Longest Increasing Subsequence in DSA Java - 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 changes as the input list grows.

How does the number of steps grow when the list gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

public int lengthOfLIS(int[] nums) {
    int n = nums.length;
    int[] dp = new int[n];
    Arrays.fill(dp, 1);
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (nums[i] > nums[j]) {
                dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
    }
    int max = 1;
    for (int length : dp) max = Math.max(max, length);
    return max;
}

This code finds the length of the longest increasing subsequence in an array using dynamic programming.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops comparing pairs of elements to update subsequence lengths.
• 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 elements.

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, so doubling the input makes the work about four times bigger.

Final Time Complexity

Time Complexity: O(n²)

This means the time needed grows roughly with the square of the number of elements in the input list.

Common Mistake

[X] Wrong: "The algorithm only looks at each element once, so it must be O(n)."

[OK] Correct: Each element is compared with many others inside nested loops, so the total steps grow much faster than just once per element.

Interview Connect

Understanding this time complexity helps you explain why some solutions are slower and shows your skill in analyzing nested loops, a common interview topic.

Self-Check

"What if we used a binary search approach to build the subsequence? 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 with all equal numbers

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

C. The longest subsequence with alternating increasing and decreasing numbers

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

Solution

1. Step 1: Understand the problem 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, while others describe different patterns.

3. Final Answer:

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

4. Quick Check:

   LIS = strictly increasing subsequence [OK]

Hint: Look for strictly increasing order in subsequence [OK]

Common Mistakes:

• Confusing subsequence with substring
• Thinking LIS means non-decreasing instead of strictly increasing
• Mixing up increasing with decreasing sequences

2. Which of the following Java code snippets correctly initializes an array to store the length of LIS ending at each index for an input array arr of size n?

easy

A. int[] lis = new int[n]; lis.fill(1);

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

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

D. int lis[] = new int[n]; Arrays.fill(lis, 0);

Solution

1. Step 1: Understand LIS initialization

   Each element starts with LIS length 1 because the smallest subsequence is the element itself.

2. Step 2: Check code correctness

   int[] lis = new int[n]; for(int i=0; ifill on array. int lis[] = new int[n]; Arrays.fill(lis, 0); fills with 0, incorrect for LIS.

3. Final Answer:

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

4. Quick Check:

   Initialize LIS array with 1s [OK]

Hint: Initialize LIS array with 1 for each element [OK]

Common Mistakes:

• Initializing LIS array with zeros
• Using non-existent array methods like fill()
• Not initializing LIS array before DP

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

int[] arr = {10, 22, 9, 33, 21, 50, 41, 60};
int n = arr.length;
int[] lis = new int[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 = 0;
for (int i = 0; i < n; i++) {
  if (lis[i] > max) max = lis[i];
}
System.out.println(max);

medium

A. 5

B. 4

C. 6

D. 3

Solution

1. Step 1: Calculate LIS array values

   For each element, find LIS ending there: - arr[0]=10: lis=1 - arr[1]=22 >10: lis=2 - arr[2]=9: lis=1 - arr[3]=33 >22: lis=3 - arr[4]=21 >10 but <22: lis=2 - arr[5]=50 >33: lis=4 - arr[6]=41 >33 but <50: lis=4 - arr[7]=60 >50: lis=5

2. Step 2: Find max LIS length

   Maximum value in lis array is 5.

3. Final Answer:

   5 -> Option A

4. Quick Check:

   Max LIS length = 5 [OK]

Hint: Trace LIS array updates step-by-step [OK]

Common Mistakes:

• Confusing subsequence length with subarray length
• Not updating lis[i] correctly inside nested loops
• Forgetting to find max LIS after filling array

4. Identify the bug in the following Java code snippet that attempts to compute the length of the Longest Increasing Subsequence for array arr:

int[] lis = new int[arr.length];
for (int i = 0; i < arr.length; i++) {
  lis[i] = 0;
}
for (int i = 1; i < arr.length; 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 = 0;
for (int i = 0; i < lis.length; i++) {
  if (lis[i] > max) max = lis[i];
}
System.out.println(max);

medium

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

B. The inner loop should start from i+1 instead of 0

C. The comparison should be arr[i] < arr[j] instead of arr[i] > arr[j]

D. The max variable should be initialized to Integer.MIN_VALUE

Solution

1. Step 1: Check LIS array initialization

   LIS array must start with 1 for each element because each element alone forms a subsequence of length 1. Initializing with 0 causes incorrect LIS lengths.

2. Step 2: Verify loops and conditions

   Inner loop and comparison are correct. Max initialization to 0 is fine since LIS lengths are positive.

3. Final Answer:

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

4. Quick Check:

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

Hint: Always start LIS array values at 1 [OK]

Common Mistakes:

• Initializing LIS array with zeros
• Changing loop ranges incorrectly
• Misunderstanding LIS condition comparison

5. Given the array arr = {3, 10, 2, 1, 20}, which of the following sequences is the Longest Increasing Subsequence (LIS)?

hard

A. [1, 2, 3]

B. [3, 10, 20]

C. [10, 20]

D. [3, 2, 1, 20]

Solution

1. Step 1: Identify increasing subsequences

   Check each option for strictly increasing order: - [3, 10, 20]: 3 < 10 < 20 (valid) - [3, 2, 1, 20]: 3 > 2 > 1 (not increasing) - [10, 20]: 10 < 20 (valid but shorter) - [1, 2, 3]: 1 < 2 < 3 (valid but not a subsequence in original order)

2. Step 2: Compare lengths of valid subsequences

   [3, 10, 20] length is 3, [10, 20] length is 2, [1, 2, 3] length is 3 but elements not in original order (subsequence must follow original order). [1, 2, 3] is invalid as subsequence because 1 comes after 3 in original array.

3. Final Answer:

   [3, 10, 20] -> Option B

4. Quick Check:

   LIS must be increasing and follow original order [OK]

Hint: Check subsequence order matches original array [OK]

Common Mistakes:

• Choosing subsequence not in original order
• Confusing increasing with non-decreasing
• Ignoring subsequence length