Written on: March 10, 2023<\/p>\n
Author: Algorithm Master<\/p>\n<\/header>\n Given an integer array and an integer K, this problem requires us to find the remainder when each element of the array is divided by K and to compute the sum of these remainders. To solve this problem, we need to traverse the array, take the remainder for each element divided by K, and then accumulate these remainders for summation.<\/p>\n To solve this problem, we will adopt the following strategy.<\/p>\n Below is the code implemented in C++.<\/p>\n Here is a detailed explanation of each part of the code.<\/p>\n The time complexity of this algorithm is O(N), where N represents the number of elements in the array. It shows linear performance as it traverses each element once to compute the remainder and accumulates these values.<\/p>\n Below are additional test cases designed to verify results for various inputs.<\/p>\n Output:<\/strong> 0 (The remainder of all numbers divided by 1 is 0)\n <\/li>\n Output:<\/strong> 6 (5 % 3 = 2, 6 % 3 = 0, 7 % 3 = 1, 8 % 3 = 2, total sum = 2 + 0 + 1 + 2 = 5)\n <\/li>\nProblem Description<\/h2>\n
Problem Summary<\/h2>\n
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arr<\/code> and integer K<\/code><\/li>\nInput Example<\/h2>\n
[\n 3, 7, 2, 9, 10\n]\nK = 4\n<\/code><\/pre>\nOutput Example<\/h2>\n
Sum of remainders: 4\n<\/code><\/pre>\nProblem Solving Strategy<\/h2>\n
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C++ Code Implementation<\/h2>\n
#include <iostream>\n#include <vector>\n\nusing namespace std;\n\nint main() {\n vector<int> arr = {3, 7, 2, 9, 10}; \/\/ Input array\n int K = 4; \/\/ Value of K\n int remainderSum = 0; \/\/ Variable to store the sum of remainders\n \n \/\/ Traverse the array\n for (int num : arr) {\n remainderSum += num % K; \/\/ Calculate and accumulate the remainders\n }\n \n \/\/ Print the result\n cout << \"Sum of remainders: \" << remainderSum << endl;\n return 0;\n}\n<\/code><\/pre>\nMain Code Explanation<\/h2>\n
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#include <iostream><\/code>: Includes the iostream library for input and output functionality.<\/li>\n#include <vector><\/code>: Includes the vector library to use dynamic arrays.<\/li>\nusing namespace std;<\/code>: Uses the std namespace for cleaner code.<\/li>\nvector<int> arr = {3, 7, 2, 9, 10};<\/code>: Initializes the integer array to be used for calculations.<\/li>\nint K = 4;<\/code>: Declares the integer K for calculating remainders.<\/li>\nint remainderSum = 0;<\/code>: Initializes the variable to store the sum of remainders.<\/li>\nfor (int num : arr)<\/code>: Traverses all elements of the array.<\/li>\nremainderSum += num % K;<\/code>: Stores the remainder of each element when divided by K.<\/li>\ncout << \"Sum of remainders: \" << remainderSum << endl;<\/code>: Outputs the final result.<\/li>\n<\/ul>\nTime Complexity Analysis<\/h2>\n
Test Cases<\/h2>\n
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arr = [1, 2, 3], K = 1<\/code><\/pre>\narr = [5, 6, 7, 8], K = 3<\/code><\/pre>\narr = [10, 20, 30], K = 10<\/code><\/pre>\n