LeetCode #3483 — EASY

Unique 3-Digit Even Numbers

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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.

Example 1:

Input: digits = [1,2,3,4]

Output: 12

Explanation: The 12 distinct 3-digit even numbers that can be formed are 124, 132, 134, 142, 214, 234, 312, 314, 324, 342, 412, and 432. Note that 222 cannot be formed because there is only 1 copy of the digit 2.

Example 2:

Input: digits = [0,2,2]

Output: 2

Explanation: The only 3-digit even numbers that can be formed are 202 and 220. Note that the digit 2 can be used twice because it appears twice in the array.

Example 3:

Input: digits = [6,6,6]

Output: 1

Explanation: Only 666 can be formed.

Example 4:

Input: digits = [1,3,5]

Output: 0

Explanation: No even 3-digit numbers can be formed.

Constraints:

3 <= digits.length <= 10
0 <= digits[i] <= 9

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[1,2,3,4]

Example 2

[0,2,2]

Example 3

[6,6,6]

Core Insight

What unlocks the optimal approach

Use brute force to try all possibilities

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 #3483: Unique 3-Digit Even Numbers
class Solution {
    public int totalNumbers(int[] digits) {
        Set<Integer> s = new HashSet<>();
        int n = digits.length;
        for (int i = 0; i < n; ++i) {
            if (digits[i] % 2 == 1) {
                continue;
            }
            for (int j = 0; j < n; ++j) {
                if (i == j) {
                    continue;
                }
                for (int k = 0; k < n; ++k) {
                    if (digits[k] == 0 || k == i || k == j) {
                        continue;
                    }
                    s.add(digits[k] * 100 + digits[j] * 10 + digits[i]);
                }
            }
        }
        return s.size();
    }
}

// Accepted solution for LeetCode #3483: Unique 3-Digit Even Numbers
func totalNumbers(digits []int) int {
	s := make(map[int]struct{})
	for i, a := range digits {
		if a%2 == 1 {
			continue
		}
		for j, b := range digits {
			if i == j {
				continue
			}
			for k, c := range digits {
				if c == 0 || k == i || k == j {
					continue
				}
				s[c*100+b*10+a] = struct{}{}
			}
		}
	}
	return len(s)
}

# Accepted solution for LeetCode #3483: Unique 3-Digit Even Numbers
class Solution:
    def totalNumbers(self, digits: List[int]) -> int:
        s = set()
        for i, a in enumerate(digits):
            if a & 1:
                continue
            for j, b in enumerate(digits):
                if i == j:
                    continue
                for k, c in enumerate(digits):
                    if c == 0 or k in (i, j):
                        continue
                    s.add(c * 100 + b * 10 + a)
        return len(s)

// Accepted solution for LeetCode #3483: Unique 3-Digit Even Numbers
fn total_numbers(digits: Vec<i32>) -> i32 {
    use std::collections::HashSet;

    let len = digits.len();
    let mut s = HashSet::new();
    for i in 0..len {
        if digits[i] == 0 {
            continue;
        }
        for j in 0..len {
            if i == j {
                continue;
            }

            for k in 0..len {
                if k == i || k == j || digits[k] % 2 == 1 {
                    continue;
                }
                s.insert(digits[i] * 100 + digits[j] * 10 + digits[k]);
            }
        }
    }
    s.len() as i32
}

fn main() {
    let digits = vec![6, 6, 6];
    let ret = total_numbers(digits);
    println!("ret={ret}");
}

#[test]
fn test() {
    {
        let digits = vec![1, 2, 3, 4];
        let ret = total_numbers(digits);
        assert_eq!(ret, 12);
    }
    {
        let digits = vec![0, 2, 2];
        let ret = total_numbers(digits);
        assert_eq!(ret, 2);
    }
    {
        let digits = vec![6, 6, 6];
        let ret = total_numbers(digits);
        assert_eq!(ret, 1);
    }
    {
        let digits = vec![1, 3, 5];
        let ret = total_numbers(digits);
        assert_eq!(ret, 0);
    }
}

// Accepted solution for LeetCode #3483: Unique 3-Digit Even Numbers
function totalNumbers(digits: number[]): number {
    const s = new Set<number>();
    const n = digits.length;
    for (let i = 0; i < n; ++i) {
        if (digits[i] % 2 === 1) {
            continue;
        }
        for (let j = 0; j < n; ++j) {
            if (i === j) {
                continue;
            }
            for (let k = 0; k < n; ++k) {
                if (digits[k] === 0 || k === i || k === j) {
                    continue;
                }
                s.add(digits[k] * 100 + digits[j] * 10 + digits[i]);
            }
        }
    }
    return s.size;
}

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^3)

Space

O(n^3)

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.