LeetCode 550: Game Play Analysis IV Solution in Python – A Step-by-Step Guide

What Is LeetCode 550: Game Play Analysis IV?

In LeetCode 550: Game Play Analysis IV, you’re given a list of player activity records as a 2D array activities where each entry is [player_id, event_date], and your task is to return the fraction (rounded to 2 decimal places) of players who logged in on the day immediately following their first login. For example, with activities = [[1, "2016-03-01"], [1, "2016-03-02"], [2, "2016-05-01"], [2, "2016-05-03"]], the fraction is 0.50 because player 1 logged in the next day, but player 2 did not. This problem builds on LeetCode 511: Game Play Analysis I for basic data grouping but adds consecutive-day logic.

Problem Statement

• Input: activities (List[List[str]])—array of [player_id, event_date] pairs.
• Output: Float—fraction of players logging in the next day after their first login, rounded to 2 decimals.
• Rules: Dates in "YYYY-MM-DD" format; count players with consecutive logins from first date.

Constraints

• 1 <= activities.length <= 10⁴
• 1 <= player_id <= 10⁵
• Dates range from "1980-01-01" to "2025-12-31".
• Multiple logins per player possible.

Examples

• Input: activities = [[1,"2016-03-01"],[1,"2016-03-02"],[2,"2016-05-01"],[2,"2016-05-03"]]
• Output: 0.50
• Player 1: First "2016-03-01", next "2016-03-02" (yes); Player 2: First "2016-05-01", next "2016-05-03" (no); 1/2 = 0.50.
• Input: activities = [[1,"2017-01-01"],[1,"2017-01-03"],[2,"2017-01-01"],[2,"2017-01-02"]]
• Output: 0.50
• Player 1: No next day; Player 2: Yes; 1/2 = 0.50.
• Input: activities = [[1,"2016-01-01"]]
• Output: 0.00
• Single login, no next day.

Understanding the Problem: Analyzing Consecutive Logins

To solve LeetCode 550: Game Play Analysis IV in Python, we need to calculate the fraction of players who logged in on the day after their first login, given a list of activity records. Each player’s first login date must be identified, then checked against their next login date (if any) for a 1-day difference. A naive approach might sort and scan all entries per player (O(n log n)), but with up to 10⁴ records, we can optimize using dictionaries or efficient grouping. We’ll explore:

• Best Solution (Dictionary with Two-Pass Tracking): O(n) time, O(n) space—efficient and clear.
• Alternative Solution (Sorting with Single Pass): O(n log n) time, O(1) space—intuitive but slower.

Let’s dive into the dictionary solution with a friendly breakdown!

Best Solution: Dictionary with Two-Pass Tracking

Why Dictionary Wins

The dictionary with two-pass tracking is the best for LeetCode 550 because it processes the activity list in O(n) time and O(n) space by using dictionaries to track each player’s first login and check for consecutive logins in two passes, avoiding sorting overhead. It’s like keeping a logbook for each player and quickly flipping through to spot next-day returns!

How It Works

Think of this as a player-login detective:

• Step 1: First Pass—Find First Logins:
• Use a dictionary to store each player’s earliest login date.
• Step 2: Second Pass—Check Next Day:
• For each record, if it’s the day after the player’s first login, mark them.
• Step 3: Calculate Fraction:
• Count players with next-day logins, divide by total unique players, round to 2 decimals.
• Step 4: Return Result:
• Float value of fraction.
• Why It Works:
• Two passes ensure accurate first-date tracking and consecutive check.
• Dictionary lookups are O(1), keeping time linear.

It’s like a login-fraction calculator!

Step-by-Step Example

Example: activities = [[1,"2016-03-01"],[1,"2016-03-02"],[2,"2016-05-01"],[2,"2016-05-03"]]

• Step 1: First Pass:
• first_login = {1: "2016-03-01", 2: "2016-05-01"}.
• Step 2: Second Pass:
• [1, "2016-03-01"]: First, skip.
• [1, "2016-03-02"]: Next day (1 day diff), next_day = {1}.
• [2, "2016-05-01"]: First, skip.
• [2, "2016-05-03"]: 2 days diff, no mark.
• next_day = {1}.
• Step 3: Fraction:
• Total players = 2, Next-day players = 1, Fraction = 1/2 = 0.50.
• Result: 0.50.

Example: activities = [[1,"2017-01-01"],[1,"2017-01-03"]]

• First Pass: first_login = {1: "2017-01-01"}.
• Second Pass: No next-day login (2 days diff), next_day = set().
• Fraction: 0/1 = 0.00.
• Result: 0.00.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codefrom datetime import datetime, timedelta

class Solution:
    def fractionNextDayLogin(self, activities: List[List[str]]) -> float:
        # Step 1: First pass - find first login per player
        first_login = {}
        for player_id, date in activities:
            date_obj = datetime.strptime(date, "%Y-%m-%d")
            if player_id not in first_login or date_obj < first_login[player_id]:
                first_login[player_id] = date_obj

        # Step 2: Second pass - check next-day logins
        next_day_players = set()
        for player_id, date in activities:
            date_obj = datetime.strptime(date, "%Y-%m-%d")
            if player_id in first_login:
                first_date = first_login[player_id]
                if date_obj == first_date + timedelta(days=1):
                    next_day_players.add(player_id)

        # Step 3: Calculate fraction
        total_players = len(first_login)
        if total_players == 0:
            return 0.0
        fraction = len(next_day_players) / total_players
        return round(fraction, 2)

• Line 5: First pass builds first_login dictionary with earliest dates.
• Lines 8-10: Parse date, update if earlier.
• Lines 13-18: Second pass checks for next-day logins, adds to set.
• Lines 21-25: Compute fraction, round to 2 decimals.
• Time Complexity: O(n)—two linear passes.
• Space Complexity: O(n)—dictionary and set storage.

It’s like a login-data analyst!

Alternative Solution: Sorting with Single Pass

Why an Alternative Approach?

The sorting with single pass solution sorts the activity list by date and player ID, then scans once to track first and next logins, running in O(n log n) time and O(1) space (excluding input). It’s intuitive but less efficient due to sorting, making it a good alternative for those who prefer a single-pass mindset.

How It Works

Picture this as a login-timeline reviewer:

• Step 1: Sort by player ID and date.
• Step 2: Single pass to track first and next logins.
• Step 3: Count next-day players and total.
• Step 4: Return fraction.

It’s like a login-sorting sleuth!

Step-by-Step Example

Example: activities = [[1,"2016-03-01"],[1,"2016-03-02"],[2,"2016-05-01"]]

• Step 1: Sort: [[1,"2016-03-01"], [1,"2016-03-02"], [2,"2016-05-01"]].
• Step 2: Scan:
• Player 1: First "2016-03-01", Next "2016-03-02" (yes), next_day = 1.
• Player 2: First "2016-05-01", no next, total = 2.
• Step 3: Fraction: 1/2 = 0.50.
• Result: 0.50.

Code for Sorting Approach

Copy codefrom datetime import datetime, timedelta

class Solution:
    def fractionNextDayLogin(self, activities: List[List[str]]) -> float:
        if not activities:
            return 0.0

        # Step 1: Sort by player ID and date
        activities.sort(key=lambda x: (x[0], datetime.strptime(x[1], "%Y-%m-%d")))

        # Step 2: Single pass to track logins
        total_players = 0
        next_day_players = 0
        i = 0
        while i < len(activities):
            player_id = activities[i][0]
            first_date = datetime.strptime(activities[i][1], "%Y-%m-%d")
            total_players += 1
            i += 1
            if i < len(activities) and activities[i][0] == player_id:
                next_date = datetime.strptime(activities[i][1], "%Y-%m-%d")
                if next_date == first_date + timedelta(days=1):
                    next_day_players += 1
            while i < len(activities) and activities[i][0] == player_id:
                i += 1

        # Step 3: Calculate fraction
        if total_players == 0:
            return 0.0
        return round(next_day_players / total_players, 2)

• Line 9: Sort by ID and date.
• Lines 12-23: Scan, track first and next logins, count players.
• Lines 26-29: Compute fraction, round.
• Time Complexity: O(n log n)—sorting dominates.
• Space Complexity: O(1)—no extra space.

It’s a sorted login checker!

Comparing the Two Solutions

• Dictionary (Best):
• Pros: O(n), O(n), fast.
• Cons: Extra space.
• Sorting (Alternative):
• Pros: O(n log n), O(1), space-efficient.
• Cons: Slower due to sort.

Dictionary wins for speed!

Additional Examples and Edge Cases

• []: 0.00.
• [[1,"2020-01-01"]]: 0.00.
• [[1,"2020-01-01"],[1,"2020-01-02"]]: 1.00.

Dictionary handles them all!

Complexity Recap

• Dictionary: Time O(n), Space O(n).
• Sorting: Time O(n log n), Space O(1).

Dictionary’s the efficiency champ!

Key Takeaways

• Dictionary: Fast tracking—learn at Python Basics!
• Sorting: Timeline clarity.
• Data: Logins are fun.
• Python: Dicts or sort, your pick!

Final Thoughts: Analyze That Game!

LeetCode 550: Game Play Analysis IV in Python is a delightful data challenge. Dictionary with two-pass tracking is your fast track, while sorting offers a clear alternative. Want more? Try LeetCode 511: Game Play Analysis I or LeetCode 128: Longest Consecutive Sequence. Ready to crunch? Head to Solve LeetCode 550 on LeetCode and analyze those logins today!