350, Intersection of Two Arrays II

I Problem

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?

Related Topics

• Array
• Hash Table
• Two Pointers
• Binary Search
• Sorting

II Solution

Approach 1: Brute Force

pub fn intersect(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
    let mut res = vec![];

    for i in 0..nums1.len() {
        for j in 0..nums2.len() {
            if nums1[i] == nums2[j] {
                res.push(nums1[i]);
                nums2[j] = i32::MIN;
                break;
            }
        }
    }

    res
}

public int[] intersect(int[] nums1, int[] nums2) {
    List<Integer> res = new ArrayList<>();

    for (int num1 : nums1) {
        for (int j = 0; j < nums2.length; j++) {
            if (num1 == nums2[j]) {
                res.add(num1);
                nums2[j] = Integer.MIN_VALUE;
                break;
            }
        }
    }

    return res.stream().mapToInt(Integer::intValue).toArray();
}

Approach 2: Use Hash

pub fn intersect(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
    let mut map =
        nums1
            .into_iter()
            .fold(HashMap::with_capacity(nums2.len()), |mut map, num| {
                map.entry(num).and_modify(|v| *v += 1).or_insert(1);
                map
            });

    nums2
        .into_iter()
        .filter(|num| match map.get_mut(num) {
            None => false,
            Some(v) => {
                let count = *v;
                *v -= 1;
                count > 0
            }
        })
        .collect()
}

public int[] intersect(int[] nums1, int[] nums2) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int num: nums1) {
        map.put(num, map.getOrDefault(num, 0) + 1);
    }

    return Arrays.stream(nums2).filter(num -> {
        Integer count = map.getOrDefault(num, 0);
        map.put(num, count - 1);
        return count > 0;
    }).toArray();
}

Approach 3: Sorting + Two Pointers

pub fn intersect(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<i32> {
    nums1.sort_unstable();
    nums2.sort_unstable();

    let mut res = vec![];
    let mut i1 = 0;
    let mut i2 = 0;
    while i1 < nums1.len() && i2 < nums2.len() {
        if nums1[i1] > nums2[i2] {
            i2 += 1;
        } else if nums1[i1] < nums2[i2] {
            i1 += 1;
        } else {
            res.push(nums1[i1]);
            i1 += 1;
            i2 += 1;
        }
    }

    res
}

public int[] intersect(int[] nums1, int[] nums2) {
    Arrays.sort(nums1);
    Arrays.sort(nums2);

    List<Integer> res = new ArrayList<>();
    int i1 = 0, i2 = 0;
    while (i1 < nums1.length && i2 < nums2.length) {
        if (nums1[i1] < nums2[i2]) {
            i1++;
        } else if (nums1[i1] > nums2[i2]) {
            i2++;
        } else {
            res.add(nums1[i1]);
            i1++;
            i2++;
        }
    }

    return res.stream().mapToInt(Integer::intValue).toArray();
}