LeetCode #2094 — EASY

Finding 3-Digit Even Numbers

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

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The Problem

Problem Statement

You are given an integer array digits, where each element is a digit. The array may contain duplicates.

You need to find all the unique integers that follow the given requirements:

The integer consists of the concatenation of three elements from digits in any arbitrary order.
The integer does not have leading zeros.
The integer is even.

For example, if the given digits were [1, 2, 3], integers 132 and 312 follow the requirements.

Return a sorted array of the unique integers.

Example 1:

Input: digits = [2,1,3,0]
Output: [102,120,130,132,210,230,302,310,312,320]
Explanation: All the possible integers that follow the requirements are in the output array. 
Notice that there are no odd integers or integers with leading zeros.

Example 2:

Input: digits = [2,2,8,8,2]
Output: [222,228,282,288,822,828,882]
Explanation: The same digit can be used as many times as it appears in digits. 
In this example, the digit 8 is used twice each time in 288, 828, and 882.

Example 3:

Input: digits = [3,7,5]
Output: []
Explanation: No even integers can be formed using the given digits.

Constraints:

3 <= digits.length <= 100
0 <= digits[i] <= 9

Step 01

Brute Force Baseline

Problem summary: You are given an integer array digits, where each element is a digit. The array may contain duplicates. You need to find all the unique integers that follow the given requirements: The integer consists of the concatenation of three elements from digits in any arbitrary order. The integer does not have leading zeros. The integer is even. For example, if the given digits were [1, 2, 3], integers 132 and 312 follow the requirements. Return a sorted array of the unique integers.

Baseline thinking

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

Pattern signal: Array · Hash Map

Example 1

[2,1,3,0]

Example 2

[2,2,8,8,2]

Example 3

[3,7,5]

Core Insight

What unlocks the optimal approach

The range of possible answers includes all even numbers between 100 and 999 inclusive. Could you check each possible answer to see if it could be formed from the digits in the array?

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 #2094: Finding 3-Digit Even Numbers
class Solution {
    public int[] findEvenNumbers(int[] digits) {
        int[] cnt = new int[10];
        for (int x : digits) {
            ++cnt[x];
        }
        List<Integer> ans = new ArrayList<>();
        for (int x = 100; x < 1000; x += 2) {
            int[] cnt1 = new int[10];
            for (int y = x; y > 0; y /= 10) {
                ++cnt1[y % 10];
            }
            boolean ok = true;
            for (int i = 0; i < 10 && ok; ++i) {
                ok = cnt[i] >= cnt1[i];
            }
            if (ok) {
                ans.add(x);
            }
        }
        return ans.stream().mapToInt(i -> i).toArray();
    }
}

// Accepted solution for LeetCode #2094: Finding 3-Digit Even Numbers
func findEvenNumbers(digits []int) (ans []int) {
	cnt := [10]int{}
	for _, x := range digits {
		cnt[x]++
	}
	for x := 100; x < 1000; x += 2 {
		cnt1 := [10]int{}
		for y := x; y > 0; y /= 10 {
			cnt1[y%10]++
		}
		ok := true
		for i := 0; i < 10 && ok; i++ {
			ok = cnt[i] >= cnt1[i]
		}
		if ok {
			ans = append(ans, x)
		}
	}
	return
}

# Accepted solution for LeetCode #2094: Finding 3-Digit Even Numbers
class Solution:
    def findEvenNumbers(self, digits: List[int]) -> List[int]:
        cnt = Counter(digits)
        ans = []
        for x in range(100, 1000, 2):
            cnt1 = Counter()
            y = x
            while y:
                y, v = divmod(y, 10)
                cnt1[v] += 1
            if all(cnt[i] >= cnt1[i] for i in range(10)):
                ans.append(x)
        return ans

// Accepted solution for LeetCode #2094: Finding 3-Digit Even Numbers
struct Solution;

impl Solution {
    fn find_even_numbers(digits: Vec<i32>) -> Vec<i32> {
        let mut all = [0; 10];
        for d in digits {
            all[d as usize] += 1;
        }
        let mut res = vec![];
        let even = vec![0, 2, 4, 6, 8];
        for i in 1..=9 {
            for j in 0..=9 {
                for &k in &even {
                    let mut count = [0; 10];
                    count[i] += 1;
                    count[j] += 1;
                    count[k] += 1;
                    if count[i] <= all[i] && count[j] <= all[j] && count[k] <= all[k] {
                        res.push((i * 100 + j * 10 + k) as i32);
                    }
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let digits = vec![2, 1, 3, 0];
    let res = vec![102, 120, 130, 132, 210, 230, 302, 310, 312, 320];
    assert_eq!(Solution::find_even_numbers(digits), res);
    let digits = vec![2, 2, 8, 8, 2];
    let res = vec![222, 228, 282, 288, 822, 828, 882];
    assert_eq!(Solution::find_even_numbers(digits), res);
    let digits = vec![3, 7, 5];
    let res: Vec<i32> = vec![];
    assert_eq!(Solution::find_even_numbers(digits), res);
}

// Accepted solution for LeetCode #2094: Finding 3-Digit Even Numbers
function findEvenNumbers(digits: number[]): number[] {
    const cnt: number[] = Array(10).fill(0);
    for (const x of digits) {
        ++cnt[x];
    }
    const ans: number[] = [];
    for (let x = 100; x < 1000; x += 2) {
        const cnt1: number[] = Array(10).fill(0);
        for (let y = x; y; y = Math.floor(y / 10)) {
            ++cnt1[y % 10];
        }
        let ok = true;
        for (let i = 0; i < 10 && ok; ++i) {
            ok = cnt[i] >= cnt1[i];
        }
        if (ok) {
            ans.push(x);
        }
    }
    return ans;
}

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.

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.