README_EN.md

350. Intersection of Two Arrays II

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Description

Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.

 

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]
Explanation: [9,4] is also accepted.

 

Constraints:

• 1 <= nums1.length, nums2.length <= 1000
• 0 <= nums1[i], nums2[i] <= 1000

 

Follow up:

• What if the given array is already sorted? How would you optimize your algorithm?
• What if nums1's size is small compared to nums2's size? Which algorithm is better?
• What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

Solutions

Python3

class Solution:
    def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
        counter = Counter(nums1)
        res = []
        for num in nums2:
            if counter[num] > 0:
                res.append(num)
                counter[num] -= 1
        return res

Java

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Map<Integer, Integer> counter = new HashMap<>();
        for (int num : nums1) {
            counter.put(num, counter.getOrDefault(num, 0) + 1);
        }
        List<Integer> t = new ArrayList<>();
        for (int num : nums2) {
            if (counter.getOrDefault(num, 0) > 0) {
                t.add(num);
                counter.put(num, counter.get(num) - 1);
            }
        }
        int[] res = new int[t.size()];
        for (int i = 0; i < res.length; ++i) {
            res[i] = t.get(i);
        }
        return res;
    }
}

JavaScript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersect = function (nums1, nums2) {
    const counter = {};
    for (const num of nums1) {
        counter[num] = (counter[num] || 0) + 1;
    }
    let res = [];
    for (const num of nums2) {
        if (counter[num] > 0) {
            res.push(num);
            counter[num] -= 1;
        }
    }
    return res;
};

C++

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> counter;
        for (int num : nums1) ++counter[num];
        vector<int> res;
        for (int num : nums2) {
            if (counter[num] > 0) {
                --counter[num];
                res.push_back(num);
            }
        }
        return res;
    }
};

Go

func intersect(nums1 []int, nums2 []int) []int {
	counter := make(map[int]int)
	for _, num := range nums1 {
		counter[num]++
	}
	var res []int
	for _, num := range nums2 {
		if counter[num] > 0 {
			counter[num]--
			res = append(res, num)
		}
	}
	return res
}

TypeScript

function intersect(nums1: number[], nums2: number[]): number[] {
    const map = new Map<number, number>();
    for (const num of nums1) {
        map.set(num, (map.get(num) ?? 0) + 1);
    }

    const res = [];
    for (const num of nums2) {
        if (map.has(num) && map.get(num) !== 0) {
            res.push(num);
            map.set(num, map.get(num) - 1);
        }
    }
    return res;
}

Rust

use std::collections::HashMap;
impl Solution {
    pub fn intersect(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
        let mut map = HashMap::new();
        for num in nums1.iter() {
            *map.entry(num).or_insert(0) += 1;
        }

        let mut res = vec![];
        for num in nums2.iter() {
            if map.contains_key(num) && map.get(num).unwrap() != &0 {
                map.insert(num, map.get(&num).unwrap() - 1);
                res.push(*num);
            }
        }
        res
    }
}

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