequence (LCS) tabulation.

DSA C++

for (int i = 1; i <= m; i++) {
    for (int j = 1; j <= n; j++) {
        if (s1[i-1] == s2[j-1]) {
            dp[i][j] = dp[i-1][j-1] [1] 1;
        } else {
            dp[i][j] = std::max(dp[i-1][j], dp[i][j-1]) ;
        }
    }
}

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

A+

B-

D*

Attempts:

3 left

💡 Hint

Common Mistakes

Subtracting or multiplying instead of adding 1.

Adding or subtracting in else case causes wrong results.

Add 1 when characters match; else take max without change.

Practice

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

easy

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

B. Solve the problem by recursion without storing results

C. Use random guesses to find the solution

D. Start solving from the largest problem and break it down

Solution

1. Step 1: Understand bottom-up approach

   Tabulation means filling a table starting from the smallest subproblems, so we solve all smaller parts first.

2. Step 2: Compare with other methods

   Unlike recursion without memory, 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 A

4. Quick Check:

   Bottom-up DP = Build table from small to big [OK]

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

Common Mistakes:

• Confusing bottom-up with recursion without memoization
• Thinking it starts from the largest problem
• Ignoring the use of a table to store results

2. Which of the following is the correct way to declare a DP table for Fibonacci numbers using tabulation in C++?

easy

A. int dp[]; dp[1] = 1;

B. int dp[100] = {0}; dp[1] = 1;

C. int dp[100]; dp[0] = 1;

D. int dp[100] = {1}; dp[2] = 1;

Solution

1. Step 1: Check array initialization

   int dp[100] = {0}; dp[1] = 1; initializes dp with zeros and sets dp[1] = 1, which is correct for Fibonacci base cases.

2. Step 2: Identify errors in other options

   int dp[]; dp[1] = 1; has no size for dp array, invalid in C++. int dp[100]; dp[0] = 1; sets dp[0] = 1 which is incorrect base case. int dp[100] = {1}; dp[2] = 1; initializes dp with 1 at dp[0], which is wrong.

3. Final Answer:

   int dp[100] = {0}; dp[1] = 1; -> Option B

4. Quick Check:

   Correct base initialization = int dp[100] = {0}; dp[1] = 1; [OK]

Hint: Initialize dp array with zeros and set base cases explicitly [OK]

Common Mistakes:

• Not specifying array size
• Wrong base case initialization
• Assuming default values without initialization

3. What will be the output of the following C++ code using bottom-up DP to find the 5th Fibonacci number?

int fib(int n) {
  int dp[6] = {0};
  dp[1] = 1;
  for (int i = 2; i <= n; i++) {
    dp[i] = dp[i-1] + dp[i-2];
  }
  return dp[n];
}

int main() {
  std::cout << fib(5);
  return 0;
}

medium

A. 5

B. 8

C. 3

D. 13

Solution

1. Step 1: Trace dp array filling

   dp[0]=0, dp[1]=1; then dp[2]=1, dp[3]=2, dp[4]=3, dp[5]=5 by adding previous two.

2. Step 2: Return dp[5]

   dp[5] is 5, so output 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 indexing
• Confusing dp[0] and dp[1] values
• Returning wrong dp index

4. Identify the error in this bottom-up DP code for computing factorial of n:

int factorial(int n) {
  int dp[100];
  dp[0] = 1;
  for (int i = 1; i <= n; i++) {
    dp[i] = dp[i-1] * i;
  }
  return dp[n];
}

medium

A. dp array not initialized to zero

B. Loop should start from i = 0

C. dp array size might be too small for large n

D. No error, code is correct

Solution

1. Step 1: Check dp array size

   dp array size is fixed at 100, so if n > 99, dp[i] access is out of bounds causing error.

2. Step 2: Analyze other parts

   Initialization and loop logic are correct; loop starts at 1 and dp[0] = 1 is base case.

3. Final Answer:

   dp array size might be too small for large n -> Option C

4. Quick Check:

   Array size must fit n [OK]

Hint: Ensure dp array size >= n to avoid overflow [OK]

Common Mistakes:

• Assuming dp size is always enough
• Starting loop at wrong index
• Ignoring base case initialization

5. Given the problem to find the minimum number of coins to make amount 11 using coins {1, 3, 5}, which bottom-up DP table entry dp[11] will hold after running this code?

int coins[] = {1, 3, 5};
int n = 11;
int dp[12];
dp[0] = 0;
for (int i = 1; i <= n; i++) {
  dp[i] = 1000; // large number
  for (int c : coins) {
    if (i - c >= 0) {
      dp[i] = std::min(dp[i], dp[i-c] + 1);
    }
  }
}
std::cout << dp[n];

hard

A. 6

B. 5

C. 4

D. 3

Solution

1. Step 1: Understand DP logic for coin change

   dp[i] stores minimum coins for amount i. For 11, check coins 1,3,5 and pick minimum.

2. Step 2: Calculate dp values up to 11

   dp[0]=0; dp[1]=1; dp[2]=2; dp[3]=1; dp[4]=2; dp[5]=1; dp[6]=2; dp[7]=3; dp[8]=2; dp[9]=3; dp[10]=2; dp[11]=3 (e.g., 5+5+1 or 3+3+5)

3. Final Answer:

   3 -> Option D

4. Quick Check:

   Minimum coins for 11 = 3 [OK]

Hint: Build dp from 0 to n, pick min coins at each step [OK]

Common Mistakes:

• Not initializing dp[0] = 0
• Using max int instead of large number
• Forgetting to check i - coin >= 0