LeetCode 674: Longest Continuous Increasing Subsequence Solution in Python – A Step-by-Step Guide

What Is LeetCode 674: Longest Continuous Increasing Subsequence?

In LeetCode 674: Longest Continuous Increasing Subsequence, you’re given an integer array nums, and your task is to find the length of the longest continuous increasing subsequence (LCIS)—a contiguous subarray where each element is strictly greater than the previous one. Return this length as an integer. For example, with nums = [1, 3, 5, 4, 7], the LCIS is [1, 3, 5], with a length of 3, because it’s the longest stretch where numbers rise continuously. Unlike the classic longest increasing subsequence (LIS), this must be a continuous segment. This problem tests your ability to track consecutive increases efficiently.

Problem Statement

• Input:
• nums: List of integers.
• Output: An integer—length of the longest continuous increasing subsequence.
• Rules:
• Continuous: Subsequence must be a contiguous subarray.
• Increasing: Each element > previous element.
• Return maximum length found.

Constraints

• 0 ≤ nums.length ≤ 10⁴.
• -10⁹ ≤ nums[i] ≤ 10⁹.

Examples

• Input: nums = [1, 3, 5, 4, 7]
• Output: 3 (LCIS: [1, 3, 5])
• Input: nums = [2, 2, 2, 2, 2]
• Output: 1 (LCIS: [2], no increase possible)
• Input: nums = [1, 3, 5, 7]
• Output: 4 (LCIS: [1, 3, 5, 7])

These examples set the array—let’s solve it!

Understanding the Problem: Finding the Longest Continuous Run

To solve LeetCode 674: Longest Continuous Increasing Subsequence in Python, we need to find the length of the longest contiguous subarray in nums where each element is strictly greater than the one before it. A brute-force approach—checking all possible contiguous subarrays—would be O(n²), inefficient for n ≤ 10⁴. Since we’re looking for continuous increases, we can optimize by scanning the array once or using DP. We’ll use:

• Best Solution (Single-Pass Greedy): O(n) time, O(1) space—fast and simple.
• Alternative Solution (Dynamic Programming): O(n) time, O(n) space—formal but unnecessary complexity.

Let’s start with the greedy solution, breaking it down for beginners.

Best Solution: Single-Pass Greedy

Why This Is the Best Solution

The single-pass greedy approach is the top choice for LeetCode 674 because it’s highly efficient—O(n) time with O(1) space—and straightforward: it scans the array once, tracking the current run of increasing numbers and updating the maximum length whenever the run breaks. It fits constraints (n ≤ 10⁴) perfectly and is intuitive once you see the counting logic. It’s like walking through the array and keeping a tally of how long the numbers keep climbing!

How It Works

Think of this as a run tracker:

• Step 1: Initialize Variables:
• curr_len: Current length of increasing run, starts at 1.
• max_len: Maximum length found, starts at 1 (or 0 if empty).
• Step 2: Scan Array:
• For each pair nums[i] and nums[i+1]:
• If nums[i] < nums[i+1], increment curr_len.
• Else, reset curr_len to 1 (new run starts).
• Update max_len = max(max_len, curr_len).
• Step 3: Handle Edge Cases:
• If array empty, return 0.
• Step 4: Return Result:
• Return max_len.

It’s like a streak counter—track and update!

Step-by-Step Example

Example: nums = [1, 3, 5, 4, 7]

• Step 1: curr_len = 1, max_len = 1.
• Step 2: Scan:
• i=0: 1 < 3, curr_len = 2, max_len = 2.
• i=1: 3 < 5, curr_len = 3, max_len = 3.
• i=2: 5 > 4, curr_len = 1, max_len = 3.
• i=3: 4 < 7, curr_len = 2, max_len = 3.
• Step 3: Not empty, proceed.
• Step 4: Return 3.
• Output: 3.

Example: nums = [2, 2, 2, 2, 2]

• Step 1: curr_len = 1, max_len = 1.
• Step 2: Scan:
• i=0: 2 = 2, curr_len = 1, max_len = 1.
• i=1: 2 = 2, curr_len = 1, max_len = 1.
• i=2: 2 = 2, curr_len = 1, max_len = 1.
• i=3: 2 = 2, curr_len = 1, max_len = 1.
• Step 4: Return 1.
• Output: 1.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained clearly:

Copy codeclass Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        # Step 1: Handle empty array
        if not nums:
            return 0

        # Step 2: Initialize variables
        curr_len = 1
        max_len = 1

        # Step 3: Scan array
        for i in range(len(nums) - 1):
            if nums[i] < nums[i + 1]:
                curr_len += 1
                max_len = max(max_len, curr_len)
            else:
                curr_len = 1

        # Step 4: Return maximum length
        return max_len

• Lines 4-6: Handle: Empty array returns 0.
• Lines 9-10: Init: Current and max lengths start at 1.
• Lines 13-18: Scan:
• Compare pairs, increment curr_len if increasing, reset if not, update max_len.
• Line 21: Return max_len.
• Time Complexity: O(n)—single pass over array.
• Space Complexity: O(1)—two variables.

This is like a run spotter—count and maximize!

Alternative Solution: Dynamic Programming

Why an Alternative Approach?

The dynamic programming approach uses an array to track lengths—O(n) time, O(n) space. It’s more formal, building the LCIS ending at each position, but overcomplicates this simple problem compared to the greedy method. It’s like keeping a detailed log of every step when a quick tally would do!

How It Works

Picture this as a length logger:

• Step 1: Init DP array:
• dp[i]: Length of LCIS ending at nums[i], starts at 1.
• Step 2: Fill DP:
• For each i:
• If nums[i-1] < nums[i], dp[i] = dp[i-1] + 1.
• Else, dp[i] = 1.
• Step 3: Find max length:
• Return max(dp).
• Step 4: Handle edge cases.

It’s like a DP tracker—log and max!

Step-by-Step Example

Example: nums = [1, 3, 5, 4, 7]

• Step 1: dp = [1, 1, 1, 1, 1].
• Step 2: Fill:
• i=1: 1 < 3, dp[1] = dp[0] + 1 = 2.
• i=2: 3 < 5, dp[2] = dp[1] + 1 = 3.
• i=3: 5 > 4, dp[3] = 1.
• i=4: 4 < 7, dp[4] = dp[3] + 1 = 2.
• Final: dp = [1, 2, 3, 1, 2].
• Step 3: Max = 3.
• Step 4: Return 3.
• Output: 3.

Code for DP Approach

Copy codeclass Solution:
    def findLengthOfLCIS(self, nums: List[int]) -> int:
        # Step 1: Handle empty array
        if not nums:
            return 0

        # Step 2: Initialize DP array
        n = len(nums)
        dp = [1] * n

        # Step 3: Fill DP array
        for i in range(1, n):
            if nums[i - 1] < nums[i]:
                dp[i] = dp[i - 1] + 1

        # Step 4: Return maximum length
        return max(dp)

• Lines 4-6: Handle: Empty array returns 0.
• Lines 9-11: Init: DP array with 1s.
• Lines 14-16: Fill:
• If previous < current, extend length.
• Line 19: Return max(dp).
• Time Complexity: O(n)—single pass.
• Space Complexity: O(n)—DP array.

It’s a length builder—log and maximize!

Comparing the Two Solutions

• Greedy (Best):
• Pros: O(n) time, O(1) space, simple and fast.
• Cons: Less formal structure.
• DP (Alternative):
• Pros: O(n) time, O(n) space, structured approach.
• Cons: Extra space, slightly more complex.

Greedy wins for simplicity.

Additional Examples and Edge Cases

• Input: []
• Output: 0.
• Input: [1, 2, 3, 4]
• Output: 4.

Greedy handles these well.

Complexity Breakdown

• Greedy: Time O(n), Space O(1).
• DP: Time O(n), Space O(n).

Greedy is optimal.

Key Takeaways

• Greedy: Run tracking—smart!
• DP: Length logging—clear!
• Sequences: Counting is fun.
• Python Tip: Loops rock—see Python Basics.

Final Thoughts: Spot That Sequence

LeetCode 674: Longest Continuous Increasing Subsequence in Python is a fun sequence challenge. Single-pass greedy offers simplicity and speed, while DP provides a formal alternative. Want more? Try LeetCode 300: Longest Increasing Subsequence or LeetCode 53: Maximum Subarray. Ready to scan? Head to Solve LeetCode 674 on LeetCode and find that longest run today!