2438. Range Product Queries of Powers

Problem Description

The problem presents us with two tasks:

Construct an array called powers from a given positive integer n, where the array consists of the smallest set of powers of 2 that add up to n. It's important to note that the array is indexed starting from 0, sorted in non-decreasing order, and has a unique possible configuration for any given n.
Process multiple queries described by the queries 2D array, where each subarray [left, right] means we need to calculate the product of powers[j] for all values of j from left to right inclusive.

The array of products, one for each query, must be computed and returned, and to manage potentially large numbers, the final products should be computed modulo 10^9 + 7.

An example can illustrate this better: Suppose n = 10, which in binary is 1010. The powers array would consist of the powers of 2 that sum up to n: [2^1, 2^3] or [2, 8]. For a query [0, 1] which means calculate the product powers[0] * powers[1], the answer would be 2 * 8 = 16.

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Intuition

To address the problem systematically, we can approach it as follows:

Creating the `powers` Array

Find the minimum set of powers of 2 that sum up to n:
- We iterate through the binary representation of n from the least significant bit to the most significant bit.
- For each bit that is set (i.e., is 1), we calculate the corresponding power of 2. This is achieved by utilizing the operation n & -n, which isolates the least significant bit that is set and yields the smallest power of 2 contributing to n.
- After finding the current smallest power of 2, we subtract it from n, and continue the process until n becomes 0.
The powers array is constructed implicitly in reverse order (from larger to smaller powers of 2) because we're starting with the least significant bit (smallest powers of 2) in n's binary representation.

Processing the Queries

Now, with the powers array ready, we loop through each query:
- Initialize a variable to accumulate the product (starting with 1) within the range [left, right].
- Multiply the components of the powers array that are within the range specified by each query. Since we want to avoid integer overflow issues, we take the modulus of the product with 10^9 + 7 at each step.
- Append the resulting product to our answer list.

The intuition behind modulo operation at each multiplication step is to ensure that we stay within the maximum limit for integer values and correctly handle cases where the product may be very large, as the final product must be returned modulo 10^9 + 7.

Learn more about Prefix Sum patterns.

Solution Approach

The solution to this LeetCode problem involves both bitwise manipulation to construct the powers array and modular arithmetic to calculate and store the query results. Here’s how the implementation goes step-by-step, corresponding to the given solution code:

Initializing the powers array: We need a list powers to store the powers of 2 that make up the number n. Since a positive integer can be represented as a sum of unique powers of 2 (according to binary representation), we find these powers and store them.
Finding the powers of 2 from n: We use a while loop that continues until n becomes 0. Inside this loop, we use x = n & -n to get the lowest power of 2 in n. This operation works because in binary, -n is n with all bits inverted plus 1 (two's complement representation), so n & -n isolates the least significant 1-bit. We append x to the powers list and subtract x from n using n -= x.
Modular exponentiation: We take note that when dealing with large numbers, we should use modular arithmetic to avoid overflow. Modulo operations are distributive over multiplication, which is crucial for our solution. We set mod to 10**9 + 7.
Processing each query: We need to iterate over the range of powers specified by each query:
- We initialize a variable x = 1 for the product result of the current query.
- The slice powers[l : r + 1] gets the relevant segment of the powers array based on the current query range.
- For each element y in the above slice, we update x as x = (x * y) % mod to get the product of the current segment, applying the modulus at each multiplication step to ensure the result stays within bounds of the integer limits.
Storing the result: For each query, after we calculate the product, we append it to the answer list ans. The list ans is our final result array that will be returned.

This solution utilizes simple bitwise operations to deconstruct a number into powers of 2, and then employs modular arithmetic with a simple iteration to answer the range product queries.

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Example Walkthrough

Let's take n = 10 and a single query [0, 1] to illustrate the solution approach step-by-step.

Initializing the powers array: We begin by creating an empty array powers to store powers of 2. Since we are provided with n = 10 (which is 1010 in binary), we need to find which powers of 2 add up to 10.
Finding the powers of 2 from n: We loop while n is not zero,
- In the first iteration, n is 10. We do x = n & -n, which gives us 2 (10 in binary), because 10 & -10 isolates the least significant bit which represents the power of 2.
- We append 2 to the powers array, which now becomes [2].
- We subtract 2 from n, making n = 10 - 2 = 8.
- In the next iteration, n is 8. Again x = n & -n gives us 8 because 8 is a power of 2.
- We append 8 to the powers array, which now becomes [2, 8].
- Subtract 8 from n, n becomes 0. The loop ends here.

After these steps, the powers array represents the powers of 2 that sum up to n: [2, 8].

Modular exponentiation: Set the variable mod to 10^9 + 7 to ensure all operations are performed modulo this number to prevent integer overflow.
Processing each query: With the powers array ready, we process the query [0, 1]:
- We initialize x = 1 which will hold our product.
- Based on the query, we need the slice powers[0 : 1+1], which is the entire powers array [2, 8].
- Loop through the slice and for each element y, update x as (x * y) % mod.
- First, x = (1 * 2) % mod = 2. Next, x = (2 * 8) % mod = 16.
- No more elements in the slice, so the final product of this query is 16.
Storing the result: We append 16 to the answer list ans. If there were more queries, each product would be appended to ans as well.

In this example, the answer list ans contains just one element [16], because there was only one query. This final array is what the function would return.

Solution Implementation

1class Solution:
2    def product_queries(self, n: int, queries: List[List[int]]) -> List[int]:
3        # Initialize an array to store the powers of 2 found in the binary representation of n
4        powers_of_two = []
5      
6        # Extract powers of 2 from n by finding the rightmost set bit continuously
7        while n:
8            # x is the largest power of 2 in n, given by the bitwise AND of n and its two's complement
9            x = n & -n
10            # Add x to the list
11            powers_of_two.append(x)
12            # Remove the largest power of 2 from n
13            n -= x
14      
15        # Set modulo as per the problem statement to avoid large integer overflow
16        mod = 10**9 + 7
17      
18        # Initialize an array to store the results of the queries
19        results = []
20      
21        # Loop through each query, which is a pair of indices (l, r)
22        for l, r in queries:
23            # Start with a product result of 1 for each query
24            product_result = 1
25          
26            # Multiply the values from powers_of_two[l] to powers_of_two[r]
27            # and take the modulo to prevent overflow
28            for power in powers_of_two[l : r + 1]:
29                product_result = (product_result * power) % mod
30          
31            # Append the product result to the results list
32            results.append(product_result)
33
34        # Return the final list of results for all queries
35        return results
36

1class Solution {
2    // Define the mod constant for the problem (10^9 + 7)
3    private static final int MOD = (int) 1e9 + 7;
4
5    // Method to solve the product queries on a range of powers of two that compose n
6    public int[] productQueries(int n, int[][] queries) {
7        // Count the number of set bits in n to determine the size of the powers array
8        int[] powers = new int[Integer.bitCount(n)];
9      
10        // Extract the powers of two which compose n
11        for (int i = 0; n > 0; ++i) {
12            int lowestOneBit = n & -n; // Get the lowest set bit (the rightmost one)
13            powers[i] = lowestOneBit; // Store the power of two in the array
14            n -= lowestOneBit; // Remove the extracted power of two from n
15        }
16      
17        // Create an array to store the answers to the queries
18        int[] ans = new int[queries.length];
19      
20        // Process each query
21        for (int i = 0; i < ans.length; ++i) {
22            long product = 1; // Store the product as a long to prevent overflow
23            int left = queries[i][0], right = queries[i][1]; // Get the left and right indices
24          
25            // Calculate the product of the powers from left to right index inclusive
26            for (int j = left; j <= right; ++j) {
27                product = (product * powers[j]) % MOD; // Multiply the current power and take mod
28            }
29          
30            // Store the result as an int after casting from long
31            ans[i] = (int) product;
32        }
33      
34        // Return the array containing the results for all the queries
35        return ans;
36    }
37}
38

1class Solution {
2public:
3    // Constants should be capitalized and use static constexpr for compile time initialization
4    static constexpr int MOD = 1e9 + 7;
5
6    // This function takes an integer and a set of queries and returns the product of powers for each query
7    vector<int> productQueries(int n, vector<vector<int>>& queries) {
8        vector<int> powersOfTwo;
9
10        // Extract powers of 2 from n and store them in the powersOfTwo vector
11        while (n > 0) {
12            // Get the rightmost set bit (largest power of 2 not greater than n)
13            int largestPowerOfTwo = n & -n;
14            // Add the power of 2 to the list
15            powersOfTwo.emplace_back(largestPowerOfTwo);
16            // Subtract the power of 2 from n to remove that bit
17            n -= largestPowerOfTwo;
18        }
19
20        vector<int> answer;
21        // Iterate through each query
22        for (auto& query : queries) {
23            int start = query[0], end = query[1];
24            long long product = 1;  // Define the product as a long long to prevent integer overflow
25
26            // Calculate the product of powers from start to end index
27            for (int i = start; i <= end; ++i) {
28                product = (product * powersOfTwo[i]) % MOD;
29            }
30
31            // Store the result in answer using modulo to fit within integer range
32            answer.emplace_back(static_cast<int>(product));
33        }
34        // Return the final answers for all queries
35        return answer;
36    }
37};
38

1// Define module-level constants in uppercase with `const` (no static needed in global context)
2const MOD = 1e9 + 7;
3
4// This function takes an integer and an array of queries and returns the product of powers for each query
5function productQueries(n: number, queries: number[][]): number[] {
6    let powersOfTwo: number[] = [];
7
8    // Extract powers of 2 from n and store them in the powersOfTwo array
9    while (n > 0) {
10        // Get the rightmost set bit (largest power of 2 not greater than n)
11        let largestPowerOfTwo = n & -n;
12        // Add the power of 2 to the array
13        powersOfTwo.push(largestPowerOfTwo);
14        // Subtract the power of 2 from n to remove that bit
15        n -= largestPowerOfTwo;
16    }
17
18    let answers: number[] = [];
19    // Iterate through each query
20    queries.forEach(query => {
21        let start = query[0], end = query[1];
22        // Define the product as a bigint to prevent integer overflow and use BigInt literals for calculations
23        let product: bigint = BigInt(1);
24
25        // Calculate the product of powers from start to end indices
26        for (let i = start; i <= end; ++i) {
27            product = (product * BigInt(powersOfTwo[i])) % BigInt(MOD);
28        }
29
30        // Store the result in answers using modulo to fit within integer range and convert BigInt to number
31        answers.push(Number(product));
32    });
33
34    // Returns the final answers for all queries
35    return answers;
36}
37

Time and Space Complexity

Time Complexity

The given Python code consists of two main parts: extracting powers of 2 from n, and computing the product of subsets as specified by queries.

Extracting Powers of 2 from n:
- This is done with a loop that runs until n becomes 0.
- For each iteration, we perform a bitwise AND of n and -n to isolate the lowest power of 2 in n.
- Since each number can have at most O(log n) bits, where log refers to the logarithm base 2, the loop runs for O(log n) iterations.
- The operation within each loop is O(1).
- Therefore, this step has a time complexity of O(log n).
Processing Queries:
- There is a loop iterating over each query in queries. Let's denote the number of queries as q.
- Inside this loop, there is another loop over the powers from index l to r, which in the worst case includes all the powers extracted previously.
- The number of powers is at most O(log n), as previously derived.
- Multiplying and taking the modulo has a constant time complexity O(1).
- Therefore, processing all queries has a worst-case time complexity of O(q * log n).

Overall, the time complexity of the entire function is O(log n) + O(q * log n) = O((q + 1) * log n).

Space Complexity

The space complexity of the code is determined by the extra space used apart from the input:

Storing Powers:
- We store each power of 2 that exists in the binary representation of n.
- This takes O(log n) space.
Answer List:
- We store one integer for each query, so this takes O(q) space.
Temporary Variables:
- Temporary variables such as x, y, and mod are O(1) space.

Overall, the resultant space complexity is O(log n) + O(q) = O(log n + q).

Learn more about how to find time and space complexity quickly using problem constraints.

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