#2080

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Range Frequency Queries

Design a data structure to handle efficient frequency queries for subarrays, focusing on hash-based lookups and efficient queries.

array hash table binary search design segment tree

Problem Statement

You are tasked with designing a data structure that supports querying the frequency of a specific value in a given subarray. The data structure should allow for efficient querying of the frequency of a value in any subarray specified by two indices, left and right.

The data structure must support the RangeFreqQuery class, which allows the following operations: initialize the data structure with an array of integers, and perform range queries where you ask for the frequency of a given value within a specified subarray range.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

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

Solution Approach

Brute-force Array Scanning

In this approach, you scan the array for each query and count the occurrences of the value within the given range. This simple solution is straightforward but inefficient for larger inputs, potentially leading to time-limit exceeded errors if used excessively.

Hash Map Preprocessing

A more efficient approach involves preprocessing the array by creating a hash map where the key is the array value and the value is a list of indices where that value occurs. This allows you to quickly count occurrences of the value in any subarray using binary search, which improves query performance.

Segment Tree with Hash Map

For even better performance, a segment tree combined with a hash map can be used. The segment tree allows for fast range queries, and the hash map stores the frequency of each value within each segment. This approach ensures optimal query times even for large arrays and frequent queries.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the approach used. Brute-force scanning results in O(n) per query. Using hash maps with binary search reduces it to O(log n) per query after O(n) preprocessing. The segment tree approach typically allows O(log n) queries after O(n log n) preprocessing. The space complexity also varies based on the preprocessing approach, with the segment tree and hash map solution requiring more space compared to brute-force.

What Interviewers Usually Probe

If the candidate suggests brute force, assess their awareness of time constraints and large input sizes.
Look for familiarity with hash-based data structures and their use in optimizing range queries.
Evaluate whether the candidate is comfortable with segment trees and understands their role in optimizing query performance.

Common Pitfalls or Variants

Common pitfalls

Using brute-force solutions without considering time limits will result in inefficiency for larger test cases.
Not handling edge cases, such as when the subarray has only one element or when the value isn't present at all.
Improper index management or off-by-one errors in calculating subarray boundaries.

Follow-up variants

What if the query range is always the entire array? You could preprocess frequency counts for the entire array to quickly answer queries.
What if the array is immutable and the query range doesn't change? A simple frequency map can be sufficient without advanced data structures.
What if the queries involve different types of range queries, such as sum or maximum instead of frequency? The solution would need to be adjusted accordingly.

FAQ

What is the optimal way to solve Range Frequency Queries?

The optimal approach uses a combination of preprocessing with hash maps or segment trees to answer queries in O(log n) time after O(n log n) preprocessing.

Why does brute force fail for this problem?

Brute force scanning leads to time-limit exceeded errors, as it takes too long to process large arrays and multiple queries.

How can segment trees be used to improve this problem?

Segment trees allow for efficient range queries by storing frequency counts for each segment of the array, reducing query time to O(log n).

What is the trade-off between using a hash map and a segment tree?

A hash map offers simpler implementation with preprocessing time O(n), but a segment tree offers better performance in handling range queries with O(log n) query time.

Can this problem be solved with just a frequency array?

Yes, a frequency array can be used for smaller cases where the array is static and the queries involve the entire array, but more advanced methods are needed for efficient handling of arbitrary subarrays.

terminal

Solution

Solution 1: Hash Table + Binary Search

We use a hash table $g$ to store the array of indices corresponding to each value. In the constructor, we traverse the array $\textit{arr}$, adding the index corresponding to each value to the hash table.

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class RangeFreqQuery:

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

    def query(self, left: int, right: int, value: int) -> int:
        idx = self.g[value]
        l = bisect_left(idx, left)
        r = bisect_left(idx, right + 1)
        return r - l


# Your RangeFreqQuery object will be instantiated and called as such:
# obj = RangeFreqQuery(arr)
# param_1 = obj.query(left,right,value)

Continue Practicing

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