Why Longest Increasing Subsequence in DSA C++?

The Big Idea

Discover how to find the longest rising pattern in a list without endless guessing!

The Scenario

Imagine you have a list of numbers representing daily temperatures, and you want to find the longest stretch of days where the temperature keeps rising.

Doing this by checking every possible combination manually is like trying to find a needle in a haystack.

The Problem

Manually checking all increasing sequences means looking at every possible group of days, which grows very fast as the list gets longer.

This takes too much time and is easy to mess up, especially with long lists.

The Solution

The Longest Increasing Subsequence (LIS) method helps find the longest rising sequence efficiently without checking every possibility.

It uses smart steps to remember and build the best sequences as it goes, saving time and effort.

Before vs After

✗ Before

int longestIncreasingSequence(vector<int> nums) {
  int maxLength = 1;
  for (int i = 0; i < nums.size(); i++) {
    for (int j = i + 1; j < nums.size(); j++) {
      // check if nums[j] > nums[i] and count length
    }
  }
  return maxLength;
}

✓ After

int longestIncreasingSubsequence(vector<int> nums) {
  vector<int> dp(nums.size(), 1);
  for (int current = 1; current < nums.size(); current++) {
    for (int previous = 0; previous < current; previous++) {
      if (nums[current] > nums[previous]) {
        dp[current] = max(dp[current], dp[previous] + 1);
      }
    }
  }
  return *max_element(dp.begin(), dp.end());
}

What It Enables

This concept makes it possible to quickly find the longest rising pattern in any list, unlocking insights from data that would be too slow to find manually.

Real Life Example

In stock market analysis, finding the longest increasing subsequence helps identify the longest period when stock prices kept going up, guiding better investment decisions.

Key Takeaways

Manual checking is slow and error-prone for long lists.

LIS uses smart memory to build solutions step-by-step.

It helps find longest rising patterns quickly and reliably.

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