2215. Find the Difference of Two Arrays Easy

Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

• answer[0] is a list of all distinct integers in nums1 which are not present in nums2.
• answer[1] is a list of all distinct integers in nums2 which are not present in nums1.

Note that the integers in the lists may be returned in any order.

Example 1:
Input: nums1 = [1,2,3], nums2 = [2,4,6]
Output: [[1,3],[4,6]]
Explanation:
For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].
For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].

Example 2:
Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
Output: [[3],[]]
Explanation:
For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].
Every integer in nums2 is present in nums1. Therefore, answer[1] = [].

Approach

Input: Two integer arrays nums1, nums2

Output: Find the numbers that do not intersect between the two arrays, return a list answer of length 2

This is a Hash Table problem.

We can use set to remove duplicates from the arrays and store them: s1, s2 = set(nums1), set(nums2).

Separately iterate through these two sets. As long as the iterated value does not appear in the other set, it meets our criteria.

Implementation

python

from typing import List

class Solution:
    def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]:
        # Convert the two arrays into sets to remove duplicates and make existence checks fast
        s1, s2 = set(nums1), set(nums2)

        # Store elements unique to nums1
        only1 = []
        for n1 in s1:
            if n1 not in s2:          # If n1 is not in s2, it means it's unique to nums1
                only1.append(n1)

        # Store elements unique to nums2
        only2 = []
        for n2 in s2:
            if n2 not in s1:          # If n2 is not in s1, it means it's unique to nums2
                only2.append(n2)
        
        # Return the result in the format: [list of elements unique to nums1, list of elements unique to nums2]
        return [only1, only2]

javascript

/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[][]}
 */
var findDifference = function(nums1, nums2) {
    const s1 = new Set(nums1);
    const s2 = new Set(nums2);

    const only1 = [];
    for (let x of s1) {
        if (!s2.has(x)) only1.push(x);
    }

    const only2 = [];
    for (let x of s2) {
        if (!s1.has(x)) only2.push(x);
    }

    return [only1, only2];
};

Complexity Analysis

• Time Complexity: O(n + m) where n is length of nums1 and m is length of nums2
• Space Complexity: O(n + m) for the two hash sets

Links

2215. Find the Difference of Two Arrays (English)2215. 找出两数组的不同 (Chinese)