17. Count Pairs whose sum is equal to X

The problem can be found at the following link: Question Link

My Approach

To solve this problem, I utilize a hash set to store the elements of the first linked list. Then, I iterate through the second linked list, checking if the difference between the target sum and the current element exists in the hash set. If it does, I increment the count of pairs. Finally, I return the count of pairs found.

Time and Auxiliary Space Complexity

• Time Complexity: O(n + m), where n is the number of elements in the first linked list and m is the number of elements in the second linked list.

• Auxiliary Space Complexity: O(n), where n is the number of elements in the first linked list.

Code (C++)

class Solution {
public:
    int countPairs(struct Node* head1, struct Node* head2, int x) {
        unordered_set<int> st;
        while (head1) {
            st.insert(head1->data);
            head1 = head1->next;
        }
        int cnt = 0;
        while (head2) {
            if (st.find(x - head2->data) != st.end())
                ++cnt;
            head2 = head2->next;
        }
        return cnt;
    }
};

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