2215. Find the Difference of Two Arrays

Description of Problem

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] = [].

Constraints:

  • 1 <= nums1.length, nums2.length <= 1000
  • -1000 <= nums1[i], nums2[i] <= 1000

Code (Rust) - Use Rust built-in functions

impl Solution {
    pub fn find_difference(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<Vec<i32>> {
        use std::collections::HashSet;
        use std::iter::FromIterator;
        let a : HashSet<_> = nums1.into_iter().collect();
        let b : HashSet<_> = nums2.into_iter().collect();
        let left_diff: Vec<_> = a.difference(&b).map(|&x| x).collect();
        let rigth_diff: Vec<_> = b.difference(&a).map(|&x| x).collect();
        return vec![left_diff, rigth_diff];
    }
}

Complexity

  • \(m\) is nums1.len()
  • \(n\) is nums2.len()

Time Complexity

  • \(T(m,n) = O(m+n)\)
    • Create HashSet for both arrays.
    • Create left-difference list and right-difference list

Auxiliary Space

  • \(S(m,n) = O(m+n)\)
    • Create HashSet for both arrays.