# 2032. Two Out of Three

## Description

Given three integer arrays nums1, nums2, and nums3, return a distinct array containing all the values that are present in at least two out of the three arrays. You may return the values in any order.

Example 1:

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

## Solutions

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.

The time complexity is $O(n_1 + n_2 + n_3)$, and the space complexity is $O(n_1 + n_2 + n_3)$. Here, $n_1, n_2, n_3$ are the lengths of the arrays nums1, nums2, and nums3, respectively.

• class Solution {
public List<Integer> twoOutOfThree(int[] nums1, int[] nums2, int[] nums3) {
int[] s1 = get(nums1), s2 = get(nums2), s3 = get(nums3);
List<Integer> ans = new ArrayList<>();
for (int i = 1; i <= 100; ++i) {
if (s1[i] + s2[i] + s3[i] > 1) {
}
}
return ans;
}

private int[] get(int[] nums) {
int[] s = new int[101];
for (int num : nums) {
s[num] = 1;
}
return s;
}
}

• class Solution {
public:
vector<int> twoOutOfThree(vector<int>& nums1, vector<int>& nums2, vector<int>& nums3) {
auto get = [](vector<int>& nums) {
vector<int> cnt(101);
for (int& v : nums) cnt[v] = 1;
return cnt;
};
auto s1 = get(nums1), s2 = get(nums2), s3 = get(nums3);
vector<int> ans;
for (int i = 1; i <= 100; ++i) {
if (s1[i] + s2[i] + s3[i] > 1) {
ans.emplace_back(i);
}
}
return ans;
}
};

• 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]


• func twoOutOfThree(nums1 []int, nums2 []int, nums3 []int) (ans []int) {
get := func(nums []int) (s [101]int) {
for _, v := range nums {
s[v] = 1
}
return
}
s1, s2, s3 := get(nums1), get(nums2), get(nums3)
for i := 1; i <= 100; i++ {
if s1[i]+s2[i]+s3[i] > 1 {
ans = append(ans, i)
}
}
return
}

• function twoOutOfThree(nums1: number[], nums2: number[], nums3: number[]): number[] {
const count = new Array(101).fill(0);
new Set(nums1).forEach(v => count[v]++);
new Set(nums2).forEach(v => count[v]++);
new Set(nums3).forEach(v => count[v]++);
const ans = [];
count.forEach((v, i) => {
if (v >= 2) {
ans.push(i);
}
});
return ans;
}


• use std::collections::HashSet;
impl Solution {
pub fn two_out_of_three(nums1: Vec<i32>, nums2: Vec<i32>, nums3: Vec<i32>) -> Vec<i32> {
let mut count = vec![0; 101];
nums1
.into_iter()
.collect::<HashSet<i32>>()
.iter()
.for_each(|&v| {
count[v as usize] += 1;
});
nums2
.into_iter()
.collect::<HashSet<i32>>()
.iter()
.for_each(|&v| {
count[v as usize] += 1;
});
nums3
.into_iter()
.collect::<HashSet<i32>>()
.iter()
.for_each(|&v| {
count[v as usize] += 1;
});
let mut ans = Vec::new();
count
.iter()
.enumerate()
.for_each(|(i, v)| {
if *v >= 2 {
ans.push(i as i32);
}
});
ans
}
}