How do you solve Longest Continuous Increasing Subsequence on LeetCode?

Longest Continuous Increasing Subsequence (LeetCode #674) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The sequence must be strictly increasing, meaning duplicates reset the count.

What is the brute force solution for Longest Continuous Increasing Subsequence?

The brute force approach for Longest Continuous Increasing Subsequence is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray to find the longest continuous increasing subsequence. This means we will compare each element with every other element that follows it to see if they form an increasing sequence. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Continuous Increasing Subsequence?

Longest Continuous Increasing Subsequence uses the following concepts: Array. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Longest Continuous Increasing Subsequence?

Clearly explain your thought process as you write the code. Consider edge cases, like arrays with all identical elements. Practice dry-running your solution with examples to ensure correctness.

What are common mistakes when solving Longest Continuous Increasing Subsequence?

Not resetting the current length when the sequence breaks. Confusing increasing subsequences with non-continuous subsequences.

#674

Longest Continuous Increasing Subsequence

Easy

ArrayArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a single pass through the array to track the length of the current increasing sequence. This reduces the time complexity significantly by avoiding nested loops.

⚙️

Algorithm

4 steps

1Step 1: Initialize a variable to keep track of the current length of the increasing sequence and the maximum length found.
2Step 2: Iterate through the array starting from the second element.
3Step 3: If the current element is greater than the previous one, increment the current length; otherwise, reset the current length to 1.
4Step 4: Update the maximum length if the current length exceeds it.

solution.py12 lines

1def longest_increasing_subsequence(nums):
2    if not nums:
3        return 0
4    max_length = 1
5    current_length = 1
6    for i in range(1, len(nums)):
7        if nums[i] > nums[i - 1]:
8            current_length += 1
9        else:
10            current_length = 1
11        max_length = max(max_length, current_length)
12    return max_length

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the array. The space complexity is O(1) since we are using a fixed number of variables.

1The sequence must be strictly increasing, meaning duplicates reset the count.
2Continuous means elements must be adjacent in the array.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.