What is the time complexity of Range Product Queries of Powers?

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

What is the space complexity of Range Product Queries of Powers?

The space complexity is O(log n).

What data structures are used to solve Range Product Queries of Powers?

This problem uses Array, Bit Manipulation, Prefix Sum. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Range Product Queries of Powers?

Insight 1: Representing a number as a sum of distinct powers of 2 is equivalent to finding the set bits in its binary representation. This directly gives us the `powers` array. Insight 2: Calculating prefix products allows us to efficiently compute the product within a range by dividing the prefix product at the `right` index by the prefix product at the `left - 1` index. However, since we need to perform modulo arithmetic, division is replaced by multiplying with the modular inverse.

Which companies ask LeetCode 2438. Range Product Queries of Powers in interviews?

This problem is frequently asked at Goldman Sachs, IBM.

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Range Product Queries of Powers - LeetCode 2438 Solution

Q: How to solve Range Product Queries of Powers?

The solution involves three main steps: 1) Construct the `powers` array by iterating through the bits of `n`. If a bit is set (i.e., equal to 1), the corresponding power of 2 is added to the `powers` array. 2) Calculate the prefix products of the `powers` array. This allows us to efficiently compute the product of elements within a range. 3) For each query `[left, right]`, calculate the product of the elements from `powers[left]` to `powers[right]` using the prefix products and modular arithmetic. Specifically, the product is equal to `prefix_products[right + 1] / prefix_products[left]`, which is computed as `prefix_products[right + 1] * (prefix_products[left])^(MOD-2) mod MOD`.

Range Product Queries of Powers - Complete Solution Guide

Range Product Queries of Powers is LeetCode problem 2438, 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

Given a positive integer n , there exists a 0-indexed array called powers , composed of the minimum number of powers of 2 that sum to n . The array is sorted in non-decreasing order, and there is only one way to form the array. You are also given a 0-indexed 2D integer array queries , where queries[i] = [left i , right i ] . Each queries[i] represents a query where you have to find the product of all powers[j] with left i <= j <= right i . Return an array answers , equal in length to queries , w

Detailed Explanation

The problem asks us to find the product of powers of 2 within specified ranges. Given an integer `n`, we first decompose it into the minimum number of powers of 2. For example, if `n = 15`, the powers of 2 are `[1, 2, 4, 8]` (1 + 2 + 4 + 8 = 15). Then, given a set of queries, where each query `[left, right]` represents a range within the powers array, we calculate the product of the powers within that range. Since the product can be very large, we need to return the result modulo `10^9 + 7`.

Solution Approach

The solution involves three main steps: 1) Construct the `powers` array by iterating through the bits of `n`. If a bit is set (i.e., equal to 1), the corresponding power of 2 is added to the `powers` array. 2) Calculate the prefix products of the `powers` array. This allows us to efficiently compute the product of elements within a range. 3) For each query `[left, right]`, calculate the product of the elements from `powers[left]` to `powers[right]` using the prefix products and modular arithmetic. Specifically, the product is equal to `prefix_products[right + 1] / prefix_products[left]`, which is computed as `prefix_products[right + 1] * (prefix_products[left])^(MOD-2) mod MOD`.

Step-by-Step Algorithm

Step 1: Create an empty array called `powers` to store the powers of 2.
Step 2: Iterate through the bits of the integer `n` (from the least significant bit to the most significant bit). This can be done by checking `(n >> i) & 1` where `i` ranges from 0 to 30.
Step 3: If the `i`-th bit is set, calculate 2^i and add it to the `powers` array.
Step 4: Create a `prefix_products` array. The first element is 1. Then calculate each element by multiplying the previous element with the corresponding element in powers[i], modulo MOD. `prefix_products[i + 1] = (prefix_products[i] * powers[i]) % MOD`
Step 5: Iterate through the queries array. For each query `[left, right]`, retrieve `prefix_products[right + 1]` and `prefix_products[left]`.
Step 6: Calculate the modular inverse of `prefix_products[left]` using Fermat's Little Theorem: `pow(prefix_products[left], MOD - 2, MOD)`. Assign to `inv_denominator`.
Step 7: Multiply `prefix_products[right + 1]` with `inv_denominator` modulo `MOD`: `(prefix_products[right + 1] * inv_denominator) % MOD`. Add the result to the `answers` array.
Step 8: Return the `answers` array.

Key Insights

Insight 1: Representing a number as a sum of distinct powers of 2 is equivalent to finding the set bits in its binary representation. This directly gives us the `powers` array.
Insight 2: Calculating prefix products allows us to efficiently compute the product within a range by dividing the prefix product at the `right` index by the prefix product at the `left - 1` index. However, since we need to perform modulo arithmetic, division is replaced by multiplying with the modular inverse.
Insight 3: Utilizing modular exponentiation with Fermat's Little Theorem optimizes the computation of the modular inverse, as direct division is not possible under modulo.
Insight 4: In the Java/C/C++ implementation exponents themselves are used for calculating the total exponent, then a power function (fast exponentiation) is called to avoid overflow.

Complexity Analysis

Time Complexity: O(m + log n)

Space Complexity: O(log n)

Topics

This problem involves: Array, Bit Manipulation, Prefix Sum.

Companies

Asked at: Goldman Sachs, IBM.

Range Product Queries of Powers | LeetCode 2438

Medium

Array Bit Manipulation Prefix Sum

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

Solution Code

Available in:

class Solution {
    public int[] productQueries(int n, int[][] queries) {
        final int MOD = (int)1e9 + 7;

        List<Integer> exponents = new ArrayList<>();
        for (int i = 0; i < 32; i++) {
            if (((n >> i) & 1) == 1) {
                exponents.add(i);
            }
        }

        int num_exponents = exponents.size();
        int[] prefix_sum_exponents = new int[num_exponents + 1];
        prefix_sum_exponents[0] = 0;
        for (int i = 0; i < num_exponents; i++) {
            prefix_sum_exponents[i+1] = prefix_sum_exponents[i] + exponents.get(i);
        }

        int[] answers = new int[queries.length];
        for (int i = 0; i < queries.length; i++) {
            int left = queries[i][0];
            int right = queries[i][1];
            int total_exponent = prefix_sum_exponents[right + 1] - prefix_sum_exponents[left];

            long result = power(2, total_exponent, MOD);
            answers[i] = (int) result;
        }

        return answers;
    }
    
    private long power(long base, long exp, int mod) {
        long res = 1;
        base %= mod;
        while (exp > 0) {
            if (exp % 2 == 1) res = (res * base) % mod;
            base = (base * base) % mod;
            exp /= 2;
        }
        return res;
    }
}

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Problem Info

Description

Problem 2438: Range Product Queries of Powers - Detailed explanation and solution approaches.

Topics

Array Bit Manipulation Prefix Sum

Companies

Goldman SachsIBM

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2438

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

8 steps

1
Step 1: Create an empty array called `powers` to store the powers of 2.
2
Step 2: Iterate through the bits of the integer `n` (from the least significant bit to the most significant bit). This can be done by checking `(n >> i) & 1` where `i` ranges from 0 to 30.
3
Step 3: If the `i`-th bit is set, calculate 2^i and add it to the `powers` array.
4
Step 4: Create a `prefix_products` array. The first element is 1. Then calculate each element by multiplying the previous element with the corresponding element in powers[i], modulo MOD. `prefix_products[i + 1] = (prefix_products[i] * powers[i]) % MOD`
5
Step 5: Iterate through the queries array. For each query `[left, right]`, retrieve `prefix_products[right + 1]` and `prefix_products[left]`.
6
Step 6: Calculate the modular inverse of `prefix_products[left]` using Fermat's Little Theorem: `pow(prefix_products[left], MOD - 2, MOD)`. Assign to `inv_denominator`.
7
Step 7: Multiply `prefix_products[right + 1]` with `inv_denominator` modulo `MOD`: `(prefix_products[right + 1] * inv_denominator) % MOD`. Add the result to the `answers` array.
8
Step 8: Return the `answers` array.

💡

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(m + log n)

💾

Space Complexity

O(log n)

🎯 Logarithmic space - Very memory efficient!

📝

Why?

The time complexity is dominated by the loop through the queries and the generation of powers array. The powers array can have at most log2(n) elements. Modular exponentiation has O(log MOD) complexity, where MOD is 10^9 + 7. The space complexity is determined by the size of the `powers` and `prefix_products` arrays, both of which are O(log n).

📖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

Without Prefix Products

Instead of using prefix products, we can directly calculate the product for each query by iterating through the `powers` array from `left` to `right` and multiplying the elements modulo `MOD`.

Time: O(m * k), where 'm' is the number of queries and 'k' is the average range size (right - left + 1).Space: O(log n)

Trade-offs: This approach has a higher time complexity if the range sizes are large. The prefix product approach is more efficient when range sizes are significant.

✨

Key Insights

1Insight 1: Representing a number as a sum of distinct powers of 2 is equivalent to finding the set bits in its binary representation. This directly gives us the `powers` array.
2Insight 2: Calculating prefix products allows us to efficiently compute the product within a range by dividing the prefix product at the `right` index by the prefix product at the `left - 1` index. However, since we need to perform modulo arithmetic, division is replaced by multiplying with the modular inverse.
3Insight 3: Utilizing modular exponentiation with Fermat's Little Theorem optimizes the computation of the modular inverse, as direct division is not possible under modulo.
4Insight 4: In the Java/C/C++ implementation exponents themselves are used for calculating the total exponent, then a power function (fast exponentiation) is called to avoid overflow.

📚

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: n = 1. The `powers` array will be `[1]`. The code correctly handles this case.
Edge case 2: queries = [[0, 0]]. The code correctly retrieves the single power of 2 at index 0.
Edge case 3: Large input values of n leading to overflow if modular arithmetic is not applied correctly. The code effectively applies modular operations at each step to prevent overflow.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not applying modular arithmetic correctly, leading to integer overflow.

✗Common mistake 2: Incorrectly calculating the modular inverse. Fermat's Little Theorem is crucial for computing the modular inverse in this scenario.

✗Common mistake 3: Assuming the `powers` array is pre-sorted. The problem statement guarantees this but it's a good practice to keep it in mind while implementing.

✗Common mistake 4: Forgetting that `n` can be as large as 10^9, which might lead to incorrect bitwise operations if not handled carefully (e.g., assuming a smaller maximum value for i when iterating through the bits).

✅

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: Efficiently querying data within ranges. Similar to database queries where you need to calculate aggregates (like product, sum, etc.) within specific ranges of data. Pre-computing prefix products can speed up these computations.

🏦Real-world scenario 2: Cryptography. The modular arithmetic concepts and techniques used are fundamental in cryptographic algorithms, particularly in scenarios involving large numbers and secure computations.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ 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: September 7, 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

❓

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 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 Product Queries of Powers"?

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 bit manipulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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