382. Linked List Random Node 🌟🌟

Given a singly linked list, return a random node’s value from the linked list. Each node must have the same probability of being chosen.

Implement the Solution class:

• Solution(ListNode* head) Initializes the object with the integer array nums.
• int getRandom() Chooses a node randomly from the list and returns its value. All the nodes of the list should be equally likely to be chosen.

Naive Solution

• We can convert linked list into an array and return a random element from the array.
• TC: O(N)
• SC: O(N)

Reservoir Sampling Solution

• In order to do random sampling over a population of unknown size with constant space, the answer is reservoir sampling.

  The reservoir sampling algorithm is intended to sample k elements from an population of unknown size. In our case, the k happens to be one.

Code

class Solution {
private:
    ListNode* root;

public:
    Solution(ListNode* head){
        this->root = head;
    }

    int getRandom()
    {
        ListNode* curr = root;
        int ans = 0, x = 1;
        while (curr) {
            if (rand() % x == 0) ans = curr->val;
            x++;
            curr = curr->next;
        }
        return ans;
    }
};

• rand() output has range of [0,RAND_MAX) so the i variable here is kinda like curr_depth+1 so so at each step its [0,depth+1)