Intersection Of Two Arrays Problem Statement

Given two lists of numbers, find their intersection.

Example One

{
"a": [4, 2, 2, 3, 1],
"b": [2, 2, 2, 3, 3]
}

Output:

[2, 2, 3]

Example Two

{
"a": [6, 2, 4],
"b": [1, 5, 3, 7]
}

Output:

[]

Notes

• The order of elements in the output list does not matter.
• The frequency of any number in the intersection must be equal to the minimum of the frequency of that number in both the given lists.

Constraints:

• 1 <= size of the input lists <= 10⁵
• -10⁶ <= any element of the input lists <= 10⁶

Code For Intersection Of Two Arrays Solution 1: List

"""
Asymptotic complexity in terms of the length of the given list `a` and `b`:
* Time: O(n * m).
* Auxiliary space: O(m).
* Total space: O(n + m).
"""

def get_intersection_with_maintained_frequency(a, b):
    """
    Args:
     a(list_int32): First list of integers.
     b(list_int32): Second list of integers.
    Returns:
     list_int32: List of integers representing the intersection with maintained frequency.
    """
    # Initialize a list to store the intersection.
    intersection = []

    # Copy list b to avoid mutating it while iterating.
    b_copy = b.copy()

    # Iterate over each element in list a.
    for element in a:
        # If the element is in list b, append it to the intersection and remove it from the copy of list b.
        if element in b_copy:
            intersection.append(element)
            b_copy.remove(element)

    # Return the intersection list.
    return intersection

Code For Intersection Of Two Arrays Solution 2: Dictionary

"""
Asymptotic complexity in terms of the length of the given list `a` and `b`:
* Time: O(n + m).
* Auxiliary space: O(m).
* Total space: O(n + m).
"""

def get_intersection_with_maintained_frequency(a, b):
    """
    Args:
     a(list_int32): First list of integers.
     b(list_int32): Second list of integers.
    Returns:
     list_int32: List of integers representing the intersection with maintained frequency.
    """

    # Create a dictionary for each list
    dict_a = {}
    for i in a:
        if i in dict_a:
            dict_a[i] += 1
        else:
            dict_a[i] = 1

    dict_b = {}
    for i in b:
        if i in dict_b:
            dict_b[i] += 1
        else:
            dict_b[i] = 1

    # Determine the length of each dictionary
    len_dict_a = len(dict_a)
    len_dict_b = len(dict_b)

    # Identify the smallest and largest dictionaries
    if len_dict_a < len_dict_b:
        smallest_dict = dict_a
        largest_dict = dict_b
    else:
        smallest_dict = dict_b
        largest_dict = dict_a

    # Initialize the intersection list
    intersection = []

    # Iterate through the items in the smallest dictionary
    for value, count_small in smallest_dict.items():
        # Check if the value is in the largest dictionary
        if value in largest_dict:
            # Get the count from the largest dictionary
            count_large = largest_dict[value]
            # Add the value to the intersection list, maintaining the frequency
            intersection.extend([value] * min(count_small, count_large))

    # Return the intersection list
    return intersection

Code For Intersection Of Two Arrays Solution 3: Hashmap

/*
Asymptotic complexity in terms of the length of list `a` (= `n`) and the length of list `b` (= `m`):
* Time: O(n + m).
* Auxiliary space: O(min(n, m)).
* Total space: O(n + m).
*/
vector<int> get_intersection_with_maintained_frequency(vector<int> &a, vector<int> &b) {

    if (a.size() > b.size()) {
        return get_intersection_with_maintained_frequency(b, a);
    }
    unordered_map<int, int> frequency_a;
    for (int i = 0; i < a.size(); i++) {
        frequency_a[a[i]]++;
    }

    vector<int> result;
    for (int i = 0; i < b.size(); i++) {
        if (frequency_a[b[i]] > 0) {
            result.push_back(b[i]);
            frequency_a[b[i]]--;
        }
    }
    return result;
}

We hope that these solutions to intersection of two arrays problem have helped you level up your coding skills. You can expect problems like these at top tech companies like Amazon and Google.

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