How do you solve Range Product Queries of Powers on LeetCode?

Range Product Queries of Powers (LeetCode #2438) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + q) time complexity and O(n) space complexity. Key insight: Understanding binary representation helps in forming the powers array efficiently.

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

The optimal solution for Range Product Queries of Powers runs in O(n + q) time and uses O(n) space. The time complexity is O(n + q) because we spend O(n) to create the powers and prefix products, and O(q) to answer each query in constant time.

What is the brute force solution for Range Product Queries of Powers?

The brute force approach for Range Product Queries of Powers is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach calculates the product for each query by directly multiplying the elements in the specified range of the powers array. This is straightforward but inefficient for large inputs. The optimal Optimal Solution approach improves this to O(n + q).

What algorithm or data structure is used to solve Range Product Queries of Powers?

Range Product Queries of Powers uses the following concepts: Array, Bit Manipulation, Prefix Sum. The recommended patterns to study are: Prefix Sum, Modular Arithmetic.

What are the interview tips for Range Product Queries of Powers?

Always clarify the constraints and edge cases before diving into coding. Discuss your thought process and approach before writing code to ensure clarity. Practice explaining your code and logic as if teaching someone else.

What are common mistakes when solving Range Product Queries of Powers?

Not considering the modulo operation during multiplication, leading to overflow. Failing to handle edge cases where queries might cover the entire range.

#2438

Range Product Queries of Powers

Medium

ArrayBit ManipulationPrefix SumPrefix SumModular Arithmetic

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

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

The optimal approach leverages prefix products to compute the product of any subarray in constant time after an initial setup. This reduces the time complexity significantly.

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Algorithm

3 steps

1Step 1: Create the powers array from the binary representation of n.
2Step 2: Build a prefix product array where each entry at index i contains the product of all elements in powers up to index i.
3Step 3: For each query, compute the product using the prefix product array to get the result in constant time.

solution.py10 lines

1def rangeProductQueries(n, queries):
2    powers = [1 << i for i in range(n.bit_length()) if (n & (1 << i))]
3    prefix_products = [1] * (len(powers) + 1)
4    for i in range(1, len(powers) + 1):
5        prefix_products[i] = (prefix_products[i - 1] * powers[i - 1]) % (10**9 + 7)
6    answers = []
7    for left, right in queries:
8        product = (prefix_products[right + 1] * pow(prefix_products[left], -1, 10**9 + 7)) % (10**9 + 7)
9        answers.append(product)
10    return answers

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Complexity note: The time complexity is O(n + q) because we spend O(n) to create the powers and prefix products, and O(q) to answer each query in constant time.

1Understanding binary representation helps in forming the powers array efficiently.
2Using prefix products allows for quick calculations of subarray products.

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