Problem Description

You are given an array of digits called digits. Your task is to determine the number of distinct three-digit even numbers that can be formed using these digits.

Note: Each copy of a digit can only be used once per number, and there may not be leading zeros.

Intuition

To solve this problem, the main idea is to systematically create all possible three-digit even numbers by using the digits provided. An even number must end with an even digit. Therefore, our first step would be to identify digits that can be used as the last digit of our number.

Here's how the solution progresses:

1. Use a set: Utilize a set to store numbers to ensure all numbers are distinct, as sets automatically manage duplicates.

2. Iteration Strategy:

   • First, loop through each digit to decide which can be the last digit, ensuring it is even. This will greatly reduce potential candidates.
   • For each even digit picked as the ending digit, choose two more digits out of the remaining to construct the hundreds and tens place.
   • Ensure no digit is repeated within any number, i.e., keep track of indices used when forming a number to avoid using the same digit more than once.

3. Avoid Leading Zero: When forming the number, ensure the hundreds place (first digit) is not zero, which disqualifies it from being a valid three-digit number.

By following this organized and methodical way of generating numbers while using a set to filter out duplicates, we efficiently find all valid combinations.

Learn more about Recursion patterns.

Solution Approach

The solution implements a method to systematically create all potential three-digit even numbers from the given digits list. This approach efficiently ensures that only valid and distinct numbers are counted. Here’s a detailed breakdown of the solution:

1. Hash Set Usage:

   • We employ a hash set s to store the numbers we generate. The purpose of using a set is to automatically handle duplicates, meaning only unique numbers are recorded.

2. Enumeration of Digits:

   • Iterate over each digit in the digits list to determine which can be the last digit of the number. For a number to be even, it needs to end in an even digit (0, 2, 4, 6, 8). This serves as our starting point for constructing a number.

3. Nested Loops for Number Construction:

   • For each possible last digit a (must be even), utilize two more loops to select different digits for the hundreds and tens places. The constraints include not reusing any digit that serves as the last digit while choosing the middle and first digits.
   • Specifically:
     • The first loop selects the digit b for the tens place.
     • The second loop chooses the digit c for the hundreds place.

4. Ensure Valid Three-Digit Number:

   • We ensure that the three-digit number does not start with zero (c should not be zero), thereby disallowing invalid numbers like 012.
   • If these conditions are met, form the number using the formula c * 100 + b * 10 + a and insert it into the set.

5. Result:

   • The size of the set (len(s)) gives the total count of distinct three-digit even numbers that can be formed as desired.

This approach is clear and concise, leveraging the set for uniqueness constraint and nested loops for systematic enumeration, all while adhering to number formation rules.

Example Walkthrough

Let's walk through an example using the solution approach to better understand how it works.

Suppose you are given the array of digits: [1, 2, 3, 4].

1. Set Up the Hash Set: Initialize an empty set s to store distinct three-digit even numbers.

2. Identify Even Digits for the Last Position:

   • The potential even digits from our list for the last digit are 2 and 4.

3. Construct Numbers Using Nested Loops:

   • Choose 2 as the last digit:

     • Try 3 as the tens digit and 1 as the hundreds digit. Form the number 132, which is valid, and add it to the set.
     • Try 1 as the tens digit and 3 as the hundreds digit. Form the number 312, which is valid, and add it to the set.
     • Other combinations include 312 and 132 again, but they would already exist in the set.

   • Choose 4 as the last digit:

     • Try 3 as the tens digit and 1 as the hundreds digit. Form the number 134, which is valid, and add it to the set.
     • Try 1 as the tens digit and 3 as the hundreds digit. Form the number 314, which is valid, and add it to the set.
     • Any other permutations that result in the same number are naturally ignored due to the set's mechanics.

4. Result Calculation:

   • The set by now contains the numbers: 132, 312, 134, 314.
   • The total number of distinct three-digit even numbers is given by len(s), resulting in 4.

This walkthrough demonstrates how the problem is tackled with a systematic approach using sets and nested loops to achieve the desired distinct outcomes.

Solution Implementation

Time and Space Complexity

The time complexity of the code is O(n^3) because the solution involves three nested loops, each iterating over the entire list digits. Each loop represents an index selection from the list, resulting in a cubic relationship relative to the length n of the array.

The space complexity can be considered O(n^3) in the worst-case scenario where all possible three-digit combinations are unique and added to the set s. However, the space complexity could be seen as smaller depending on the input constraints and the distinctiveness of the digits since a set naturally handles duplicates. Nonetheless, as per the reference, considering every possibility leads to this space complexity.

Learn more about how to find time and space complexity quickly.