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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`

` ````
```class Solution {
public:
vector> findDifference(vector& nums1, vector& nums2) {
// Create unordered sets to store unique elements from nums1 and nums2
unordered_set set1(nums1.begin(), nums1.end());
unordered_set set2(nums2.begin(), nums2.end());
// Initialize a 2D vector to store the differences (elements in nums1 not in nums2 and vice versa)
vector> answer(2);
// Iterate through unique elements in set1
for (int num : set1) {
// If the element is not present in set2, add it to the first vector of differences (elements in nums1 not in nums2)
if (set2.count(num) == 0) {
answer[0].push_back(num);
}
}
// Iterate through unique elements in set2
for (int num : set2) {
// If the element is not present in set1, add it to the second vector of differences (elements in nums2 not in nums1)
if (set1.count(num) == 0) {
answer[1].push_back(num);
}
}
// Return the 2D vector containing the differences
return answer;
}
};

In this program, we have to find the difference between two sets represented by vectors (‘nums1’ and ‘nums2’). The sets are implemented using unordered sets.

An unordered set (std::unordered_set) is a container that contains a set of unique elements. It uses hashing to achieve constant time-average complexity for insertions and deletions.

Let’s see the step-by-step approach to the solution:

- Let’s make two unordered sets ‘set1’ and ‘set2’ both of ‘int’ type in which they contain all the elements of ‘nums1’ and ‘nums2’ from beginning to end.

- We have to make a 2D vector ( since it is given that we have to return a 2D array) ‘answer’ with two vectors inside it ‘answer[ 0 ]’ will store the elements that are in ‘nums1’ but not in ‘nums2’, and ‘answer[ 1 ] ‘ will store the elements that are in ‘nums2’ but not in ‘nums1’.

- Now, we will start the iteration through the elements of ‘set1’ and check if each element is not present in ‘set2’. If they are not present, add it to ‘answer[ 0 ]’.

- Similarly, we will iterate through the elements of ‘set2’ and check if each element is not present in ‘set1’. If not present, add it to ‘answer[ 1]’.

- Now ‘push_back’ function is used to add elements to vectors within a 2D vector (‘answer’).

Finally, return ’ answer ‘.

**Summary:-** the solution used unordered sets to find and store the differences between two sets represented by vectors. The use of unordered sets ensures that the resulting differences are unique.

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