LeetCode #3220 — MEDIUM

Odd and Even Transactions

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

The Problem

Problem Statement

Table: transactions

+------------------+------+
| Column Name      | Type | 
+------------------+------+
| transaction_id   | int  |
| amount           | int  |
| transaction_date | date |
+------------------+------+
The transactions_id column uniquely identifies each row in this table.
Each row of this table contains the transaction id, amount and transaction date.

Write a solution to find the sum of amounts for odd and even transactions for each day. If there are no odd or even transactions for a specific date, display as 0.

Return the result table ordered by transaction_date in ascending order.

The result format is in the following example.

Example:

Input:

transactions table:

+----------------+--------+------------------+
| transaction_id | amount | transaction_date |
+----------------+--------+------------------+
| 1              | 150    | 2024-07-01       |
| 2              | 200    | 2024-07-01       |
| 3              | 75     | 2024-07-01       |
| 4              | 300    | 2024-07-02       |
| 5              | 50     | 2024-07-02       |
| 6              | 120    | 2024-07-03       |
+----------------+--------+------------------+

Output:

+------------------+---------+----------+
| transaction_date | odd_sum | even_sum |
+------------------+---------+----------+
| 2024-07-01       | 75      | 350      |
| 2024-07-02       | 0       | 350      |
| 2024-07-03       | 0       | 120      |
+------------------+---------+----------+

Explanation:

For transaction dates:
- 2024-07-01:
  - Sum of amounts for odd transactions: 75
  - Sum of amounts for even transactions: 150 + 200 = 350
- 2024-07-02:
  - Sum of amounts for odd transactions: 0
  - Sum of amounts for even transactions: 300 + 50 = 350
- 2024-07-03:
  - Sum of amounts for odd transactions: 0
  - Sum of amounts for even transactions: 120

Note: The output table is ordered by transaction_date in ascending order.

Step 01

Brute Force Baseline

Problem summary: Table: transactions +------------------+------+ | Column Name | Type | +------------------+------+ | transaction_id | int | | amount | int | | transaction_date | date | +------------------+------+ The transactions_id column uniquely identifies each row in this table. Each row of this table contains the transaction id, amount and transaction date. Write a solution to find the sum of amounts for odd and even transactions for each day. If there are no odd or even transactions for a specific date, display as 0. Return the result table ordered by transaction_date in ascending order. 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":{"transactions":["transaction_id","amount","transaction_date"]},"rows":{"transactions":[[1,150,"2024-07-01"],[2,200,"2024-07-01"],[3,75,"2024-07-01"],[4,300,"2024-07-02"],[5,50,"2024-07-02"],[6,120,"2024-07-03"]]}}

Step 02

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 #3220: Odd and Even Transactions
// Auto-generated Java example from py.
class Solution {
    public void exampleSolution() {
    }
}
// Reference (py):
// # Accepted solution for LeetCode #3220: Odd and Even Transactions
// import pandas as pd
// 
// 
// def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
//     transactions["odd_sum"] = transactions["amount"].where(
//         transactions["amount"] % 2 == 1, 0
//     )
//     transactions["even_sum"] = transactions["amount"].where(
//         transactions["amount"] % 2 == 0, 0
//     )
// 
//     result = (
//         transactions.groupby("transaction_date")
//         .agg(odd_sum=("odd_sum", "sum"), even_sum=("even_sum", "sum"))
//         .reset_index()
//     )
// 
//     result = result.sort_values("transaction_date")
// 
//     return result

// Accepted solution for LeetCode #3220: Odd and Even Transactions
// Auto-generated Go example from py.
func exampleSolution() {
}
// Reference (py):
// # Accepted solution for LeetCode #3220: Odd and Even Transactions
// import pandas as pd
// 
// 
// def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
//     transactions["odd_sum"] = transactions["amount"].where(
//         transactions["amount"] % 2 == 1, 0
//     )
//     transactions["even_sum"] = transactions["amount"].where(
//         transactions["amount"] % 2 == 0, 0
//     )
// 
//     result = (
//         transactions.groupby("transaction_date")
//         .agg(odd_sum=("odd_sum", "sum"), even_sum=("even_sum", "sum"))
//         .reset_index()
//     )
// 
//     result = result.sort_values("transaction_date")
// 
//     return result

# Accepted solution for LeetCode #3220: Odd and Even Transactions
import pandas as pd


def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
    transactions["odd_sum"] = transactions["amount"].where(
        transactions["amount"] % 2 == 1, 0
    )
    transactions["even_sum"] = transactions["amount"].where(
        transactions["amount"] % 2 == 0, 0
    )

    result = (
        transactions.groupby("transaction_date")
        .agg(odd_sum=("odd_sum", "sum"), even_sum=("even_sum", "sum"))
        .reset_index()
    )

    result = result.sort_values("transaction_date")

    return result

// Accepted solution for LeetCode #3220: Odd and Even Transactions
// 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 #3220: Odd and Even Transactions
// import pandas as pd
// 
// 
// def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
//     transactions["odd_sum"] = transactions["amount"].where(
//         transactions["amount"] % 2 == 1, 0
//     )
//     transactions["even_sum"] = transactions["amount"].where(
//         transactions["amount"] % 2 == 0, 0
//     )
// 
//     result = (
//         transactions.groupby("transaction_date")
//         .agg(odd_sum=("odd_sum", "sum"), even_sum=("even_sum", "sum"))
//         .reset_index()
//     )
// 
//     result = result.sort_values("transaction_date")
// 
//     return result

// Accepted solution for LeetCode #3220: Odd and Even Transactions
// Auto-generated TypeScript example from py.
function exampleSolution(): void {
}
// Reference (py):
// # Accepted solution for LeetCode #3220: Odd and Even Transactions
// import pandas as pd
// 
// 
// def sum_daily_odd_even(transactions: pd.DataFrame) -> pd.DataFrame:
//     transactions["odd_sum"] = transactions["amount"].where(
//         transactions["amount"] % 2 == 1, 0
//     )
//     transactions["even_sum"] = transactions["amount"].where(
//         transactions["amount"] % 2 == 0, 0
//     )
// 
//     result = (
//         transactions.groupby("transaction_date")
//         .agg(odd_sum=("odd_sum", "sum"), even_sum=("even_sum", "sum"))
//         .reset_index()
//     )
// 
//     result = result.sort_values("transaction_date")
// 
//     return result

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.