Formatted question description: https://leetcode.ca/all/2032.html

2032. Two Out of Three (Easy)

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

Solution 1.

// OJ: https://leetcode.com/problems/two-out-of-three/
// Time: O(N)
// Space: O(R) where `R` is the range of the numbers in arrays.
class Solution {
public:
    vector<int> twoOutOfThree(vector<int>& A, vector<int>& B, vector<int>& C) {
        unordered_set<int> a(begin(A), end(A)), b(begin(B), end(B)), c(begin(C), end(C));
        unordered_map<int, int> cnt;
        for (int n : a) cnt[n]++;
        for (int n : b) cnt[n]++;
        for (int n : c) cnt[n]++;
        vector<int> ans;
        for (auto [n, c] : cnt) {
            if (c >= 2) ans.push_back(n);
        }
        return ans;
    }
};

Or

// OJ: https://leetcode.com/problems/two-out-of-three/
// Time: O(N)
// Space: O(R) where `R` is the range of the numbers in arrays.
class Solution {
public:
    vector<int> twoOutOfThree(vector<int>& A, vector<int>& B, vector<int>& C) {
        bool seen[3][101] = {};
        for (int n : A) seen[0][n] = true;
        for (int n : B) seen[1][n] = true;
        for (int n : C) seen[2][n] = true;
        vector<int> ans;
        for (int i = 1; i <= 100; ++i) {
            if (seen[0][i] + seen[1][i] + seen[2][i] >= 2) ans.push_back(i);
        }
        return ans;
    }
};

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