Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Minimum operations to make GCD of array a multiple of k in C++
Suppose we have an array arr and another value k. We have to find a minimum number of operations to make the GCD of the array equal to the multiple of k. In this case, the operation is increasing or decreasing the value. Suppose the array is like {4, 5, 6}, and k is 5. We can increase 4 by 1, and decrease 6 by 1, so it becomes 5. Here a number of operations is 2.
We have to follow these steps to get the result −
Steps −
- for all elements e in the array, follow steps 2 and 3
- if e is not 1, and e > k, then increase result as min of (e mod k) and (k – e mod k).
- otherwise, result will be result + k – e
- return result
Example
#include <iostream>
using namespace std;
int countMinOp(int arr[], int n, int k) {
int result = 0;
for (int i = 0; i < n; ++i) {
if (arr[i] != 1 && arr[i] > k) {
result = result + min(arr[i] % k, k - arr[i] % k);
} else {
result = result + k - arr[i];
}
}
return result;
}
int main() {
int arr[] = { 4, 5, 6 };
int n = sizeof(arr) / sizeof(arr[0]);
int k = 5;
cout << "Minimum operation required: " << countMinOp(arr, n, k);
}Output
Minimum operation required: 2
Updated on: 21-Oct-2019
72 Views
Advertisements