Instagram link:- / reelcoding
/ @reelcoding
Approach 1:-
Use a HashMap to store the count of each element in the first array (nums1).
Traverse the second array (nums2), and for each element, check if it exists in the HashMap and has a count greater than 0.
If it does, add the element to the result list and decrement its count in the HashMap.
Finally, convert the result list to an array and return it.
Time and Space Complexity:
Time Complexity:
Building the HashMap: O(n) where n is the length of nums1.
Traversing nums2 and building the result list: O(m) where m is the length of nums2.
Overall time complexity: O(n + m).
Space Complexity:
The HashMap requires O(n) space to store the counts of elements in nums1.
The result list requires O(min(n, m)) space in the worst case, where all elements intersect.
Overall space complexity: O(n + min(n, m)).
Approach 2:-
Sort both arrays.
Use two pointers to traverse through both arrays simultaneously.
Compare elements at both pointers:
If the element in nums1 is smaller, move the pointer in nums1.
If the element in nums2 is smaller, move the pointer in nums2.
If elements are equal, add the element to the result list and move both pointers.
Convert the result list to an array and return it.
Time and Space Complexity:
Time Complexity:
Sorting both arrays: O(n log n + m log m) where n is the length of nums1 and m is the length of nums2.
Traversing both arrays: O(n + m).
Overall time complexity: O(n log n + m log m).
Space Complexity:
The result list requires O(min(n, m)) space in the worst case, where all elements intersect.
Overall space complexity: O(min(n, m)).
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