What is the time complexity of Count Pairs With XOR in a Range?

The optimal solution has a time complexity of O(n*log(m)).

What is the space complexity of Count Pairs With XOR in a Range?

The space complexity is O(n*log(m)).

What data structures are used to solve Count Pairs With XOR in a Range?

This problem uses Array, Bit Manipulation, Trie. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Count Pairs With XOR in a Range?

Insight 1: The problem can be solved efficiently using a Trie data structure to store the numbers in their binary representation. Insight 2: We can break the range [low, high] into two subproblems: count pairs with XOR less than or equal to `high` and count pairs with XOR less than `low`. The desired result is the difference between these two counts.

Which companies ask LeetCode 1803. Count Pairs With XOR in a Range in interviews?

This problem is frequently asked at Vimeo.

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Count Pairs With XOR in a Range - LeetCode 1803 Solution

Q: How to solve Count Pairs With XOR in a Range?

The solution uses a Trie (prefix tree) to store the binary representations of the numbers in the array. The core idea is to use the Trie to efficiently count the number of pairs whose XOR value is less than a given value `k`. The `countPairs` function computes the result by calling a helper function `count_less_than(k)` twice: once for `high + 1` and once for `low`, and then returns the difference. The `count_less_than(k)` function iterates through the `nums` array, inserting each number into the Trie and, at the same time, querying the Trie to count the number of existing numbers that, when XORed with the current number, result in a value less than `k`.

Count Pairs With XOR in a Range - Complete Solution Guide

Count Pairs With XOR in a Range is LeetCode problem 1803, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given a (0-indexed) integer array nums and two integers low and high , return the number of nice pairs . A nice pair is a pair (i, j) where 0 <= i < j < nums.length and low <= (nums[i] XOR nums[j]) <= high . Example 1: Input: nums = [1,4,2,7], low = 2, high = 6 Output: 6 Explanation: All nice pairs (i, j) are as follows: - (0, 1): nums[0] XOR nums[1] = 5 - (0, 2): nums[0] XOR nums[2] = 3 - (0, 3): nums[0] XOR nums[3] = 6 - (1, 2): nums[1] XOR nums[2] = 6 - (1, 3): nums[1] XOR nums[3] = 3 - (2, 3

Detailed Explanation

The problem asks us to find the number of pairs (i, j) in a given integer array `nums` such that 0 <= i < j < nums.length and `low` <= (nums[i] XOR nums[j]) <= `high`. In other words, we need to count how many pairs of elements in the array have an XOR value that falls within the specified range [low, high].

Solution Approach

The solution uses a Trie (prefix tree) to store the binary representations of the numbers in the array. The core idea is to use the Trie to efficiently count the number of pairs whose XOR value is less than a given value `k`. The `countPairs` function computes the result by calling a helper function `count_less_than(k)` twice: once for `high + 1` and once for `low`, and then returns the difference. The `count_less_than(k)` function iterates through the `nums` array, inserting each number into the Trie and, at the same time, querying the Trie to count the number of existing numbers that, when XORed with the current number, result in a value less than `k`.

Step-by-Step Algorithm

Step 1: Define a Trie data structure with a `count` field to keep track of how many numbers share that prefix.
Step 2: Implement the `count_less_than(k)` function. This function initializes an empty Trie.
Step 3: Iterate through the `nums` array. For each number `num`:
Step 4: Initialize a `node` pointer to the root of the Trie and a `current_count` to 0.
Step 5: Iterate from the most significant bit (MAX_BIT = 14) to the least significant bit (0).
Step 6: Extract the `num_bit` (the i-th bit of `num`) and the `k_bit` (the i-th bit of `k`).
Step 7: If `k_bit` is 1, this means that to have an XOR value less than `k`, we can have either a 0 or 1 in the XOR result. If the number present in the Trie have the same bit as num (num_bit), the XOR will be '0'. Then we add all the node count to the current count
Step 8: If `k_bit` is 0, this means the XOR result must also be '0'.
Step 9: Update total_pairs by adding current_count
Step 10: Insert the number `num` into the Trie. Traverse the Trie based on the bits of `num`, incrementing the `count` at each node.
Step 11: Return the `total_pairs`.
Step 12: Calculate and return the final result: `count_less_than(high + 1) - count_less_than(low)`.

Key Insights

Insight 1: The problem can be solved efficiently using a Trie data structure to store the numbers in their binary representation.
Insight 2: We can break the range [low, high] into two subproblems: count pairs with XOR less than or equal to `high` and count pairs with XOR less than `low`. The desired result is the difference between these two counts.
Insight 3: To handle the constraints, the maximum number of bits needed to represent each number is determined (15 bits, since 2 * 10^4 < 2^15). This defines the depth of the Trie.

Complexity Analysis

Time Complexity: O(n*log(m))

Space Complexity: O(n*log(m))

Topics

This problem involves: Array, Bit Manipulation, Trie.

Companies

Asked at: Vimeo.

Count Pairs With XOR in a Range | LeetCode 1803

Hard

Array Bit Manipulation Trie

Time: O(n*log(m))Space: O(n*log(m))

Solution Code

Available in:

import java.util.HashMap;
import java.util.Map;

class Solution {
    public int countPairs(int[] nums, int low, int high) {
        
        return countLessThan(nums, high + 1) - countLessThan(nums, low);
    }
    
    private int countLessThan(int[] nums, int k) {
        if (k <= 0) {
            return 0;
        }
        
        Map<Integer, Object> root = new HashMap<>();
        int totalPairs = 0;
        // Constraints: nums[i], high <= 2 * 10^4 < 2^15.
        // So 15 bits (indices 0 to 14) are sufficient.
        int MAX_BIT = 14;
        
        for (int num : nums) {
            Map<Integer, Object> node = root;
            int currentCount = 0;
            for (int i = MAX_BIT; i >= 0; i--) {
                if (node == null) {
                    break;
                }
                
                int numBit = (num >> i) & 1;
                int kBit = (k >> i) & 1;
                
                if (kBit == 1) {
                    // If k's bit is 1, any pair with XOR bit 0 at this position
                    // will be smaller than k.
                    // XOR bit is 0 if the number in the trie has the same bit as num.
                    if (node.containsKey(numBit)) {
                        currentCount += ((Map<Integer, Integer>)node.get(numBit)).get("count");
                    }
                    
                    // We continue the search on the path where the XOR bit is 1.
                    node = (Map<Integer, Object>) node.get(1 - numBit);
                } else {
                    // If k's bit is 0, the XOR bit must also be 0.
                    // This happens if the trie number's bit is the same as num's bit.
                    node = (Map<Integer, Object>) node.get(numBit);
                }
            }
            
            totalPairs += currentCount;
            
            node = root;
            for (int i = MAX_BIT; i >= 0; i--) {
                int bit = (num >> i) & 1;
                if (node.get(bit) == null) {
                    Map<Integer, Integer> newMap = new HashMap<>();
                    newMap.put("count", 0);
                    node.put(bit, newMap);
                }
                node = (Map<Integer, Object>) node.get(bit);
                ((Map<Integer, Integer>) node).put("count", ((Map<Integer, Integer>) node).get("count") + 1);
            }
        }
        
        return totalPairs;
    }
}

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Problem Info

Description

Problem 1803: Count Pairs With XOR in a Range - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation Trie

Companies

Vimeo

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1803

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

12 steps

1
Step 1: Define a Trie data structure with a `count` field to keep track of how many numbers share that prefix.
2
Step 2: Implement the `count_less_than(k)` function. This function initializes an empty Trie.
3
Step 3: Iterate through the `nums` array. For each number `num`:
4
Step 4: Initialize a `node` pointer to the root of the Trie and a `current_count` to 0.
5
Step 5: Iterate from the most significant bit (MAX_BIT = 14) to the least significant bit (0).
6
Step 6: Extract the `num_bit` (the i-th bit of `num`) and the `k_bit` (the i-th bit of `k`).
7
Step 7: If `k_bit` is 1, this means that to have an XOR value less than `k`, we can have either a 0 or 1 in the XOR result. If the number present in the Trie have the same bit as num (num_bit), the XOR will be '0'. Then we add all the node count to the current count
8
Step 8: If `k_bit` is 0, this means the XOR result must also be '0'.
9
Step 9: Update total_pairs by adding current_count
10
Step 10: Insert the number `num` into the Trie. Traverse the Trie based on the bits of `num`, incrementing the `count` at each node.
11
Step 11: Return the `total_pairs`.
12
Step 12: Calculate and return the final result: `count_less_than(high + 1) - count_less_than(low)`.

💡

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(n*log(m))

💾

Space Complexity

O(n*log(m))

📝

Why?

The time complexity is derived from iterating through the `nums` array (O(n)) and, for each number, performing Trie operations that take O(log(m)) time. The space complexity arises from storing the Trie, which, in the worst case, can have a node for each bit of each number in `nums`.

📖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

Brute Force

Iterate through all possible pairs (i, j) in `nums` and check if `low` <= (nums[i] XOR nums[j]) <= `high`.

Time: O(n^2)Space: O(1)

Trade-offs: This approach is simple to implement but very inefficient for larger input sizes due to its quadratic time complexity. It's suitable for small `n` only.

✨

Key Insights

1Insight 1: The problem can be solved efficiently using a Trie data structure to store the numbers in their binary representation.
2Insight 2: We can break the range [low, high] into two subproblems: count pairs with XOR less than or equal to `high` and count pairs with XOR less than `low`. The desired result is the difference between these two counts.
3Insight 3: To handle the constraints, the maximum number of bits needed to represent each number is determined (15 bits, since 2 * 10^4 < 2^15). This defines the depth of the Trie.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: `low` is 0. The algorithm correctly handles this by subtracting `count_less_than(0)`, which will return 0, thus not affecting the final result.
Edge case 2: `high` is very large (close to the maximum allowed value). The algorithm is designed to handle values up to 2 * 10^4 so it handles this correctly.
Edge case 3: All numbers in the `nums` array are the same. The Trie will still be constructed and traversed correctly.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering the constraint 0 <= i < j < nums.length. The brute force approach may incorrectly count pairs where i >= j.

✗Common mistake 2: Integer overflow when calculating the XOR value. Although XOR itself doesn't lead to overflow, be mindful if further arithmetic operations are performed on the XOR result.

✗Common mistake 3: Incorrectly implementing the Trie insertion or query logic, leading to incorrect counts.

✅

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: Network routing. XOR operations are used in routing protocols to efficiently calculate network addresses and determine the shortest path for data packets. This algorithm can be adapted for range-based routing decisions.

🏦Real-world scenario 2: Data compression. XOR operations are used in data compression algorithms to identify redundant data and reduce storage space. Finding pairs with specific XOR properties can be applied in compression techniques.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

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

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: June 4, 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

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

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

@DevExpert92Jun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Pairs With XOR in a Range"?

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

🤖 Need help with this problem?