Why Longest Increasing Subsequence in DSA C?

The Big Idea

Discover how to find the longest rising pattern in a sea of numbers without getting lost!

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 the longest path in a maze by guessing every route.

The Problem

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

This takes a lot of time and is easy to make mistakes, especially with bigger lists.

The Solution

The Longest Increasing Subsequence (LIS) method smartly finds the longest rising sequence without checking every possibility.

It uses a step-by-step approach to build solutions from smaller parts, saving time and avoiding errors.

Before vs After

✗ Before

int longest_increasing_subsequence(int arr[], int n) {
    int max_length = 1;
    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {
            // check if arr[j] > arr[i] and count length
        }
    }
    return max_length;
}

✓ After

int longest_increasing_subsequence(int arr[], int n) {
    int lis[n];
    for (int i = 0; i < n; i++) lis[i] = 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 = 1;
    for (int i = 0; i < n; i++) if (lis[i] > max) max = lis[i];
    return max;
}

What It Enables

This concept lets you quickly find the longest rising pattern in data, helping with predictions and understanding trends.

Real Life Example

In stock market analysis, LIS helps find the longest period where stock prices keep increasing, guiding investment decisions.

Key Takeaways

Manual checking is slow and error-prone for finding longest increasing sequences.

LIS uses smart step-by-step building to find the answer efficiently.

This method is useful for trend analysis and pattern recognition in data.

Practice

1. What does the Longest Increasing Subsequence (LIS) problem ask you to find in a sequence of numbers?

easy

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

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

C. The longest subsequence where numbers are equal

D. The shortest subsequence with increasing numbers

Solution

1. Step 1: Understand the LIS definition

   The Longest Increasing Subsequence is a subsequence where every next number is strictly greater than the previous one.

2. Step 2: Compare options with the definition

   The longest subsequence where each number is larger than the previous one matches the definition exactly. Options B, C, and D do not describe LIS correctly.

3. Final Answer:

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

4. Quick Check:

   LIS = longest increasing subsequence [OK]

Hint: LIS means strictly increasing sequence, not equal or decreasing [OK]

Common Mistakes:

• Confusing increasing with non-decreasing
• Thinking LIS means equal numbers
• Mixing up subsequence with substring

2. Which of the following C code snippets correctly initializes the array to store LIS lengths for each element in an array arr of size n?

easy

A. int lis[n] = {0};

B. int lis[n]; for(int i=0; i

C. int lis[n]; for(int i=0; i

D. int lis[n]; memset(lis, 1, n*sizeof(int));

Solution

1. Step 1: Understand LIS initialization

   Each element starts with LIS length 1 because each element alone is a subsequence.

2. Step 2: Check code correctness

   int lis[n]; for(int i=0; i

3. Final Answer:

   int lis[n]; for(int i=0; i -> Option B

4. Quick Check:

   Initialize LIS array with 1 [OK]

Hint: Initialize LIS array elements to 1, not 0 [OK]

Common Mistakes:

• Using memset to set integers to 1 (wrong)
• Initializing LIS array with zeros
• Not initializing LIS array at all

3. What is the output of the following C code snippet that calculates the length of LIS for array arr = {10, 22, 9, 33, 21, 50, 41, 60}?

int arr[] = {10, 22, 9, 33, 21, 50, 41, 60};
int n = 8;
int lis[8];
for(int i=0; i arr[j] && lis[i] < lis[j] + 1) {
      lis[i] = lis[j] + 1;
    }
  }
}
int max = lis[0];
for(int i=1; i max) max = lis[i];
}
printf("%d", max);

medium

A. 4

B. 6

C. 5

D. 3

Solution

1. Step 1: Calculate LIS values step-by-step

   Start with lis = {1,1,1,1,1,1,1,1} - i=1: arr[1]=22 > arr[0]=10, lis[1]=max(1,1+1)=2 - i=2: arr[2]=9 no smaller than previous, lis[2]=1 - i=3: arr[3]=33 > 22 and 10, lis[3]=max(1,2+1)=3 - i=4: arr[4]=21 > 10 but <22, lis[4]=max(1,1+1)=2 - i=5: arr[5]=50 > 33, 22, 21, lis[5]=max(1,3+1)=4 - i=6: arr[6]=41 > 33, 22, 21, lis[6]=max(1,3+1)=4 - i=7: arr[7]=60 > 50, 41, lis[7]=max(1,4+1)=5

2. Step 2: Find maximum LIS length

   Maximum value in lis array is 5.

3. Final Answer:

   5 -> Option C

4. Quick Check:

   LIS length for given array = 5 [OK]

Hint: Trace LIS array updates carefully for each element [OK]

Common Mistakes:

• Counting non-increasing subsequences
• Forgetting to update lis[i] only if condition met
• Mixing up indices in nested loops

4. Identify the bug in the following C code snippet that attempts to compute the LIS length:

int arr[] = {3, 10, 2, 1, 20};
int n = 5;
int lis[5];
for(int i=0; i arr[j] && lis[i] < lis[j] + 1) {
      lis[i] = lis[j] + 1;
    }
  }
}
int max = lis[0];
for(int i=1; i max) max = lis[i];
}
printf("%d", max);

medium

A. LIS array should be initialized with 1, not 0

B. The nested loops should start from i=0

C. The condition arr[i] > arr[j] should be arr[i] >= arr[j]

D. The max variable should be initialized to 0

Solution

1. Step 1: Check LIS array initialization

   LIS array elements represent length of subsequence ending at each index. Each element alone forms subsequence of length 1, so initialize lis[i] = 1.

2. Step 2: Analyze effect of initialization

   Initializing lis[i] = 0 causes all comparisons lis[i] < lis[j] + 1 to fail or produce wrong results, leading to incorrect max LIS length.

3. Final Answer:

   LIS array should be initialized with 1, not 0 -> Option A

4. Quick Check:

   Initialize LIS array with 1 to avoid zero-length subsequences [OK]

Hint: Always start LIS lengths at 1, not 0 [OK]

Common Mistakes:

• Initializing LIS array with zeros
• Changing comparison operators incorrectly
• Starting loops at wrong indices

5. Given an array arr = {5, 7, 4, 8, 6, 9}, write a C function to find the length of the Longest Increasing Subsequence using dynamic programming. What is the length of the LIS for this array?

hard

A. 6

B. 3

C. 5

D. 4

Solution

1. Step 1: Initialize LIS array with 1 for each element

   lis = {1,1,1,1,1,1}

2. Step 2: Update LIS values by comparing each element with previous ones

   - i=1 (7): 7 > 5, lis[1] = max(1, 1+1) = 2 - i=2 (4): 4 > 5? No, lis[2] = 1 - i=3 (8): 8 > 5, lis[3] = max(1,1+1)=2; 8 > 7, lis[3] = max(2,2+1)=3; 8 > 4, lis[3] = max(3,1+1)=3 - i=4 (6): 6 > 5, lis[4] = max(1,1+1)=2; 6 > 7? No; 6 > 4, lis[4] = max(2,1+1)=2; 6 > 8? No - i=5 (9): 9 > 5, lis[5] = max(1,1+1)=2; 9 > 7, lis[5] = max(2,2+1)=3; 9 > 4, lis[5] = max(3,1+1)=3; 9 > 8, lis[5] = max(3,3+1)=4; 9 > 6, lis[5] = max(4,2+1)=4

3. Step 3: Find maximum LIS length

   Maximum lis value is 4.

4. Final Answer:

   4 -> Option D

5. Quick Check:

   LIS length for given array = 4 [OK]

Hint: Update LIS by checking all previous smaller elements [OK]

Common Mistakes:

• Not updating LIS correctly for all previous elements
• Confusing subsequence with substring
• Forgetting to initialize LIS array with 1