How do you solve Range Frequency Queries on LeetCode?

Range Frequency Queries (LeetCode #2080) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + q log k) time complexity and O(n) space complexity. Key insight: Using a hashmap allows for quick access to indices, making queries efficient.

What is the brute force solution for Range Frequency Queries?

The brute force approach for Range Frequency Queries is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks every element in the specified subarray to count occurrences of the target value. This is straightforward but inefficient for large arrays or many queries. The optimal Optimal Solution approach improves this to O(n + q log k).

What are the interview tips for Range Frequency Queries?

Always discuss your thought process and potential optimizations during the interview. Consider the constraints and how they affect your approach. Practice explaining your code clearly, as communication is key in interviews.

What are common mistakes when solving Range Frequency Queries?

Not considering edge cases where the value does not exist in the array. Failing to optimize for multiple queries, leading to time limit exceeded errors.

#2080

Range Frequency Queries

Medium

ArrayHash TableBinary SearchDesignSegment TreeHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + q log k)
Space	O(1)	O(n)

💡

Intuition

Time O(n + q log k)Space O(n)

By using a hashmap to store the indices of each value, we can quickly determine how many times a value appears in any subarray using binary search. This drastically reduces the time complexity for each query.

⚙️

Algorithm

3 steps

1Step 1: Create a hashmap where keys are the values from the array and values are lists of indices where these values occur.
2Step 2: For each query, retrieve the list of indices for the target value.
3Step 3: Use binary search to find the count of indices that fall within the specified left and right bounds.

solution.py16 lines

1from collections import defaultdict
2import bisect
3
4class RangeFreqQuery:
5    def __init__(self, arr):
6        self.index_map = defaultdict(list)
7        for i, num in enumerate(arr):
8            self.index_map[num].append(i)
9
10    def query(self, left, right, value):
11        if value not in self.index_map:
12            return 0
13        indices = self.index_map[value]
14        left_index = bisect.bisect_left(indices, left)
15        right_index = bisect.bisect_right(indices, right)
16        return right_index - left_index

ℹ

Complexity note: The time complexity is O(n) for preprocessing the array into a hashmap and O(log k) for each query, where k is the number of occurrences of the value. This is efficient for multiple queries.

1Using a hashmap allows for quick access to indices, making queries efficient.
2Binary search is a powerful tool for finding boundaries in sorted data.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.