LeetCode #550 — MEDIUM

Game Play Analysis IV

Move from brute-force thinking to an efficient approach using core interview patterns strategy.

The Problem

Problem Statement

Table: Activity

+--------------+---------+
| Column Name  | Type    |
+--------------+---------+
| player_id    | int     |
| device_id    | int     |
| event_date   | date    |
| games_played | int     |
+--------------+---------+
(player_id, event_date) is the primary key (combination of columns with unique values) of this table.
This table shows the activity of players of some games.
Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on someday using some device.

Write a solution to report the fraction of players that logged in again on the day after the day they first logged in, rounded to 2 decimal places. In other words, you need to determine the number of players who logged in on the day immediately following their initial login, and divide it by the number of total players.

The result format is in the following example.

Example 1:

Input: 
Activity table:
+-----------+-----------+------------+--------------+
| player_id | device_id | event_date | games_played |
+-----------+-----------+------------+--------------+
| 1         | 2         | 2016-03-01 | 5            |
| 1         | 2         | 2016-03-02 | 6            |
| 2         | 3         | 2017-06-25 | 1            |
| 3         | 1         | 2016-03-02 | 0            |
| 3         | 4         | 2018-07-03 | 5            |
+-----------+-----------+------------+--------------+
Output: 
+-----------+
| fraction  |
+-----------+
| 0.33      |
+-----------+
Explanation: 
Only the player with id 1 logged back in after the first day he had logged in so the answer is 1/3 = 0.33

Step 01

Brute Force Baseline

Problem summary: Table: Activity +--------------+---------+ | Column Name | Type | +--------------+---------+ | player_id | int | | device_id | int | | event_date | date | | games_played | int | +--------------+---------+ (player_id, event_date) is the primary key (combination of columns with unique values) of this table. This table shows the activity of players of some games. Each row is a record of a player who logged in and played a number of games (possibly 0) before logging out on someday using some device. Write a solution to report the fraction of players that logged in again on the day after the day they first logged in, rounded to 2 decimal places. In other words, you need to determine the number of players who logged in on the day immediately following their initial login, and divide it by the number of total players. The result format is in the following example.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

{"headers":{"Activity":["player_id","device_id","event_date","games_played"]},"rows":{"Activity":[[1,2,"2016-03-01",5],[1,2,"2016-03-02",6],[2,3,"2017-06-25",1],[3,1,"2016-03-02",0],[3,4,"2018-07-03",5]]}}

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #550: Game Play Analysis IV
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #550: Game Play Analysis IV
// import pandas as pd
// 
// 
// def gameplay_analysis(activity: pd.DataFrame) -> pd.DataFrame:
//     activity["first"] = activity.groupby("player_id").event_date.transform(min)
//     activity_2nd_day = activity[
//         activity["first"] + pd.DateOffset(1) == activity["event_date"]
//     ]
// 
//     return pd.DataFrame(
//         {"fraction": [round(len(activity_2nd_day) / activity.player_id.nunique(), 2)]}
//     )

// Accepted solution for LeetCode #550: Game Play Analysis IV
// Auto-generated Go example from py.
func exampleSolution() {
}
// Reference (py):
// # Accepted solution for LeetCode #550: Game Play Analysis IV
// import pandas as pd
// 
// 
// def gameplay_analysis(activity: pd.DataFrame) -> pd.DataFrame:
//     activity["first"] = activity.groupby("player_id").event_date.transform(min)
//     activity_2nd_day = activity[
//         activity["first"] + pd.DateOffset(1) == activity["event_date"]
//     ]
// 
//     return pd.DataFrame(
//         {"fraction": [round(len(activity_2nd_day) / activity.player_id.nunique(), 2)]}
//     )

# Accepted solution for LeetCode #550: Game Play Analysis IV
import pandas as pd


def gameplay_analysis(activity: pd.DataFrame) -> pd.DataFrame:
    activity["first"] = activity.groupby("player_id").event_date.transform(min)
    activity_2nd_day = activity[
        activity["first"] + pd.DateOffset(1) == activity["event_date"]
    ]

    return pd.DataFrame(
        {"fraction": [round(len(activity_2nd_day) / activity.player_id.nunique(), 2)]}
    )

// Accepted solution for LeetCode #550: Game Play Analysis IV
// Rust example auto-generated from py reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (py):
// # Accepted solution for LeetCode #550: Game Play Analysis IV
// import pandas as pd
// 
// 
// def gameplay_analysis(activity: pd.DataFrame) -> pd.DataFrame:
//     activity["first"] = activity.groupby("player_id").event_date.transform(min)
//     activity_2nd_day = activity[
//         activity["first"] + pd.DateOffset(1) == activity["event_date"]
//     ]
// 
//     return pd.DataFrame(
//         {"fraction": [round(len(activity_2nd_day) / activity.player_id.nunique(), 2)]}
//     )

// Accepted solution for LeetCode #550: Game Play Analysis IV
// Auto-generated TypeScript example from py.
function exampleSolution(): void {
}
// Reference (py):
// # Accepted solution for LeetCode #550: Game Play Analysis IV
// import pandas as pd
// 
// 
// def gameplay_analysis(activity: pd.DataFrame) -> pd.DataFrame:
//     activity["first"] = activity.groupby("player_id").event_date.transform(min)
//     activity_2nd_day = activity[
//         activity["first"] + pd.DateOffset(1) == activity["event_date"]
//     ]
// 
//     return pd.DataFrame(
//         {"fraction": [round(len(activity_2nd_day) / activity.player_id.nunique(), 2)]}
//     )

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.