Given an array of length N. The task is to convert it into a sequence in which all elements are greater than or equal to K.The only operation allowed is taking two smallest elements of the sequence and replace them by their LCM. Find the minimum number of operations required.
If it is impossible to get such an array, print -1.
Examples:
Input : N = 4, K = 3 , arr=[1 4 5 5]
Output : 1
LCM of 1 and 4 is 4, hence Replace (1,4) with 4.
Now the array becomes [4,4,5].
Every element in this array is greater than or equal to K.
No of operations required is equal to 1.Input : N = 5, K = 8 , arr=[4,4,4,4,4]
Output : -1
It is not possible to convert the given array.
Approach:
- The idea is to use a priority queue(min heap) which can handle delete and insert operation in log(N) time.
- The impossible case will arise when the number of elements in the priority queue is less than 2. The answer is equal to -1 in this case.
- Otherwise, take two elements from the top of the queue and replace it by their LCM.
- Do this until the smallest number that is top of the queue is less than K.
Below is the implementation of above approach:
C++
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std; // C++ function to get // minimum operation needed int FindMinOperation(int a[], int n, int k) { // The priority queue holds a minimum // element in the top position priority_queue<int, vector<int>, greater<int> > Q; for (int i = 0; i < n; i++) // push value one by one // from the given array Q.push(a[i]); // store count of minimum operation needed int ans = 0; while (1) { // All elements are now >= k if (Q.top() >= k) break; // It is impossible to make as there are // no sufficient elements available if (Q.size() < 2) return -1; // Take two smallest elements and // replace them by their LCM // first smallest element int x = Q.top(); Q.pop(); // Second smallest element int y = Q.top(); Q.pop(); int z = (x * y) / __gcd(x, y); Q.push(z); // Increment the count ans++; } return ans; } // Driver Code int main() { int a[] = { 3, 5, 7, 6, 8 }; int k = 8; int n = sizeof(a) / sizeof(a[0]); cout << FindMinOperation(a, n, k); } |
Java
// Java implementation of above approach import java.util.PriorityQueue; class GFG { // Function to calculate gcd of two numbers static int gcd(int a, int b) { if (a == 0) return b; return gcd(b % a, a); } // Java function to get // minimum operation needed static int FindMinOperation(int[] a, int n, int k) { // The priority queue holds a minimum // element in the top position PriorityQueue<Integer> Q = new PriorityQueue<>(); for (int i = 0; i < n; i++) // push value one by one // from the given array Q.add(a[i]); // store count of minimum operation needed int ans = 0; while (true) { // All elements are now >= k if (Q.peek() >= k) break; // It is impossible to make as there are // no sufficient elements available if (Q.size() < 2) return -1; // Take two smallest elements and // replace them by their LCM // first smallest element int x = Q.peek(); Q.poll(); // Second smallest element int y = Q.peek(); Q.poll(); int z = (x * y) / gcd(x, y); Q.add(z); // Increment the count ans++; } return ans; } // Driver code public static void main(String[] args) { int[] a = { 3, 5, 7, 6, 8 }; int k = 8; int n = a.length; System.out.println(FindMinOperation(a, n, k)); } } // This code is contributed by // sanjeev2552 |
2
Time Complexity: O(NlogN)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
Recommended Posts:
- Minimum number of given operations required to make two strings equal
- Minimum changes required to make all element in an array equal
- Minimum increment/decrement operations required to make Median as X
- Minimum operations to make two numbers equal
- Minimum deletions required to make GCD of the array equal to 1
- Minimum increment operations to make K elements equal
- Minimum increment by k operations to make all elements equal
- Minimum operations to make all elements equal using the second array
- Minimum operations to make frequency of all characters equal K
- Minimum number of given operations required to reduce the array to 0 element
- Minimum number of operations required to reduce N to 1
- Minimum number of operations required to obtain a given Binary String
- Minimum number of operations required to set all elements of a binary matrix
- Minimum number of operations required to delete all elements of the array
- Minimum operations to make GCD of array a multiple of k
- Minimum operations to make sum of neighbouring elements <= X
- Minimize swaps required to maximize the count of elements replacing a greater element in an Array
- Minimum operation required to make a balanced sequence
- Minimum removals required to make ranges non-overlapping
- Minimum number of pairs required to make two strings same
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.
Improved By : sanjeev2552
