\n This problem involves finding the product corresponding to a specific interval of a given number array. \n The problem is as follows:\n <\/p>\n \nInput<\/strong>: <\/p>\n Output<\/strong>: Calculate and output the product for each element’s interval.\n <\/p><\/blockquote>\n \n To solve the problem, we need to first consider an efficient method to calculate the product of the intervals of the number array. \n To efficiently find the interval product, we can use a Cumulative Product array. \n In this case, the interval product can be obtained as follows:\n <\/p>\n \n Here, \n However, when the signs are negative, the results can vary depending on the multiplication of negative and positive numbers. \n If the count of negatives within the interval is odd, the result will be negative, and if it is even, it will be positive. \n Given the above approach, let\u2019s implement the code in Kotlin.\n <\/p>\n \n Next, I will write a function that returns the interval product.\n <\/p>\n \n Finally, let\u2019s combine the entire algorithm.\n <\/p>\n
\n For example, given the number array [2, 3, 4, 5]<\/code>,
\n the product corresponding to the interval [1, 3]<\/code> is 3 * 4 * 5<\/code>, and the result is 60<\/code>.\n <\/p>\n\n
n<\/code> : Size of the integer array (1 \u2264 n<\/code> \u2264 105<\/sup>)<\/li>\narr<\/code> : An array of size n<\/code> (1 \u2264 arr[i]<\/code> \u2264 1000)<\/li>\nm<\/code> : Number of interval requests (1 \u2264 m<\/code> \u2264 105<\/sup>)<\/li>\n[l, r]<\/code> (1 \u2264 l<\/code> \u2264 r<\/code> \u2264 n<\/code>)<\/li>\n<\/ul>\nApproach to the Problem<\/h2>\n
\n If we simply try to calculate the product for the interval [l, r]<\/code>, the worst-case time complexity can become O(n * m), which may degrade performance.
\n Therefore, we need to find a method to derive results within linear time.\n <\/p>\n1. Cumulative Product Array<\/h3>\n
\n By storing the product up to the i-th element of the array, we can compute the interval product as follows.\n <\/p>\ncumulativeProduct[i] = arr[0] * arr[1] * ... * arr[i]<\/code><\/pre>\nproduct(l, r) = cumulativeProduct[r] \/ cumulativeProduct[l-1]<\/code><\/pre>\ncumulativeProduct[0]<\/code> is initialized to 1. This allows us to generate the cumulative product array in linear time O(n), and
\n compute the interval product in O(1) for each query.\n <\/p>\n2. Handling Negatives<\/h3>\n
\n In such cases, we need to determine the count of negatives to adjust the result.\n <\/p>\n
\n In this case, additional logic may be needed to select negatives to achieve the maximum product.\n <\/p>\nImplementation Code<\/h2>\n
fun getCumulativeProducts(arr: IntArray): IntArray {\n val n = arr.size\n val cumulativeProduct = IntArray(n)\n cumulativeProduct[0] = arr[0]\n \n for (i in 1 until n) {\n cumulativeProduct[i] = cumulativeProduct[i - 1] * arr[i]\n }\n \n return cumulativeProduct\n}<\/code><\/pre>\nfun rangeProduct(cumulativeProduct: IntArray, l: Int, r: Int): Int {\n if (l == 0) return cumulativeProduct[r]\n return cumulativeProduct[r] \/ cumulativeProduct[l - 1]\n}<\/code><\/pre>\nfun main() {\n val arr = intArrayOf(2, 3, 4, 5)\n val cumulativeProduct = getCumulativeProducts(arr)\n val queries = listOf(Pair(1, 3), Pair(0, 2))\n\n for (query in queries) {\n val (l, r) = query\n val result = rangeProduct(cumulativeProduct, l, r)\n println(\"Product of interval ($l, $r): $result\")\n }\n}<\/code><\/pre>\nComplexity Analysis<\/h2>\n