Longest Increasing Subsequence in DSA C++ - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp array with 1s for LIS calculation.

DSA C++

std::vector<int> dp(n, [1]);

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

A0

B-1

Cn

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing dp with 0 causes incorrect LIS length calculation.

Using n or -1 as initial values is incorrect for LIS.

Each element is at least an increasing subsequence of length 1 by itself, so dp is initialized with 1.

2fill in blank

medium

Complete the code to check if current element can extend the subsequence ending at j.

DSA C++

if (arr[i] [1] arr[j]) {

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

A<=

B>

C<

D>=

Attempts:

3 left

💡 Hint

Common Mistakes

Using >= or <= allows non-increasing sequences.

Using < instead of > reverses the logic.

For LIS, we check if current element is greater than previous to extend the subsequence.

3fill in blank

hard

Fix the error in updating dp[i] when a longer subsequence is found.

DSA C++

dp[i] = std::max(dp[i], dp[[1]] + 1);

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

An

Bi

Cj

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using dp[i] + 1 causes no change or infinite loop.

Using dp[n] or dp[0] is out of context.

We update dp[i] based on dp[j] where j < i and arr[i] > arr[j].

4fill in blank

hard

Fill both blanks to complete the nested loops for LIS calculation.

DSA C++

for (int [1] = 1; [1] < n; [1]++) {
    for (int [2] = 0; [2] < [1]; [2]++) {

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

Ai

Bj

Ck

Dx

Attempts:

3 left

💡 Hint

Common Mistakes

Using same variable name for both loops causes errors.

Using indices out of range breaks the logic.

Outer loop uses i from 1 to n-1, inner loop uses j from 0 to i-1 to compare elements.

5fill in blank

hard

Fill all three blanks to find the maximum LIS length after dp is computed.

DSA C++

int max_len = 0;
for (int [1] = 0; [1] < n; [1]++) {
    max_len = std::max(max_len, dp[[2]]);
}
return [3];

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

Ai

Bj

Cmax_len

Ddp[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using different variables for loop and dp access causes errors.

Returning dp[i] instead of max_len returns partial result.

We iterate i from 0 to n-1, update max_len with dp[i], and return max_len.

Practice

1. What does the Longest Increasing Subsequence (LIS) problem ask for in a given sequence of numbers?

easy

A. The longest subsequence where numbers are equal

B. The longest subsequence where each number is larger than the previous one

C. The longest subsequence where each number is smaller than the previous one

D. The shortest subsequence with increasing numbers

Solution

1. Step 1: Understand the problem definition

   The LIS problem looks for a subsequence where every next number is bigger than the one before it.

2. Step 2: Compare options with the definition

   The longest subsequence where each number is larger than the previous one matches the definition exactly, others describe different or incorrect sequences.

3. Final Answer:

   The longest subsequence where each number is larger than the previous one -> Option B

4. Quick Check:

   LIS means increasing subsequence = The longest subsequence where each number is larger than the previous one [OK]

Hint: LIS means strictly increasing numbers in order [OK]

Common Mistakes:

• Confusing increasing with decreasing subsequence
• Thinking LIS means contiguous subarray
• Assuming equal numbers count as increasing

2. Which of the following is the correct way to declare a vector for storing LIS lengths in C++?

easy

A. std::vector<int> lis(n, 1);

B. vector<int> lis(n);

C. std::vector lis(n, 0);

D. vector lis<int>(n, 1);

Solution

1. Step 1: Check correct vector declaration syntax

   In C++, std::vector<int> lis(n, 1); creates a vector of size n with all values initialized to 1.

2. Step 2: Validate each option

   std::vector<int> lis(n, 1); uses correct syntax and initialization. vector<int> lis(n); misses initialization, C misses template argument, D has wrong template syntax.

3. Final Answer:

   std::vector<int> lis(n, 1); -> Option A

4. Quick Check:

   Correct vector syntax with initialization = std::vector<int> lis(n, 1); [OK]

Hint: Use std::vector<int> name(size, initial_value) [OK]

Common Mistakes:

• Omitting std:: or using wrong template syntax
• Not initializing vector values
• Using wrong brackets or missing template argument

3. What is the output of the following C++ code snippet?

std::vector<int> arr = {10, 22, 9, 33, 21, 50, 41, 60};
int n = arr.size();
std::vector<int> lis(n, 1);
for (int i = 1; i < n; i++) {
  for (int j = 0; j < i; j++) {
    if (arr[i] > arr[j] && lis[i] < lis[j] + 1) {
      lis[i] = lis[j] + 1;
    }
  }
}
int max_lis = 0;
for (int i = 0; i < n; i++) {
  if (lis[i] > max_lis) max_lis = lis[i];
}
std::cout << max_lis << std::endl;

medium

A. 3

B. 4

C. 6

D. 5

Solution

1. Step 1: Trace LIS calculation for each element

   Calculate lis array step-by-step: lis = [1,2,1,3,2,4,4,5]

2. Step 2: Find maximum value in lis array

   Maximum lis value is 5, which is printed as output.

3. Final Answer:

   5 -> Option D

4. Quick Check:

   Max LIS length = 5 [OK]

Hint: Track lis array updates to find max length [OK]

Common Mistakes:

• Confusing subsequence with subarray
• Not updating lis[i] correctly
• Forgetting to find max after loops

4. Identify the bug in this LIS code snippet:

std::vector<int> arr = {3, 10, 2, 1, 20};
int n = arr.size();
std::vector<int> lis(n, 1);
for (int i = 0; i < n; i++) {
  for (int j = 0; j < i; j++) {
    if (arr[i] > arr[j]) {
      lis[i] = lis[j] + 1;
    }
  }
}
int max_lis = *max_element(lis.begin(), lis.end());
std::cout << max_lis << std::endl;

medium

A. Missing check if lis[i] < lis[j] + 1 before updating

B. Wrong initialization of lis vector

C. Incorrect loop ranges for i and j

D. Using max_element incorrectly

Solution

1. Step 1: Analyze update condition inside nested loops

   The code updates lis[i] without checking if lis[j] + 1 is greater, causing wrong LIS values.

2. Step 2: Identify missing condition

   Adding 'if (lis[i] < lis[j] + 1)' before update fixes the bug.

3. Final Answer:

   Missing check if lis[i] < lis[j] + 1 before updating -> Option A

4. Quick Check:

   Update only if new LIS is longer [OK]

Hint: Always check if new LIS length is greater before update [OK]

Common Mistakes:

• Updating lis[i] blindly overwrites correct values
• Confusing loop indices
• Incorrect use of max_element function

5. Given the sequence {5, 7, 4, 8, 6, 9}, what is the length of the Longest Increasing Subsequence if you must also ensure the subsequence elements are at even indices only (0-based)?

medium

A. 4

B. 3

C. 2

D. 1

Solution

1. Step 1: Identify even indices (0-based)

   Positions 0, 2, 4 have values 5, 4, 6.

2. Step 2: Compute LIS on {5, 4, 6}

   Possible increasing subsequences: {5, 6} or {4, 6}, both length 2. No length 3 possible (5 > 4).

3. Final Answer:

   2 -> Option C

4. Quick Check:

   LIS length on even indices only = 2 [OK]

Hint: Filter even indices first, then find LIS [OK]

Common Mistakes:

• Ignoring index restriction
• Using full sequence instead of filtered
• Assuming contiguous subsequence