Longest Increasing Subsequence in DSA C++ - Time & Space Complexity

Time Complexity: Longest Increasing Subsequence

O(n²)

Understanding Time Complexity

We want to understand how the time needed to find the longest increasing subsequence grows as the input list gets bigger.

How does the number of steps change when the input size increases?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

int lengthOfLIS(vector<int>& nums) {
    int n = nums.size();
    vector<int> dp(n, 1);
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (nums[j] < nums[i]) {
                dp[i] = max(dp[i], dp[j] + 1);
            }
        }
    }
    return *max_element(dp.begin(), dp.end());
}
    

This code finds the length of the longest increasing subsequence by checking all pairs and building up solutions.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops comparing pairs of elements.
• How many times: Outer loop runs n times; inner loop runs up to i times, so roughly n x n = n² times.

How Execution Grows With Input

As the input size grows, the number of comparisons grows much faster because each new element is compared with all previous ones.

Input Size (n)	Approx. Operations
10	About 100 comparisons
100	About 10,000 comparisons
1000	About 1,000,000 comparisons

Pattern observation: The operations grow roughly by the square of the input size.

Final Time Complexity

Time Complexity: O(n²)

This means if the input doubles, the time to run roughly quadruples because every element is compared with many others.

Common Mistake

[X] Wrong: "The algorithm runs in linear time because it just loops through the list once."

[OK] Correct: There are actually two loops nested inside each other, so the inner loop runs many times for each outer loop step, making it much slower than just one pass.

Interview Connect

Understanding this time complexity helps you explain why some solutions are slower and shows you the value of more efficient methods, which is a great skill to share in interviews.

Self-Check

"What if we used a binary search approach inside the loop to find positions? How would the time complexity change?"

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