While a hash map is a common solution, the Two Pointer approach is extremely efficient if the arrays are sorted. We sort both arrays and use two pointers to find common elements.

1. Sort nums1 and nums2.
2. Initialize i = 0, j = 0, res = [].
3. While i < len(nums1) and j < len(nums2):
   • If nums1[i] < nums2[j], increment i.
   • Else if nums1[i] > nums2[j], increment j.
   • Else:
     • Append nums1[i] to res.
     • Increment i and j.
4. Return res.

• Time Complexity: O(N log N + M log M) due to sorting.
• Space Complexity: O(N + M).

def intersect(nums1, nums2):
    nums1.sort()
    nums2.sort()
    
    i = j = 0
    res = []
    
    while i < len(nums1) and j < len(nums2):
        if nums1[i] < nums2[j]:
            i += 1
        elif nums1[i] > nums2[j]:
            j += 1
        else:
            res.append(nums1[i])
            i += 1
            j += 1
            
    return res