380. Insert Delete GetRandom O(1) [Leetcode][C++]

mayankdv

Mayank Arora

Posted on December 23, 2021

380. Insert Delete GetRandom O(1) [Leetcode][C++]

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Leetcode Problem Link: 380. Insert Delete GetRandom O(1)


Approach:

The order of elements is not important in the data structure. If we use an array as the data structure, insertion at the end will be in O(1) time but searching whether the element is already present in the array will take O(N) time. Removal of the element will require searching for the element which will take O(N) time and left shifting the subsequent array elements will also take O(N) time. Getting random element will take O(1) time using the rand() function. So, we need some additional data structure that allows us to locate the array index for removal and also to search whether the element is already present or not. This data structure should do these tasks in O(1) time. Hashmap fulfils our requirements.

C++ Code:

class RandomizedSet {
    // Optimal Solution Time O(1) & Auxiliary Space O(N)
private:
    vector<int> a; // array vector
    unordered_map<int,int> m; // Unordered Map does searching, insertion & deletion of element in O(1) time
public:
    /** Initialize your data structure here. */
    RandomizedSet() {
    // No need to initialise a & m as they are initialised automatically
    // to 0 as and when their container size increases.
    }

    /** Inserts a value to the array vector. Returns true if the array did not already contain the specified element. */
    bool insert(int val) {
        if(m.find(val)!=m.end())
            // If val is not already present in the map, find() function  
            // returns an iterator(m.end()) pointing to next memory location  
            // of the last element of the map. Otherwise, find() returns an iterator 
            // pointing to val which was already present in the map.  
            return false;
        else{
            a.push_back(val);  // insert val at the end of the array
            m[val]=a.size()-1; // m[key,value] stores the array element and 
                               // its index as key=array element & value=array element index
            return true;
        }
    }

    /** Removes a value from the array vector. Returns true if the array contained the specified element. */
    bool remove(int val) {
        if(m.find(val)==m.end()) // val not present in the array vector
            return false;
        else{
            // val present in the array vector
            // For example: a=[8,4,3,2], m={[8,0],[4,1],[3,2],[2,3]}, val=4, last=2
            // After a[m[val]]=a.back(); a=[8,2,3,2], m={[8,0],[4,1],[3,2],[2,3]}
            // After a.pop_back(); a=[8,2,3], m={[8,0],[4,1],[3,2],[2,3]}
            // After m[last]=m[val]; a=[8,2,3], m={[8,0],[4,1],[3,2],[2,1]}
            // After m.erase(val); a=[8,2,3], m={[8,0],[3,2],[2,1]}
            int last=a.back();  // back() fetches last element of the array vector
            a[m[val]]=a.back(); // m[val] locates the index of val in the array vector.
                                // Then we copy array last element value to the val location in the array
            a.pop_back();       // Delete the last element of the array 
            m[last]=m[val];     // In hashmap, assign index of val in array to the index of the last element   
            m.erase(val);       // Delete the val entry from map
            return true;
        }
    }

    /** Get a random element from the array vector */
    int getRandom() {
        // rand() function gives random value in the range of 0 to RAND_MAX(whose value is 32767). x%y gives 
        // remainder when x is divided by y and this remainder is in the range of 0 to y-1.
        // rand()%a.size() gives random value in the range of (0 to a.size()-1).
        // a[rand()%a.size()] will give random value of array in the range of a[0] to a[a.size()-1].
        return a[rand()%a.size()];
    }
};
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All suggestions are welcome. Please upvote if you like it. Thank you.

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mayankdv
Mayank Arora

Posted on December 23, 2021

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