JavaScript Coding Test Course, Calculating Interval Products

One of the problems often presented in coding interviews is the problem of finding the product of a specific range (subarray) of an array. In this course, we will define this as the Range Product Problem and explain how to solve it in detail.

Problem Definition

Given an array A and a list of queries, each query includes a range [L, R], and we need to calculate the product from A[L] to A[R]. Let’s assume the given array A and queries are as follows:

A = [2, 3, 4, 5, 6]
Queries = [(1, 3), (0, 2), (2, 4)] // Indices start from 0

We need to output the result of each query. For example, for the query (1, 3), A[1] * A[2] * A[3] = 3 * 4 * 5 = 60.

Understanding the Problem

To solve this problem, we need to understand a few conditions of the problem:

• Validation of the size of array A and the range (L, R) of each query
• How to quickly calculate the product of the range
• How to address the inefficiency issue of calculating multiple products when each element of the array is given

Solution Strategy

To efficiently calculate the range product, we can utilize the prefix product. This method generates a new array P(A) from the original array A, storing the product from index 0 to index i for each index i. Then the product for a query (L, R) can be calculated as follows.

Range Product Q(L, R) = P(R) / P(L - 1)

Here, P(i) represents the product up to index i, and P(0) is A[0].

Implementation

Now, let’s implement the above strategy in JavaScript:

function productRange(arr, queries) {
    // Array size
    const n = arr.length;

    // Initialize the prefix product array
    const prefixProduct = new Array(n);
    prefixProduct[0] = arr[0];

    // Calculate the prefix product
    for (let i = 1; i < n; i++) {
        prefixProduct[i] = prefixProduct[i - 1] * arr[i];
    }

    // Initialize the result array
    const result = [];

    // Process the queries
    queries.forEach(([L, R]) => {
        if (L === 0) {
            result.push(prefixProduct[R]);
        } else {
            result.push(prefixProduct[R] / prefixProduct[L - 1]);
        }
    });

    return result;
}

// Test
const A = [2, 3, 4, 5, 6];
const queries = [[1, 3], [0, 2], [2, 4]];
console.log(productRange(A, queries)); // Expected result: [60, 24, 120]

Analysis

The time complexity of the above code is O(N + Q), where N is the size of the array and Q is the number of queries. This is because calculating the prefix product takes O(N) time, and processing each query takes O(1) time.

This approach allows us to efficiently solve the range product problem, making it useful for query processing under various conditions.

Alternative Method

If there are changes made to the array, an efficient data structure may be needed for updates or dynamically increasing queries. In this case, a data structure such as a segment tree can be utilized. This allows both updates and query processing to be done in O(log N) time.

Conclusion

In this course, we discussed the problem of calculating range products in JavaScript. We demonstrated how to efficiently solve the problem using the prefix product and mentioned that it can be extended into more complex structures if necessary.

Try using these techniques to solve your own coding problems!