README_EN.md

comments	difficulty	edit_url	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/0300-0399/0350.Intersection%20of%20Two%20Arrays%20II/README_EN.md	Array Hash Table Two Pointers Binary Search Sorting

350. Intersection of Two Arrays II

中文文档

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

Solution 1: Hash Table

We can use a hash table $\textit{cnt}$ to count the occurrences of each element in the array $\textit{nums1}$. Then, we iterate through the array $\textit{nums2}$. If an element $x$ is in $\textit{cnt}$ and the occurrence of $x$ is greater than $0$, we add $x$ to the answer and then decrement the occurrence of $x$ by one.

After the iteration is finished, we return the answer array.

The time complexity is $O(m + n)$, and the space complexity is $O(m)$. Here, $m$ and $n$ are the lengths of the arrays $\textit{nums1}$ and $\textit{nums2}$, respectively.

Python3

class Solution:
    def intersect(self, nums1: List[int], nums2: List[int]) -> List[int]:
        cnt = Counter(nums1)
        ans = []
        for x in nums2:
            if cnt[x]:
                ans.append(x)
                cnt[x] -= 1
        return ans

Java

class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        int[] cnt = new int[1001];
        for (int x : nums1) {
            ++cnt[x];
        }
        List<Integer> ans = new ArrayList<>();
        for (int x : nums2) {
            if (cnt[x]-- > 0) {
                ans.add(x);
            }
        }
        return ans.stream().mapToInt(Integer::intValue).toArray();
    }
}

C++

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

Go

func intersect(nums1 []int, nums2 []int) (ans []int) {
	cnt := map[int]int{}
	for _, x := range nums1 {
		cnt[x]++
	}
	for _, x := range nums2 {
		if cnt[x] > 0 {
			ans = append(ans, x)
			cnt[x]--
		}
	}
	return
}

TypeScript

function intersect(nums1: number[], nums2: number[]): number[] {
    const cnt: Record<number, number> = {};
    for (const x of nums1) {
        cnt[x] = (cnt[x] || 0) + 1;
    }
    const ans: number[] = [];
    for (const x of nums2) {
        if (cnt[x]-- > 0) {
            ans.push(x);
        }
    }
    return ans;
}

Rust

use std::collections::HashMap;

impl Solution {
    pub fn intersect(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
        let mut cnt = HashMap::new();
        for &x in &nums1 {
            *cnt.entry(x).or_insert(0) += 1;
        }
        let mut ans = Vec::new();
        for &x in &nums2 {
            if let Some(count) = cnt.get_mut(&x) {
                if *count > 0 {
                    ans.push(x);
                    *count -= 1;
                }
            }
        }
        ans
    }
}

JavaScript

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

C#

public class Solution {
    public int[] Intersect(int[] nums1, int[] nums2) {
        Dictionary<int, int> cnt = new Dictionary<int, int>();
        foreach (int x in nums1) {
            if (cnt.ContainsKey(x)) {
                cnt[x]++;
            } else {
                cnt[x] = 1;
            }
        }
        List<int> ans = new List<int>();
        foreach (int x in nums2) {
            if (cnt.ContainsKey(x) && cnt[x] > 0) {
                ans.Add(x);
                cnt[x]--;
            }
        }
        return ans.ToArray();
    }
}

PHP

class Solution {
    /**
     * @param Integer[] $nums1
     * @param Integer[] $nums2
     * @return Integer[]
     */
    function intersect($nums1, $nums2) {
        $cnt = [];
        foreach ($nums1 as $x) {
            if (isset($cnt[$x])) {
                $cnt[$x]++;
            } else {
                $cnt[$x] = 1;
            }
        }

        $ans = [];
        foreach ($nums2 as $x) {
            if (isset($cnt[$x]) && $cnt[$x] > 0) {
                $ans[] = $x;
                $cnt[$x]--;
            }
        }

        return $ans;
    }
}