LeetCode 2893 - Calculate Orders Within Each Interval

The problem asks us to process an Orders table where each row contains a minute and the number of orders received during that specific minute. We are tasked with calculating the total number of orders in intervals of six consecutive minutes.

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LeetCode Problem 2893

Difficulty: 🟡 Medium
Topics: Database

Solution

Problem Understanding

The problem asks us to process an Orders table where each row contains a minute and the number of orders received during that specific minute. We are tasked with calculating the total number of orders in intervals of six consecutive minutes. For example, minutes 1 through 6 are interval 1, minutes 7 through 12 are interval 2, and so on. The output should include the interval number and the total orders in that interval, sorted by interval number.

The input is guaranteed to have a number of rows that is a multiple of six, which means we do not need to handle incomplete intervals. Each minute is unique and acts as a primary key, so there are no duplicate minutes. The main edge cases are very small or very large numbers of minutes, but the multiple-of-six guarantee simplifies handling of intervals. We also need to ensure the ordering of results by interval_no in ascending order.

In summary, we are aggregating orders per fixed-size chunk of six minutes, calculating the sum of order_count in each chunk, and returning it with a sequential interval number.

Approaches

A brute-force approach would involve explicitly iterating over the table in groups of six minutes, summing the orders for each group, and assigning the interval number manually. This works but requires row-by-row aggregation logic, which is cumbersome in SQL and unnecessary when we can use arithmetic to calculate intervals dynamically.

The key insight for an optimal solution is that we can compute the interval number directly using integer arithmetic: (minute - 1) / 6 + 1. This formula converts any minute into the corresponding interval of six minutes. Using this, we can group rows by this calculated interval number and sum the order_count for each group. SQL’s GROUP BY operation handles the aggregation efficiently.

Approach Time Complexity Space Complexity Notes
Brute Force O(n) O(n) Iterate through rows in chunks of six, sum orders manually
Optimal O(n) O(n) Use integer division to calculate interval number, then GROUP BY

Algorithm Walkthrough

  1. Compute the interval number for each row using the formula (minute - 1) / 6 + 1. This ensures minutes 1-6 map to interval 1, 7-12 to interval 2, and so on.
  2. Group the table rows by the calculated interval number. SQL GROUP BY will aggregate all rows that belong to the same interval.
  3. Compute the sum of order_count for each group using SUM(order_count). This gives the total orders per interval.
  4. Sort the results by interval_no in ascending order to satisfy the output requirements.

Why it works: Each minute belongs to exactly one interval, and the integer division formula guarantees correct grouping. SQL aggregation ensures all orders within an interval are summed correctly. Sorting ensures intervals are presented in sequence.

Python Solution

import sqlite3
from typing import List, Tuple

def calculate_orders(intervals: List[Tuple[int, int]]) -> List[Tuple[int, int]]:
    """
    intervals: List of tuples (minute, order_count)
    returns: List of tuples (interval_no, total_orders)
    """
    # Convert input to a temporary SQLite table for SQL-like operations
    conn = sqlite3.connect(":memory:")
    cursor = conn.cursor()
    cursor.execute("CREATE TABLE Orders(minute INT, order_count INT)")
    cursor.executemany("INSERT INTO Orders(minute, order_count) VALUES (?, ?)", intervals)
    
    cursor.execute("""
        SELECT 
            (minute - 1) / 6 + 1 AS interval_no,
            SUM(order_count) AS total_orders
        FROM Orders
        GROUP BY interval_no
        ORDER BY interval_no ASC
    """)
    
    results = cursor.fetchall()
    conn.close()
    return results

This Python implementation uses SQLite for simplicity. The interval_no is calculated directly in the SQL query. GROUP BY performs the aggregation, and ORDER BY ensures the intervals are returned in ascending order. The approach mirrors the algorithm walkthrough.

Go Solution

package main

import (
    "database/sql"
    _ "github.com/mattn/go-sqlite3"
    "fmt"
)

func calculateOrders(intervals [][2]int) [][2]int {
    db, _ := sql.Open("sqlite3", ":memory:")
    defer db.Close()

    db.Exec("CREATE TABLE Orders(minute INT, order_count INT)")
    for _, row := range intervals {
        db.Exec("INSERT INTO Orders(minute, order_count) VALUES (?, ?)", row[0], row[1])
    }

    rows, _ := db.Query(`
        SELECT 
            (minute - 1) / 6 + 1 AS interval_no,
            SUM(order_count) AS total_orders
        FROM Orders
        GROUP BY interval_no
        ORDER BY interval_no ASC
    `)
    defer rows.Close()

    var results [][2]int
    for rows.Next() {
        var interval_no, total_orders int
        rows.Scan(&interval_no, &total_orders)
        results = append(results, [2]int{interval_no, total_orders})
    }
    return results
}

func main() {
    data := [][2]int{{1,0},{2,2},{3,4},{4,6},{5,1},{6,4},{7,1},{8,2},{9,4},{10,1},{11,4},{12,6}}
    fmt.Println(calculateOrders(data))
}

The Go solution is similar in approach to Python. SQLite is used for aggregation, and intervals are computed using integer division. We iterate over rows returned by the query and construct the result slice.

Worked Examples

Using the example input:

minute | order_count
1      | 0
2      | 2
3      | 4
4      | 6
5      | 1
6      | 4
7      | 1
8      | 2
9      | 4
10     | 1
11     | 4
12     | 6

Step 1: Compute interval number:

minute 1-6 -> interval 1
minute 7-12 -> interval 2

Step 2: Aggregate orders:

interval 1: 0 + 2 + 4 + 6 + 1 + 4 = 17
interval 2: 1 + 2 + 4 + 1 + 4 + 6 = 18

Step 3: Output table:

interval_no | total_orders
1           | 17
2           | 18

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each row is processed once for aggregation, where n is the number of minutes
Space O(n) Space to store the temporary table and results

The complexity is linear because we only perform a single pass over the data with grouping and summation.

Test Cases

# test cases
assert calculate_orders([(1,0),(2,2),(3,4),(4,6),(5,1),(6,4),(7,1),(8,2),(9,4),(10,1),(11,4),(12,6)]) == [(1,17),(2,18)] # example
assert calculate_orders([(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)]) == [(1,30)] # single interval
assert calculate_orders([(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),(7,2),(8,2),(9,2),(10,2),(11,2),(12,2)]) == [(1,6),(2,12)] # multiple intervals
assert calculate_orders([(1,0),(2,0),(3,0),(4,0),(5,0),(6,0),(7,0),(8,0),(9,0),(10,0),(11,0),(12,0)]) == [(1,0),(2,0)] # zero orders
Test Why
example Validates basic functionality and aggregation across two intervals
single interval Tests minimal multiple-of-six scenario
multiple intervals Tests correct calculation across several intervals with different order counts
zero orders Ensures sum works with zeros and does not fail

Edge Cases

One edge case is when all order_count values are zero. This ensures that the aggregation correctly computes zero without error. Our implementation handles this naturally because summing zeros produces zero.

Another edge case is when the table has only one interval. The algorithm still calculates (minute-1)/6 + 1 correctly, producing interval 1, and the sum aggregates all six rows correctly.

A third edge case is very large numbers for order_count. Since SQL SUM can handle large integers in most databases and our Python/Go integers are arbitrary or 64-bit, this works correctly. We do not perform any intermediate type conversions that could overflow.

This approach is robust, simple, and guarantees correct results for all inputs that meet the problem constraints.