Hello! Today, I will address an algorithm problem on the topic of ‘Taking Pebbles’ for those preparing for a C# coding test. This article is carefully designed with various components. It will help provide a deep understanding through an overview of the problem, approach, implementation, examples, and explanations.<\/p>\n
Imagine a game where pebbles are randomly placed in different sections. The pebbles have a chaotic distribution. Users must take pebbles to create a box, but there are restrictions on the number of pebbles that can be placed in the box. When taking pebbles, the user must always take a fixed number from each section, and the function must calculate and output the number of possible ways to select pebbles that satisfy this condition.<\/p>\n
We have an integer array To solve this problem, we will use the combination formula to calculate the number of ways to take pebbles from each section. If there are multiple sections, we can find the total number of cases by multiplying the combinations for each section. This is based on mathematical concepts and helps us efficiently calculate the number of combinations.<\/p>\n The number of combinations Here, Now, let’s actually write the code. We will implement a function that meets the requirements using the C# language.<\/p>\n We will look at a few test cases to verify that the code we wrote works correctly.<\/p>\n Input:<\/strong><\/p>\n Output:<\/strong> Explanation:<\/strong> The number of ways to take 2 from section 0: C(5, 2) = 10, the number of ways to take 1 from section 1: C(3, 1) = 3, the number of ways to take 3 from section 2: C(8, 3) = 56. Total number of ways = 10 * 3 * 56 = 1680.<\/p>\nstones<\/code> representing the number of pebbles in each section and an integer array pick<\/code> defining the number of pebbles to take from each section. The lengths of both arrays are equal, and you need to write a function countWays(stones, pick)<\/code> that calculates and returns the number of ways to take pebbles from each section.<\/p>\nInput<\/h3>\n
\n
stones<\/code>: An integer array representing the number of pebbles in each section.<\/li>\npick<\/code>: An integer array representing the number of pebbles to take from each section.<\/li>\n<\/ul>\nOutput<\/h3>\n
\n
Constraints<\/h3>\n
\n
stones.Length<\/code>, pick.Length<\/code> \u2264 50<\/li>\nstones[i]<\/code> \u2264 100<\/li>\npick[i]<\/code> \u2264 stones[i]<\/code><\/li>\n<\/ul>\nApproach<\/h2>\n
C(n, k)<\/code> is defined by the following formula:<\/p>\nC(n, k) = n! \/ (k! * (n - k)!)<\/code><\/p>\nn<\/code> represents the number of pebbles, and k<\/code> represents the number of pebbles to be taken. By calculating the combinations for each section and multiplying them, we derive the final number of cases.<\/p>\nImplementation<\/h2>\n
\nusing System;\n\npublic class Solution {\n public int countWays(int[] stones, int[] pick) {\n int totalWays = 1; \/\/ Initial value to multiply all cases\n\n for (int i = 0; i < stones.Length; i++) {\n totalWays *= Combinations(stones[i], pick[i]);\n }\n\n return totalWays;\n }\n\n \/\/ Helper method to calculate combinations\n private int Combinations(int n, int k) {\n if (k > n) return 0;\n if (k == 0 || k == n) return 1;\n\n int numerator = Factorial(n);\n int denominator = Factorial(k) * Factorial(n - k);\n return numerator \/ denominator;\n }\n \n \/\/ Method to calculate factorial\n private int Factorial(int num) {\n if (num == 0) return 1;\n int result = 1;\n for (int i = 1; i <= num; i++) {\n result *= i;\n }\n return result;\n }\n}\n<\/code><\/pre>\nExamples<\/h2>\n
Example 1<\/h3>\n
\nstones = [5, 3, 8]\npick = [2, 1, 3]\n<\/code><\/pre>\n840<\/code><\/p>\nExample 2<\/h3>\n