350. Intersection of Two Arrays II

Posted 2022-08-03 wx62ea2466cca9a

Given two arrays, write a function to compute their intersection.

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:
Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.
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?

public class Solution 
    public int[] intersect(int[] nums1, int[] nums2)
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        ArrayList result = new ArrayList();
        for (int i = 0, j = 0; i < nums1.length && j < nums2.length; )
            if (nums1[i] == nums2[j])
                result.add(nums1[i]);
                i++;
                j++;
             else if (nums1[i] < nums2[j]) 
                i++;
             else 
                j++;
            
        
        int[] res = new int[result.size()];
        for (int i = 0; i < result.size(); i++)
            res[i] = (int) result.get(i);
        
        return res;
    

/**
 * @param number[] nums1
 * @param number[] nums2
 * @return number[]
 */
var intersect = function(nums1, nums2) 
    if (!nums1.length || !nums2.length) 
        return []
    
    const map1 = new Map()
    const res = []
    nums1.forEach(num => 
        if (map1.has(num)) 
            map1.set(num, map1.get(num) + 1)
         else 
            map1.set(num, 1)
        
    )
    nums2.forEach(num => 
        if (map1.has(num) && map1.get(num) > 0) 
            res.push(num)
            map1.set(num, map1.get(num) - 1)
        
    )
    return res
;