#2032

Easy

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Two Out of Three

Identify all elements appearing in at least two of three arrays using efficient scanning and hash lookups for fast verification.

array hash table bit manipulation

Problem Statement

Given three integer arrays nums1, nums2, and nums3, return a list of all integers that appear in at least two of the arrays. Each list may contain duplicates, but the result should contain unique numbers only. The order of the result does not matter.

For example, if nums1 = [1,1,3,2], nums2 = [2,3], and nums3 = [3], the output should be [3,2] because 3 appears in all three arrays and 2 appears in nums1 and nums2. Constraints: 1 <= nums1.length, nums2.length, nums3.length <= 100 and 1 <= nums1[i], nums2[j], nums3[k] <= 100.

Examples

Example 1

Input: nums1 = [1,1,3,2], nums2 = [2,3], nums3 = [3]

Output: [3,2]

The values that are present in at least two arrays are:

3, in all three arrays.
2, in nums1 and nums2.

Example 2

Input: nums1 = [3,1], nums2 = [2,3], nums3 = [1,2]

Output: [2,3,1]

The values that are present in at least two arrays are:

2, in nums2 and nums3.
3, in nums1 and nums2.
1, in nums1 and nums3.

Example 3

Input: nums1 = [1,2,2], nums2 = [4,3,3], nums3 = [5]

Output: []

No value is present in at least two arrays.

Constraints

1 <= nums1.length, nums2.length, nums3.length <= 100
1 <= nums1[i], nums2[j], nums3[k] <= 100

Solution Approach

Use hash sets to record each array

Convert each array into a set to remove duplicates and allow O(1) membership checks. This ensures we only count unique numbers per array when scanning for overlaps.

Count occurrences across arrays

Iterate through all unique numbers from nums1, nums2, and nums3. Use a dictionary to track in how many arrays each number appears. If the count reaches two or more, add it to the result list.

Return the aggregated result

After scanning all arrays, collect numbers that appeared in at least two arrays. This method guarantees correct output while keeping operations efficient and avoiding repeated full searches.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n1+n2+n3) because each array is scanned once and set operations are O(1). Space complexity is O(n1+n2+n3) to store unique numbers from each array in sets and a count map.

What Interviewers Usually Probe

The candidate immediately considers using hash sets for quick membership checks.
They attempt to avoid nested loops over arrays to reduce time complexity.
They clearly separate counting elements per array from aggregating the final result.

Common Pitfalls or Variants

Common pitfalls

Counting duplicates within the same array instead of across arrays.
Using nested loops over arrays, causing unnecessary O(n^2) time.
Forgetting to remove duplicates before counting, leading to incorrect results.

Follow-up variants

Find numbers present in exactly two arrays instead of at least two.
Arrays contain negative numbers, requiring careful set handling.
Arrays are very large, emphasizing memory-efficient counting methods.

FAQ

What is the main strategy for Two Out of Three?

Use hash sets for each array and count how many arrays each number appears in to find numbers in at least two arrays.

Can duplicates in a single array affect the result?

No, duplicates within the same array are ignored; only presence across multiple arrays matters.

How does the array scanning plus hash lookup pattern help?

It allows O(1) checks for element presence and avoids repeated full array searches, keeping the solution efficient.

What if all arrays are empty?

The result will be an empty list because no number appears in at least two arrays.

Are the output numbers ordered?

No, the problem only requires the list of numbers; their order does not matter.

terminal

Solution

Solution 1: Array + Enumeration

We can first put each element of the arrays into an array, then enumerate each number $i$ from $1$ to $100$, and check whether $i$ appears in at least two arrays. If so, add $i$ to the answer array.

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class Solution:
    def twoOutOfThree(
        self, nums1: List[int], nums2: List[int], nums3: List[int]
    ) -> List[int]:
        s1, s2, s3 = set(nums1), set(nums2), set(nums3)
        return [i for i in range(1, 101) if (i in s1) + (i in s2) + (i in s3) > 1]

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