carloscn
commented
1 year ago Analysis
pub fn find_difference(nums1: Vec<i32>, nums2: Vec<i32>) -> Vec<Vec<i32>>
{
let mut ret:Vec<Vec<i32>> = vec![];
if nums1.is_empty() || nums2.is_empty() {
return ret;
}
let mut v1:Vec<i32> = vec![];
for e in &nums1 {
if !nums2.contains(e) && !v1.contains(e) {
v1.push(*e);
}
}
ret.push(v1);
let mut v2:Vec<i32> = vec![];
for e in &nums2 {
if !nums1.contains(e) && !v2.contains(e){
v2.push(*e);
}
}
ret.push(v2);
return ret;
}
Description
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