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

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]

Note:

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

 

Not good enough approach

var intersect = function(nums1, nums2) {
    let r = [];
    // get larger array
    const len1 = nums1.length;
    const len2 = nums2.length;
    
    // larger & smaller
    const larger = len1 > len2 ? nums1: nums2;
    const smaller = len1 > len2 ? nums2: nums1;
    
    // conver larger array to object
    let hashed = {};
    for (let n of larger) {
        if (n in hashed) {
            hashed[n]++;
        } else {
            hashed[n] = 1;
        }
    }
 
    // loop over smaller array
    for (let n of smaller) {
        if (`${n}` in hashed) {
            r.push(n);
            hashed[n] = hashed[n]-1;
            if (hashed[n] === 0) {
                delete hashed[n];
            }
        }
    }
    
    return r;
};

 

The reason that code above is not good enough is because, \

1. we use larger array as lookup, this cause more memory usage. -- actually we need to use smaller array as lookup

2. we use 'len1, len2, samller, larger' extra storage, we can actully swap nums1 and nums by one extra function call.

var intersect = function(nums1, nums2) {

    if (nums1.length > nums2.length) {
        return intersect(nums2, nums1);
    }
    
    // conver samller array to object
    let hashed = {};
    for (let n of nums2) {
        if (n in hashed) {
            hashed[n]++;
        } else {
            hashed[n] = 1;
        }
    }
    
    let r = [];
    // loop over smaller array
    for (let n of nums1) {
        if (hashed[n] > 0) {
            r.push(n);
            hashed[n] = hashed[n]-1;
        }
    }
    
    return r;
}

 

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?

Will write another post for the follow up questions.