LeetCode 2159 - Order Two Columns Independently

The problem provides a database table named Data with two integer columns: firstcol and secondcol. Each row represents a pair of numbers, and duplicate rows are allowed.

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LeetCode Problem 2159

Difficulty: 🟡 Medium
Topics: Database

Solution

Problem Understanding

The problem provides a database table named Data with two integer columns: first_col and second_col. Each row represents a pair of numbers, and duplicate rows are allowed. The task is to return a new table where the numbers in each column are independently sorted: first_col in ascending order and second_col in descending order. Importantly, the sorting of one column does not affect the other. This means the smallest value of first_col does not have to align with the largest value of second_col in the original data.

The input represents a dataset with possibly repeated integer values. The output must preserve the original table shape but with independently reordered columns. Constraints are not explicitly stated, but since the problem is in SQL and typically runs on moderate-sized tables, we can assume there could be thousands of rows.

Important edge cases include tables with duplicate values, a single row, or columns that are already sorted. A naive implementation might mistakenly sort rows together, which is incorrect because the orderings are independent.

Approaches

A brute-force approach is straightforward: fetch all values of first_col and second_col, sort each list according to the required order, and then pair the sorted results by index. This guarantees correctness because sorting independently produces the desired independent orderings. The limitation of this approach is that it is memory-intensive in non-database contexts if the table is very large, though for most SQL queries this is handled efficiently by the database engine.

The key insight for a more optimal SQL solution is that the database allows us to sort each column independently and assign row numbers to map sorted values back into table form. Using window functions, we can assign a sequential index to each sorted column and then join the columns on these indices. This approach is efficient and leverages SQL's built-in optimizations.

Approach Time Complexity Space Complexity Notes
Brute Force O(n log n) O(n) Fetch columns, sort independently, reconstruct table
SQL Window Function O(n log n) O(n) Assign row numbers to sorted columns and join

Algorithm Walkthrough

  1. Extract first_col from the Data table and sort it in ascending order. Assign a sequential row number to each element using ROW_NUMBER(). This row number will act as a key to reassemble the column later.
  2. Similarly, extract second_col from the Data table and sort it in descending order. Assign a sequential row number to each element using ROW_NUMBER().
  3. Perform an inner join on the two sets of data using the row numbers. The join ensures that the smallest value of first_col pairs with the largest value of second_col, then the second smallest with the second largest, and so on.
  4. Select the joined columns in the final output order: first_col followed by second_col.

Why it works: The row number assignment guarantees a one-to-one mapping of positions after sorting, and the join ensures each position in first_col is paired with the corresponding position in second_col after independent sorting. This preserves the independent ordering requirement.

Python Solution

# Python solution for reference, simulating SQL behavior
from typing import List
import pandas as pd

class Solution:
    def orderColumnsIndependently(self, Data: pd.DataFrame) -> pd.DataFrame:
        # Sort first_col ascending and reset index
        first_sorted = Data[['first_col']].sort_values(by='first_col', ascending=True).reset_index(drop=True)
        # Sort second_col descending and reset index
        second_sorted = Data[['second_col']].sort_values(by='second_col', ascending=False).reset_index(drop=True)
        # Concatenate columns side by side
        result = pd.concat([first_sorted, second_sorted], axis=1)
        return result

In this implementation, we use pandas to simulate independent column sorting. We first sort first_col in ascending order and second_col in descending order, resetting the index to create a clean positional mapping. Finally, concat merges the two DataFrames column-wise, mimicking the SQL window function join.

Go Solution

package main

import (
    "sort"
)

type DataRow struct {
    FirstCol  int
    SecondCol int
}

func orderColumnsIndependently(data []DataRow) []DataRow {
    n := len(data)
    firstCol := make([]int, n)
    secondCol := make([]int, n)
    
    for i, row := range data {
        firstCol[i] = row.FirstCol
        secondCol[i] = row.SecondCol
    }
    
    sort.Ints(firstCol)
    sort.Slice(secondCol, func(i, j int) bool {
        return secondCol[i] > secondCol[j]
    })
    
    result := make([]DataRow, n)
    for i := 0; i < n; i++ {
        result[i] = DataRow{FirstCol: firstCol[i], SecondCol: secondCol[i]}
    }
    return result
}

In Go, slices are used to hold each column separately. sort.Ints sorts first_col in ascending order, and sort.Slice with a custom comparator sorts second_col in descending order. Finally, we reconstruct the result slice by pairing the sorted values.

Worked Examples

Example input:

first_col: [4, 2, 3, 1]
second_col: [2, 3, 1, 4]

Step 1: Sort first_col ascending → [1, 2, 3, 4]

Step 2: Sort second_col descending → [4, 3, 2, 1]

Step 3: Pair by index → result table:

first_col | second_col
1         | 4
2         | 3
3         | 2
4         | 1

Complexity Analysis

Measure Complexity Explanation
Time O(n log n) Sorting each column independently takes O(n log n)
Space O(n) Storing sorted columns and constructing result table requires O(n)

Sorting dominates the time complexity. Additional memory is needed for temporary sorted lists, leading to linear space usage.

Test Cases

import pandas as pd

# Provided example
df = pd.DataFrame({'first_col': [4,2,3,1], 'second_col':[2,3,1,4]})
assert Solution().orderColumnsIndependently(df).equals(
    pd.DataFrame({'first_col':[1,2,3,4], 'second_col':[4,3,2,1]})
)

# Single row
df = pd.DataFrame({'first_col':[5], 'second_col':[10]})
assert Solution().orderColumnsIndependently(df).equals(
    pd.DataFrame({'first_col':[5], 'second_col':[10]})
)

# Duplicates
df = pd.DataFrame({'first_col':[2,2,1], 'second_col':[1,3,2]})
assert Solution().orderColumnsIndependently(df).equals(
    pd.DataFrame({'first_col':[1,2,2], 'second_col':[3,2,1]})
)

# Already sorted
df = pd.DataFrame({'first_col':[1,2,3], 'second_col':[3,2,1]})
assert Solution().orderColumnsIndependently(df).equals(
    pd.DataFrame({'first_col':[1,2,3], 'second_col':[3,2,1]})
)

# All same values
df = pd.DataFrame({'first_col':[1,1,1], 'second_col':[2,2,2]})
assert Solution().orderColumnsIndependently(df).equals(
    pd.DataFrame({'first_col':[1,1,1], 'second_col':[2,2,2]})
)
Test Why
Provided example Validates normal case with independent column sorting
Single row Tests minimal input size
Duplicates Checks handling of repeated values
Already sorted Ensures algorithm does not disrupt already sorted columns
All same values Confirms algorithm handles non-varying data correctly

Edge Cases

One important edge case is when the table has only one row. Sorting a single value in either order should return the same row, which our implementation correctly handles because sorting a list of length one is a no-op.

Another edge case is when multiple rows have identical values. This could trip up algorithms that attempt to sort entire rows rather than columns. Our approach treats columns independently and does not rely on row ordering, ensuring correct results.

Finally, when both columns are already sorted in the required order, a naive implementation that recombines columns incorrectly could shuffle values. Our algorithm explicitly sorts each column independently and pairs by index, preserving the existing correct ordering.