Hello! In this post, we will explore how to solve the range product query problem. The goal of this problem is to implement an algorithm that quickly calculates the product of specific ranges in a given array. This tutorial will progress step by step, starting from problem definition, moving to approach, implementation, and performance analysis.
Problem Definition
Given an integer array nums and several queries (i, j), we need to calculate the product from index i to j of the array. Since this needs to be repeated multiple times, an efficient method is required, considering time complexity.
Example
Input:
nums = [1, 2, 3, 4, 5]
queries = [(0, 1), (1, 3), (0, 4)]
Output:
Query 0: 1 * 2 = 2
Query 1: 2 * 3 * 4 = 24
Query 2: 1 * 2 * 3 * 4 * 5 = 120
Approach
Intuitively, we can directly calculate the product from the nums array for each query. However, this requires O(n) time complexity in the worst case, making it inefficient when performing multiple queries. Therefore, we can consider the following two approaches:
- Prefix Product Array: This method involves pre-calculating the cumulative product of the array, allowing us to quickly calculate the remaining product for each query in O(1). This approach needs O(n) to prepare the prefix product array and O(1) for each query, resulting in a total of O(n + m) (where m is the number of queries).
- Segment Tree: A more general approach is to use a segment tree to efficiently calculate the product for each range. In this case, the initialization takes O(n log n) and each query takes O(log n).
In this tutorial, we will proceed with the implementation using the first method, the prefix product array.
Implementation
First, we will create a prefix product array and then calculate the product for each query accordingly.
public class RangeProduct {
public static void main(String[] args) {
int[] nums = {1, 2, 3, 4, 5};
int[][] queries = {{0, 1}, {1, 3}, {0, 4}};
int[] result = rangeProduct(nums, queries);
for (int res : result) {
System.out.println(res);
}
}
public static int[] rangeProduct(int[] nums, int[][] queries) {
int n = nums.length;
int[] prefixProduct = new int[n + 1];
// Create prefix product array
prefixProduct[0] = 1; // Initial value
for (int i = 1; i <= n; i++) {
prefixProduct[i] = prefixProduct[i - 1] * nums[i - 1];
}
int[] result = new int[queries.length];
// Process queries
for (int q = 0; q < queries.length; q++) {
int i = queries[q][0];
int j = queries[q][1];
// Calculate range product
result[q] = prefixProduct[j + 1] / prefixProduct[i]; // Product from 0 to j / Product from 0 to (i-1)
}
return result;
}
}
Performance Analysis
The above solution demonstrates a way to efficiently process queries. Creating the initial prefix product array takes O(n) time, while each query is processed in O(1). The total time complexity becomes O(n + m). This method shows efficient performance as the number of queries increases.
Conclusion
In this tutorial, we explored how to solve the range product query problem. We learned an efficient way to process queries using the prefix product array. If you are interested in more complex data structures like segment trees or Fenwick trees, look forward to the next tutorial!
Next Steps
After this tutorial, it might be beneficial to explore alternative methods to solve this problem. By learning various methodologies, you can expand your algorithmic thinking skills. If you have any questions or topics for discussion, please leave a comment!