The problem can be found at the following link: Question Link
- We create an unordered set to store all elements of array b.
- Then, we iterate through array a. For each element in a, if it's not present in set b, we push it into the output vector.
- Finally, we return the output vector containing the missing elements.
- Time Complexity: O(n + m), where n is the size of array a and m is the size of array b.
- Auxiliary Space Complexity: O(m), where m is the size of array b.
class Solution {
public:
vector<int> findMissing(int a[], int b[], int n, int m)
{
unordered_set<int> s;
vector<int> out;
for(int i = 0; i < m; ++i)
s.insert(b[i]);
for(int i = 0; i < n; ++i)
if(!s.count(a[i]))
out.push_back(a[i]);
return out;
}
};
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