Update: 2215 - Find the Difference of Two Arrays | Space complexity

Author: AHTHneeuhl

cpp/2215-find-the-difference-of-two-arrays.cpp

@@ -1,5 +1,5 @@
-// Time Complexity: O(n), where n is the maximum length between nums1 and nums2.
-// Space Complexity: O(m), where m is the length of the resulting difference vectors.
+// Time Complexity: O(m + n), we check each element of nums1Set and nums2Set
+// Space Complexity: O(m + n), where m and n are length sets in worst case.
 
 class Solution
 {

javascript/2215-find-the-difference-of-two-arrays.js

@@ -4,8 +4,8 @@
  * @return {number[][]}
  */
 
-// Time Complexity: O(n), where n is the maximum length between nums1 and nums2.
-// Space Complexity: O(m), where m is the length of the resulting difference lists.
+// Time Complexity: O(m + n), we check each element of nums1Set and nums2Set
+// Space Complexity: O(m + n), where m and n are length sets in worst case.
 
 var findDifference = function (nums1, nums2) {
     const nums1Set = new Set(nums1);

python/2215-find-the-difference-of-two-arrays.py

@@ -1,5 +1,5 @@
-# Time Complexity: O(n), where n is the maximum length between nums1 and nums2.
-# Space Complexity: O(m), where m is the length of the resulting difference lists.
+# Time Complexity: O(m + n), we check each element of nums1Set and nums2Set
+# Space Complexity: O(m + n), where m and n are length sets in worst case.
 
 from typing import List  # ignore this, just for typing
 
@@ -8,6 +8,8 @@ class Solution:
     def findDifference(self, nums1: List[int], nums2: List[int]) -> List[List[int]]:
         nums1_set = set(nums1)
         nums2_set = set(nums2)
+
         lst1 = [num for num in nums1_set if num not in nums2_set]
         lst2 = [num for num in nums2_set if num not in nums1_set]
+
         return [lst1, lst2]

typescript/2215-find-the-difference-of-two-arrays.ts

@@ -1,5 +1,5 @@
-// Time Complexity: O(n), where n is the maximum length between nums1 and nums2.
-// Space Complexity: O(m), where m is the length of the resulting difference lists.
+// Time Complexity: O(m + n), we check each element of nums1Set and nums2Set
+// Space Complexity: O(m + n), where m and n are length sets in worst case.
 
 function findDifference(nums1: number[], nums2: number[]): number[][] {
     const nums1Set: Set<number> = new Set(nums1);