[LeetCode][380. Insert Delete GetRandom O(1)] Data Structures: Thought Process from HashMap to HashMap + Array

By Long Luo

This article is the solution Data Structures: Thought Process from HashMap to HashMap + Array of Problem 380. Insert Delete GetRandom O(1).

Intuition

It’s easy to think of using a Hash Table to achieve $O(1)$ time complexity for $\texttt{insert}$ and $\texttt{remove}$ operations. However, we need $O(1)$ time complexity to complete the $\texttt{getRandom}$ operation.

The Array structure can complete the operation of obtaining random elements in $O(1)$ time complexity, but it can’t completed the $\texttt{insert}$ and $\texttt{remove}$ operations in $O(1)$ time complexity.

So How?

Aha!!!

We can combining the $\texttt{HashMap}$ and Array. The Array stores the elements, and the $\texttt{HashMap}$ stores the input value as the $\texttt{key}$ and the array subscript $\texttt{loc}$ as the value.

We need to apply for an array nums of sufficient size (the data range is $2 \times 10^5$ ), and a variable $\texttt{idx}$ to record how many space is currently used.

• $\texttt{insert}$: Judge whether the $\texttt{HashMap}$ contains the $\texttt{val}$, if not, insert the $\texttt{val}$ to the $\texttt{HashMap}$ and update the Array nums.

• $\texttt{remove}$: Judge whether the $\texttt{HashMap}$ contains the $\texttt{val}$, if not, get the location loc of the input val, replace the loc with the last value of the array, remove the $\texttt{val}$ from $\texttt{HashMap}$ and update the $\texttt{idx}$.

• $\texttt{getRandom}$: Just select a subscript at random and return the element of the array.

class RandomizedSet {
    Map<Integer, Integer> map;
    int[] nums;
    int idx;
    Random random;

    public RandomizedSet() {
        map = new HashMap<>();
        nums = new int[2_00_002];
        idx = 0;
        random = new Random();
    }

    public boolean insert(int val) {
        if (map.containsKey(val)) {
            return false;
        }

        nums[idx] = val;
        map.put(val, idx);
        idx++;
        return true;
    }

    public boolean remove(int val) {
        if (map.containsKey(val)) {
            int loc = map.get(val);
            map.remove(val);

            if (loc != idx - 1) {
                nums[loc] = nums[idx - 1];
                map.put(nums[idx - 1], loc);
            }

            idx--;
            return true;
        }

        return false;
    }

    public int getRandom() {
        int loc = random.nextInt(idx);
        return nums[loc];
    }
}

Analysis

• Time Complexity: $O(1)$.
• Space Complexity: $O(n)$.

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