Hello! In today’s lecture, we will solve an algorithm problem called “Sum of Remainders” using Java. This problem involves finding the remainders when a given integer array is divided by a specified number and then summing those remainders. This problem covers important basic concepts for preparing for coding tests.<\/p>\n
Given an integer array Print the sum of the remainders after dividing the elements of the array by To solve this problem, follow these steps:<\/p>\n Now, let’s implement the above strategy in Java. First, we will write a method to input the array and calculate the remainder for each element when divided by The time complexity of this programming problem is O(n)<\/strong>. Due to the length of the array, we need to iterate through all elements in the worst case. The space complexity is O(1)<\/strong> as there is almost no additional space required.<\/p>\n It is also good to consider a few additional test cases after implementing:<\/p>\n Today, we solved the \"Sum of Remainders\" problem using Java. Through this example, we learned how to effectively utilize array processing, loops, and remainder calculations. These types of problems can be transformed in various ways, helping to build adaptability. We hope to improve our skills by solving many different algorithm problems in the future!<\/p>\n Thank you!<\/p>\n","protected":false},"excerpt":{"rendered":"arr<\/code> and an integer m<\/code>, write a function that calculates the remainder of each element in the array when divided by m<\/code> and returns the total sum of those remainders.<\/p>\nInput<\/h3>\n
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n<\/code> (1 \u2264 n<\/code> \u2264 100,000).<\/li>\nn<\/code> integers arr[i]<\/code> (1 \u2264 arr[i]<\/code> \u2264 1,000,000) separated by spaces.<\/li>\nm<\/code> (1 \u2264 m<\/code> \u2264 1,000,000).<\/li>\n<\/ul>\nOutput<\/h3>\n
m<\/code>.<\/p>\nExample<\/h3>\n
\nInput:\n5\n1 2 3 4 5\n3\n\nOutput:\n15\n<\/pre>\n
Problem-Solving Strategy<\/h2>\n
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m<\/code>, calculate the remainders for all elements in the array using m<\/code>.<\/li>\nImplementation<\/h2>\n
m<\/code>. Below is the Java code that solves this problem:<\/p>\n\nimport java.util.Scanner;\n\npublic class RemainderSum {\n public static void main(String[] args) {\n Scanner scanner = new Scanner(System.in);\n \n \/\/ 1. Input\n int n = scanner.nextInt(); \/\/ size of the array\n int[] arr = new int[n];\n \n \/\/ Input array elements\n for (int i = 0; i < n; i++) {\n arr[i] = scanner.nextInt();\n }\n \n int m = scanner.nextInt(); \/\/ divisor\n\n \/\/ 2. Calculate sum of remainders\n long remainderSum = 0; \/\/ variable to store the result\n for (int i = 0; i < n; i++) {\n remainderSum += arr[i] % m; \/\/ add remainders\n }\n\n \/\/ 3. Print result\n System.out.println(remainderSum);\n \n scanner.close();\n }\n}\n<\/code><\/pre>\nCode Explanation<\/h2>\n
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Scanner<\/code> class is used to handle input. First, it reads the size of the array n<\/code>, then stores each element in arr<\/code>. Finally, it reads the integer m<\/code> that we want to divide by.<\/li>\nremainderSum<\/code> variable, a loop is used to add the remainders of each element in the array divided by m<\/code>. The long<\/code> type is used to safely store the sum of large numbers.<\/li>\nComplexity Analysis<\/h2>\n
Additional Example Tests<\/h2>\n
Example 1<\/h3>\n
\nInput:\n3\n5 10 15\n4\n\nOutput:\n6\n<\/pre>\n
Example 2<\/h3>\n
\nInput:\n4\n1 100 10000 1000000\n1\n\nOutput:\n0\n<\/pre>\n
Example 3<\/h3>\n
\nInput:\n5\n0 10 20 30 40\n7\n\nOutput:\n10\n<\/pre>\n
Conclusion<\/h2>\n