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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.

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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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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) {
                ans.add(i);
            }
        }
        return ans;
    }

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

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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