What is the time complexity of Range Frequency Queries?

The optimal solution has a time complexity of O(log N).

What data structures are used to solve Range Frequency Queries?

This problem uses Array, Hash Table, Binary Search, Design, Segment Tree. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Range Frequency Queries?

Insight 1: Preprocessing the array to store the indices of each value allows for efficient querying. Insight 2: Using binary search to find the first and last occurrence of the relevant indices within the specified range is crucial for optimizing the query operation.

Which companies ask LeetCode 2080. Range Frequency Queries in interviews?

This problem is frequently asked at Quora.

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Range Frequency Queries - LeetCode 2080 Solution

Range Frequency Queries - Complete Solution Guide

Range Frequency Queries is LeetCode problem 2080, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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.

Detailed Explanation

The problem asks us to design a data structure, `RangeFreqQuery`, that efficiently answers frequency queries on a given array. Specifically, we need to be able to find how many times a given `value` appears within a specified subarray `arr[left...right]`. The constructor takes an integer array `arr` as input. The `query(left, right, value)` method should return the number of times `value` appears in the subarray defined by the indices `left` and `right` (inclusive).

Solution Approach

The provided solution uses a hash map (or dictionary) to store the indices of each value in the input array. During the constructor, the code iterates through the input array and stores each value along with its indices in the hash map. The `query` function then retrieves the list of indices associated with the given value. If the value is not found, it returns 0. Otherwise, it uses binary search to find the number of indices within the range [left, right]. The difference between the rightmost and leftmost index found using binary search gives the number of times the value appears within the given range.

Step-by-Step Algorithm

Step 1: **Initialization:** Create a hash map (or dictionary) called `value_map` to store the indices of each value in the input array. The keys are the values from the array, and the values are lists of indices where those values appear.
Step 2: **Preprocessing (Constructor):** Iterate through the input array `arr`. For each element `arr[i]`, append its index `i` to the list associated with the key `arr[i]` in the `value_map`.
Step 3: **Query Handling:** In the `query(left, right, value)` function, retrieve the list of indices associated with the given `value` from the `value_map`.
Step 4: **Value Not Found:** If the `value` is not found in the `value_map` (i.e., the value doesn't exist in the original array), return 0.
Step 5: **Binary Search (Left):** Use binary search (`bisect_left` or equivalent) on the list of indices to find the index of the first element that is greater than or equal to `left`. This is the starting index of the range within the stored indices.
Step 6: **Binary Search (Right):** Use binary search (`bisect_right` or equivalent) on the list of indices to find the index of the first element that is strictly greater than `right`. This is the ending index of the range within the stored indices.
Step 7: **Calculate Frequency:** Calculate the frequency of the `value` within the given range by subtracting the `start_idx` from the `end_idx`. This result is returned.

Key Insights

Insight 1: Preprocessing the array to store the indices of each value allows for efficient querying.
Insight 2: Using binary search to find the first and last occurrence of the relevant indices within the specified range is crucial for optimizing the query operation.
Insight 3: Handling the case where the given 'value' is not present in the input array is important to avoid errors and provide a correct result.

Complexity Analysis

Time Complexity: O(log N)

Space Complexity: O(N)

Topics

This problem involves: Array, Hash Table, Binary Search, Design, Segment Tree.

Companies

Asked at: Quora.

Range Frequency Queries | LeetCode 2080

Medium

Array Hash Table Binary Search Design Segment Tree

Time: O(log N)Space: O(N)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class RangeFreqQuery {

    private Map<Integer, List<Integer>> valueMap;

    public RangeFreqQuery(int[] arr) {
        valueMap = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            int v = arr[i];
            if (!valueMap.containsKey(v)) {
                valueMap.put(v, new ArrayList<>());
            }
            valueMap.get(v).add(i);
        }
    }
    
    public int query(int left, int right, int value) {
        List<Integer> indices = valueMap.get(value);
        if (indices == null || indices.isEmpty()) {
            return 0;
        }

        int startIdx = binarySearchLeft(indices, left);
        int endIdx = binarySearchRight(indices, right);
        
        return endIdx - startIdx;
    }

    private int binarySearchLeft(List<Integer> arr, int target) {
        int low = 0;
        int high = arr.size();
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (arr.get(mid) < target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }

    private int binarySearchRight(List<Integer> arr, int target) {
        int low = 0;
        int high = arr.size();
        while (low < high) {
            int mid = low + (high - low) / 2;
            if (arr.get(mid) <= target) {
                low = mid + 1;
            } else {
                high = mid;
            }
        }
        return low;
    }
}

/**
 * Your RangeFreqQuery object will be instantiated and called as such:
 * RangeFreqQuery obj = new RangeFreqQuery(arr);
 * int param_1 = obj.query(left,right,value);
 */

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Problem Info

Description

Problem 2080: Range Frequency Queries - Detailed explanation and solution approaches.

Topics

Array Hash Table Binary Search Design Segment Tree

Companies

Quora

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2080

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: **Initialization:** Create a hash map (or dictionary) called `value_map` to store the indices of each value in the input array. The keys are the values from the array, and the values are lists of indices where those values appear.
2
Step 2: **Preprocessing (Constructor):** Iterate through the input array `arr`. For each element `arr[i]`, append its index `i` to the list associated with the key `arr[i]` in the `value_map`.
3
Step 3: **Query Handling:** In the `query(left, right, value)` function, retrieve the list of indices associated with the given `value` from the `value_map`.
4
Step 4: **Value Not Found:** If the `value` is not found in the `value_map` (i.e., the value doesn't exist in the original array), return 0.
5
Step 5: **Binary Search (Left):** Use binary search (`bisect_left` or equivalent) on the list of indices to find the index of the first element that is greater than or equal to `left`. This is the starting index of the range within the stored indices.
6
Step 6: **Binary Search (Right):** Use binary search (`bisect_right` or equivalent) on the list of indices to find the index of the first element that is strictly greater than `right`. This is the ending index of the range within the stored indices.
7
Step 7: **Calculate Frequency:** Calculate the frequency of the `value` within the given range by subtracting the `start_idx` from the `end_idx`. This result is returned.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(log N)

🚀 Logarithmic - Very efficient! Halves the problem each step.

💾

Space Complexity

O(N)

📝

Why?

The O(log N) query time is due to the usage of binary search to find the start and end indices of occurrences within a given range. The space complexity arises from storing the indices of each element in the array within a hash map (or equivalent data structure like defaultdict).

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Segment Tree

Use a segment tree to store the frequencies of each value. Each node in the segment tree represents a subarray, and stores a hash map of value frequencies within that subarray. This requires more complex construction and more memory but potentially leads to query improvements if the range spans many nodes.

Time: O(N log N) for construction, O(log N) for query.Space: O(N log N)

Trade-offs: Segment tree offers better query time especially if many queries are done. But it requires more space and more complex implementation. Use when number of queries are significantly greater than the size of the input array, and memory consumption is not a major issue.

Brute Force

Iterate through the subarray `arr[left...right]` and count the number of times `value` appears. This is the simplest approach but has the worst time complexity.

Time: O(N) for each query.Space: O(1)

Trade-offs: The brute force approach is simple to implement but has a high time complexity for each query, making it unsuitable for large arrays and many queries. It is better when the array and number of queries are small.

✨

Key Insights

1Insight 1: Preprocessing the array to store the indices of each value allows for efficient querying.
2Insight 2: Using binary search to find the first and last occurrence of the relevant indices within the specified range is crucial for optimizing the query operation.
3Insight 3: Handling the case where the given 'value' is not present in the input array is important to avoid errors and provide a correct result.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Edge Cases

Edge case 1: **Value not in array:** If the given `value` is not present in the input array, the `value_map` will not contain an entry for it. The solution correctly handles this case by checking if the `value` exists in the `value_map` and returns 0 if it does not.
Edge case 2: **Empty subarray:** If `left > right`, the intended subarray is empty. The given code doesn't explicitly handle this. However, bisect_left will return the index where left should be inserted. bisect_right will return the index where right+1 should be inserted. So the diff will be zero.
Edge case 3: **left and right equal:** If `left` and `right` are equal, the subarray contains only one element. The code handles this correctly as binary search will find either 0 or 1 occurrences depending on whether the value exists at index left.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the case when the value doesn't exist in the input array, leading to potential errors or incorrect results.

✗Common mistake 2: Implementing binary search incorrectly, resulting in incorrect start and end indices and thus an incorrect frequency count. Using library functions such as `bisect` is generally recommended to prevent such mistakes.

✗Common mistake 3: Forgetting that binary search works on sorted data. Preprocessing step to store indices must maintain sorted order (implicitly maintained due to traversal order).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: **Data Analysis:** Analyzing the frequency of specific events or data points within a time series dataset. For example, finding how many times a particular product was sold within a specific time window.

🏦Real-world scenario 2: **Database Queries:** Optimizing database queries that involve counting the occurrences of specific values within a given range of records. For instance, counting the number of customers who placed orders within a certain date range and for a specific product.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Range Frequency Queries"?

For this medium-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?