Problem statement

Given three integer arrays, return every distinct value that appears in at least two of them.

Return the values in ascending order; if there are none, return an empty array.

Example 1

Input: nums1 = [4,7,7,1], nums2 = [7,2], nums3 = [1,9]
Output: [1,7]
Explanation: 1 is in nums1 and nums3; 7 is in nums1 and nums2.

Example 2

Input: nums1 = [5], nums2 = [6], nums3 = [8]
Output: []

Example 3

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

Constraints

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

Hints for Two Out of Three

1. Hint 1

   Repeats inside one array do not count twice — only which arrays contain a value matters.

2. Hint 2

   Give each value a 3-bit mask: bit k set when the value occurs in array k.

3. Hint 3

   Values are at most 100: scan v from 1 to 100 and keep those whose mask has at least two bits set.

How to solve Two Out of Three

Record, for every value, the set of arrays it appears in as a 3-bit mask; a value qualifies when its mask has two or more bits.

Approach

1. Create mask[0..100] filled with 0.
2. For array k (0, 1, 2) and each of its values v, set mask[v] |= 1 << k.
3. Scan v from 1 to 100 and collect it if mask[v] is 3, 5, 6 or 7 (at least two bits).

Why it works

OR-ing a bit is idempotent, so duplicates within one array do not inflate the count; the mask records exactly which arrays contain v. Scanning values in increasing order produces the required ascending output with no sort.

Complexity

• Time — O(n1 + n2 + n3 + V) with V = 100
• Space — O(V)

Pitfalls

• Counting occurrences instead of arrays wrongly accepts [7,7] appearing in only one array.
• Each value must be reported once, even if it is in all three arrays.
• The output order must be ascending here.

Reference solution

Python

from typing import List

def twoOutOfThree(nums1: List[int], nums2: List[int], nums3: List[int]) -> List[int]:
    s1, s2, s3 = set(nums1), set(nums2), set(nums3)
    return sorted(v for v in s1 | s2 | s3 if (v in s1) + (v in s2) + (v in s3) >= 2)

Java

public static int[] twoOutOfThree(int[] nums1, int[] nums2, int[] nums3) {
    int[] mask = new int[101];
    for (int v : nums1) mask[v] |= 1;
    for (int v : nums2) mask[v] |= 2;
    for (int v : nums3) mask[v] |= 4;
    List<Integer> out = new ArrayList<>();
    for (int v = 1; v <= 100; v++) {
        if (Integer.bitCount(mask[v]) >= 2) out.add(v);
    }
    int[] res = new int[out.size()];
    for (int i = 0; i < res.length; i++) res[i] = out.get(i);
    return res;
}

C++

vector<int> twoOutOfThree(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3) {
    vector<int> mask(101, 0);
    for (int v : nums1) mask[v] |= 1;
    for (int v : nums2) mask[v] |= 2;
    for (int v : nums3) mask[v] |= 4;
    vector<int> out;
    for (int v = 1; v <= 100; v++) {
        int m = mask[v];
        if (m == 3 || m == 5 || m == 6 || m == 7) out.push_back(v);
    }
    return out;
}

JavaScript

var twoOutOfThree = function(nums1, nums2, nums3) {
    var mask = new Array(101).fill(0);
    var arrays = [nums1, nums2, nums3];
    for (var k = 0; k < 3; k++) {
        for (var i = 0; i < arrays[k].length; i++) mask[arrays[k][i]] |= 1 << k;
    }
    var out = [];
    for (var v = 1; v <= 100; v++) {
        var m = mask[v];
        if (m === 3 || m === 5 || m === 6 || m === 7) out.push(v);
    }
    return out;
};

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