leetcode/solution/2000-2099/2032.Two Out of Three

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2000-2099/2032.Two%20Out%20of%20Three/README_EN.md	1269	Weekly Contest 262 Q1	Bit Manipulation Array Hash Table

2032. Two Out of Three

中文文档

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.

Python3

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]

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

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
    }
}