Problem Description
This problem involves finding the product corresponding to a specific interval of a given number array.
For example, given the number array [2, 3, 4, 5],
the product corresponding to the interval [1, 3] is 3 * 4 * 5, and the result is 60.
The problem is as follows:
Input:
n: Size of the integer array (1 ≤n≤ 105)- Integer array
arr: An array of sizen(1 ≤arr[i]≤ 1000)m: Number of interval requests (1 ≤m≤ 105)- Interval request
[l, r](1 ≤l≤r≤n)Output: Calculate and output the product for each element’s interval.
Approach to the Problem
To solve the problem, we need to first consider an efficient method to calculate the product of the intervals of the number array.
If we simply try to calculate the product for the interval [l, r], the worst-case time complexity can become O(n * m), which may degrade performance.
Therefore, we need to find a method to derive results within linear time.
1. Cumulative Product Array
To efficiently find the interval product, we can use a Cumulative Product array.
By storing the product up to the i-th element of the array, we can compute the interval product as follows.
cumulativeProduct[i] = arr[0] * arr[1] * ... * arr[i]In this case, the interval product can be obtained as follows:
product(l, r) = cumulativeProduct[r] / cumulativeProduct[l-1]
Here, cumulativeProduct[0] is initialized to 1. This allows us to generate the cumulative product array in linear time O(n), and
compute the interval product in O(1) for each query.
2. Handling Negatives
However, when the signs are negative, the results can vary depending on the multiplication of negative and positive numbers.
In such cases, we need to determine the count of negatives to adjust the result.
If the count of negatives within the interval is odd, the result will be negative, and if it is even, it will be positive.
In this case, additional logic may be needed to select negatives to achieve the maximum product.
Implementation Code
Given the above approach, let’s implement the code in Kotlin.
fun getCumulativeProducts(arr: IntArray): IntArray {
val n = arr.size
val cumulativeProduct = IntArray(n)
cumulativeProduct[0] = arr[0]
for (i in 1 until n) {
cumulativeProduct[i] = cumulativeProduct[i - 1] * arr[i]
}
return cumulativeProduct
}Next, I will write a function that returns the interval product.
fun rangeProduct(cumulativeProduct: IntArray, l: Int, r: Int): Int {
if (l == 0) return cumulativeProduct[r]
return cumulativeProduct[r] / cumulativeProduct[l - 1]
}Finally, let’s combine the entire algorithm.
fun main() {
val arr = intArrayOf(2, 3, 4, 5)
val cumulativeProduct = getCumulativeProducts(arr)
val queries = listOf(Pair(1, 3), Pair(0, 2))
for (query in queries) {
val (l, r) = query
val result = rangeProduct(cumulativeProduct, l, r)
println("Product of interval ($l, $r): $result")
}
}Complexity Analysis
– Time Complexity: O(n + m) (where n is the size of the array and m is the number of requests)
– Space Complexity: O(n) (space required to store the cumulative product array)
Through this method, we can efficiently solve the problem.
By using the cumulative product array, we can calculate the product in one go, allowing for rapid results for each query.
Conclusion
In this lecture, we implemented an efficient algorithm using the cumulative product array to solve the interval product problem.
This approach can be useful for handling various array problems.
If you have any additional questions or concerns, please feel free to ask in the comments. Thank you!