# Solving Leetcode Interviews in Seconds with AI: Range Frequency Queries


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2080" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Design a data structure to find the frequency of a given value in a given subarray. The frequency of a value in a subarray is the number of occurrences of that value in the subarray. Implement the RangeFreqQuery class:  RangeFreqQuery(int[] arr) Constructs an instance of the class with the given 0-indexed integer array arr. int query(int left, int right, int value) Returns the frequency of value in the subarray arr[left...right].  A subarray is a contiguous sequence of elements within an array. arr[left...right] denotes the subarray that contains the elements of nums between indices left and right (inclusive).   Example 1:  Input ["RangeFreqQuery", "query", "query"] [[[12, 33, 4, 56, 22, 2, 34, 33, 22, 12, 34, 56]], [1, 2, 4], [0, 11, 33]] Output [null, 1, 2]  Explanation RangeFreqQuery rangeFreqQuery = new RangeFreqQuery([12, 33, 4, 56, 22, 2, 34, 33, 22, 12, 34, 56]); rangeFreqQuery.query(1, 2, 4); // return 1. The value 4 occurs 1 time in the subarray [33, 4] rangeFreqQuery.query(0, 11, 33); // return 2. The value 33 occurs 2 times in the whole array.    Constraints:  1 <= arr.length <= 105 1 <= arr[i], value <= 104 0 <= left <= right < arr.length At most 105 calls will be made to query  

	# Explanation
	Here's a solution to the Range Frequency Query problem:

*   **Preprocessing:** Store the indices of each unique value in the input array using a hash map (dictionary). This allows us to quickly find all occurrences of a given value.
*   **Binary Search:** For each query, use binary search on the sorted list of indices for the given value to find the first index greater than or equal to `left` and the last index less than or equal to `right`.
*   **Frequency Calculation:** The frequency is then the difference between these two indices plus one.

*   **Runtime Complexity:** O(n + q * log n), where n is the length of the array and q is the number of queries. Preprocessing takes O(n) time. Each query takes O(log n) time due to binary search.  **Storage Complexity:** O(n), where n is the length of the array. We store indices for each value.

	
	# Code
	```python
	from collections import defaultdict
import bisect

class RangeFreqQuery:

    def __init__(self, arr: list[int]):
        self.indices = defaultdict(list)
        for i, num in enumerate(arr):
            self.indices[num].append(i)

    def query(self, left: int, right: int, value: int) -> int:
        if value not in self.indices:
            return 0

        indices_of_value = self.indices[value]
        
        left_index = bisect.bisect_left(indices_of_value, left)
        right_index = bisect.bisect_right(indices_of_value, right)

        return right_index - left_index
	```