LeetCode 2480 - Form a Chemical Bond

The problem asks us to identify all valid chemical bonds that can form between elements in a given table. The Elements table contains three columns: symbol, type, and electrons. The type column is an enumeration of 'Metal', 'Nonmetal', and 'Noble'.

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LeetCode Problem 2480

Difficulty: 🟢 Easy
Topics: Database

Solution

Problem Understanding

The problem asks us to identify all valid chemical bonds that can form between elements in a given table. The Elements table contains three columns: symbol, type, and electrons. The type column is an enumeration of 'Metal', 'Nonmetal', and 'Noble'. The electrons column indicates how many electrons a metal can give or a nonmetal needs, while noble gases have zero electrons.

A valid chemical bond occurs only between a metal and a nonmetal. Therefore, for every element classified as 'Metal', it can potentially bond with every element classified as 'Nonmetal'. The output should return all such metal-nonmetal pairs, with the metal in the first column and the nonmetal in the second column.

The problem guarantees that each element has a unique symbol. There is no restriction on the number of metals or nonmetals, and the output can be in any order. This allows us to approach the problem without worrying about duplicates or sorting.

Important edge cases to consider include situations where there are no metals, no nonmetals, or only noble gases. In such cases, the result should simply be empty because no bonds can form.

Approaches

Brute Force Approach

The brute force approach is straightforward: iterate through each element in the table and, for every metal element, iterate through every nonmetal element to create a pair. This works because we are directly following the bonding rule. While this is simple and guarantees correctness, it requires joining every metal with every nonmetal, which is computationally unnecessary if the dataset is large.

Optimal Approach

The optimal approach leverages SQL's ability to perform a self-join with conditions. By selecting all metals and all nonmetals and performing a cross join filtered by the bonding rule, we efficiently generate all valid pairs in a single query. This avoids nested loops at the application level and fully utilizes the database engine's optimization for joins.

Approach Time Complexity Space Complexity Notes
Brute Force O(M * N) O(M * N) Iterate through all metals and nonmetals manually
Optimal O(M * N) O(M * N) Use SQL join to generate pairs efficiently

Algorithm Walkthrough

  1. Identify all elements in the Elements table with type = 'Metal'. These are our candidate metals.
  2. Identify all elements in the Elements table with type = 'Nonmetal'. These are our candidate nonmetals.
  3. Perform a cross join between the metals and nonmetals. This creates all possible pairs where the first element is a metal and the second is a nonmetal.
  4. Select the symbol from the metal as the metal column and the symbol from the nonmetal as the nonmetal column.
  5. Return the resulting table. The database engine handles the ordering and joins, and since the output order is not required, no sorting is necessary.

Why it works: The algorithm guarantees correctness because it explicitly enforces the bonding rule: only metals are paired with nonmetals. Cross joining ensures that every metal is paired with every nonmetal, which matches the problem requirement to list all possible bonds.

Python Solution

# LeetCode Python SQL solution using pandas for demonstration
import pandas as pd

def formChemicalBond(elements: pd.DataFrame) -> pd.DataFrame:
    metals = elements[elements['type'] == 'Metal']
    nonmetals = elements[elements['type'] == 'Nonmetal']
    
    # Cross join metals and nonmetals
    metals['key'] = 1
    nonmetals['key'] = 1
    result = pd.merge(metals, nonmetals, on='key').drop('key', axis=1)
    
    # Select required columns and rename
    result = result[['symbol_x', 'symbol_y']].rename(columns={'symbol_x': 'metal', 'symbol_y': 'nonmetal'})
    return result

Explanation: First, we filter the metals and nonmetals separately. We introduce a dummy column key to perform a cross join using merge. After the join, we select and rename the columns to match the expected output.

Go Solution

package main

import "database/sql"

func formChemicalBond(db *sql.DB) (*sql.Rows, error) {
    query := `
        SELECT m.symbol AS metal, n.symbol AS nonmetal
        FROM Elements m
        JOIN Elements n
        ON m.type = 'Metal' AND n.type = 'Nonmetal';
    `
    return db.Query(query)
}

Go-specific notes: In Go, we rely on SQL directly. There is no need to handle empty slices explicitly because the query returns zero rows if no metals or nonmetals exist. Unlike Python's pandas, Go's sql package performs the join at the database level.

Worked Examples

Using the example input:

Elements table:
+--------+----------+-----------+
| symbol | type     | electrons |
+--------+----------+-----------+
| He     | Noble    | 0         |
| Na     | Metal    | 1         |
| Ca     | Metal    | 2         |
| La     | Metal    | 3         |
| Cl     | Nonmetal | 1         |
| O      | Nonmetal | 2         |
| N      | Nonmetal | 3         |
+--------+----------+-----------+

Step by step:

  1. Filter metals: Na, Ca, La.
  2. Filter nonmetals: Cl, O, N.
  3. Cross join:
Metal Nonmetal
Na Cl
Na O
Na N
Ca Cl
Ca O
Ca N
La Cl
La O
La N

Result matches expected output.

Complexity Analysis

Measure Complexity Explanation
Time O(M * N) Cross join generates all combinations of M metals and N nonmetals
Space O(M * N) Need to store all possible metal-nonmetal pairs

The complexity is linear in terms of the product of the number of metals and nonmetals. Since the output requires listing all pairs, this is optimal.

Test Cases

# test cases
import pandas as pd

# Example case
elements_df = pd.DataFrame([
    {'symbol': 'He', 'type': 'Noble', 'electrons': 0},
    {'symbol': 'Na', 'type': 'Metal', 'electrons': 1},
    {'symbol': 'Ca', 'type': 'Metal', 'electrons': 2},
    {'symbol': 'La', 'type': 'Metal', 'electrons': 3},
    {'symbol': 'Cl', 'type': 'Nonmetal', 'electrons': 1},
    {'symbol': 'O', 'type': 'Nonmetal', 'electrons': 2},
    {'symbol': 'N', 'type': 'Nonmetal', 'electrons': 3},
])
result = formChemicalBond(elements_df)
assert len(result) == 9  # 3 metals * 3 nonmetals

# Edge case: no metals
elements_df = pd.DataFrame([
    {'symbol': 'He', 'type': 'Noble', 'electrons': 0},
    {'symbol': 'Cl', 'type': 'Nonmetal', 'electrons': 1},
])
result = formChemicalBond(elements_df)
assert len(result) == 0

# Edge case: no nonmetals
elements_df = pd.DataFrame([
    {'symbol': 'Na', 'type': 'Metal', 'electrons': 1},
])
result = formChemicalBond(elements_df)
assert len(result) == 0

# Edge case: only noble gases
elements_df = pd.DataFrame([
    {'symbol': 'He', 'type': 'Noble', 'electrons': 0},
])
result = formChemicalBond(elements_df)
assert len(result) == 0
Test Why
3 metals × 3 nonmetals Verifies basic functionality and cross join
No metals Tests empty result when no metals are present
No nonmetals Tests empty result when no nonmetals are present
Only noble gases Ensures nobel gases are ignored in bonding

Edge Cases

One edge case is when the table contains only noble gases. These elements cannot form bonds, so the output should be empty. Our implementation filters by type, so noble gases are ignored naturally.

Another edge case occurs when all elements are metals or all are nonmetals. Since bonding requires one of each, these cases produce no output. The implementation handles this because the cross join produces zero rows if one side is empty.

A third edge case is when the table has large numbers of metals and nonmetals. Since the algorithm produces all combinations, the output can be very large. This is expected behavior and aligns with the problem requirement, but it is important to recognize the memory and time implications when scaling to thousands of elements.