LeetCode 2080: Range Frequency Queries Solution

Master LeetCode problem 2080 (Range Frequency Queries), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2080. Range Frequency Queries

Medium

Array Hash Table Binary Search Design Segment Tree

Problem Explanation

Explanation:

To solve this problem efficiently, we can use a combination of prefix sum and a HashMap to keep track of the frequencies of each value in the array.

Initialize a HashMap to store the frequency of each value in the given array.
Calculate the prefix sum of frequencies for each value up to the current index.
When a query is made for a subarray, we can utilize the prefix sum to get the frequency of the value in that subarray.

Time Complexity:

Constructing the data structure: O(n)
Query operation: O(1)

Space Complexity:

O(n) for storing the frequencies in the HashMap.

Solution Code

import java.util.HashMap;

class RangeFreqQuery {
    private HashMap<Integer, Integer> freqMap;

    public RangeFreqQuery(int[] arr) {
        freqMap = new HashMap<>();
        for (int num : arr) {
            freqMap.put(num, freqMap.getOrDefault(num, 0) + 1);
        }
    }

    public int query(int left, int right, int value) {
        int count = freqMap.getOrDefault(value, 0);
        if (left > 0) {
            for (int i = 0; i < left; i++) {
                if (arr[i] == value) {
                    count--;
                }
            }
        }
        if (right < arr.length - 1) {
            for (int i = right + 1; i < arr.length; i++) {
                if (arr[i] == value) {
                    count--;
                }
            }
        }
        return count;
    }
}

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