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, 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

    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