Partitions possible such that the minimum element divides all the other elements of the partition
Given an integer array arr[], the task is to count the number of partitions possible such that in each partition the minimum element divides all the other elements of the partition. The partition need not be continuous.
Examples:
Input: arr[] = {10, 7, 20, 21, 13}
Output: 3
The possible partitions are {10, 20}, {7, 21} and {13}.
In each partition, all the elements are divisible by
the minimum element of the partition.Input: arr[] = {7, 6, 5, 4, 3, 2, 2, 3}
Output: 4
Approach:
- Find the minimum element in the array which is not equal to INT_MAX.
- Remove all the elements (replace by INT_MAX) from the array divisible by the minimum element.
- The number of valid minimum elements as a result of the operations is the required number of partitions.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std; // Function to return the count partitions // possible from the given array such that // the minimum element of any partition // divides all the other elements // of that partition int countPartitions(int A[], int N) { // Initialize the count variable int count = 0; for (int i = 0; i < N; i++) { // Find the minimum element int min_elem = *min_element(A, A + N); // Break if no minimum element present if (min_elem == INT_MAX) break; // Increment the count if // minimum element present count++; // Replace all the element // divisible by min_elem for (int i = 0; i < N; i++) { if (A[i] % min_elem == 0) A[i] = INT_MAX; } } return count; } // Driver code int main() { int arr[] = { 7, 6, 5, 4, 3, 2, 2, 3 }; int N = sizeof(arr) / sizeof(arr[0]); cout << countPartitions(arr, N); return 0; } |
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Java
// Java implementation of the approach class GFG { static int INT_MAX = Integer.MAX_VALUE ; static int min_element(int []A, int N) { int min = A[0]; int i; for( i = 1; i < N ; i++) { if (min > A[i]) { min = A[i]; } } return min; } // Function to return the count partitions // possible from the given array such that // the minimum element of any partition // divides all the other elements // of that partition static int countPartitions(int []A, int N) { // Initialize the count variable int count = 0; int i, j; for (i = 0; i < N; i++) { // Find the minimum element int min_elem = min_element(A, N); // Break if no minimum element present if (min_elem == INT_MAX) break; // Increment the count if // minimum element present count++; // Replace all the element // divisible by min_elem for (j = 0; j < N; j++) { if (A[j] % min_elem == 0) A[j] = INT_MAX; } } return count; } // Driver code public static void main (String[] args) { int arr[] = { 7, 6, 5, 4, 3, 2, 2, 3 }; int N = arr.length; System.out.println(countPartitions(arr, N)); } } // This code is contributed by AnkitRai01 |
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Python3
# Python3 implementation of the approach import sys INT_MAX = sys.maxsize; # Function to return the count partitions # possible from the given array such that # the minimum element of any partition # divides all the other elements # of that partition def countPartitions(A, N) : # Initialize the count variable count = 0; for i in range(N) : # Find the minimum element min_elem = min(A); # Break if no minimum element present if (min_elem == INT_MAX) : break; # Increment the count if # minimum element present count += 1; # Replace all the element # divisible by min_elem for i in range(N) : if (A[i] % min_elem == 0) : A[i] = INT_MAX; return count; # Driver code if __name__ == "__main__" : arr = [ 7, 6, 5, 4, 3, 2, 2, 3 ]; N = len(arr); print(countPartitions(arr, N)); # This code is contributed by AnkitRai01 |
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C#
// C# implementation of the approach using System; class GFG { static int INT_MAX = int.MaxValue ; static int min_element(int []A, int N) { int min = A[0]; int i; for( i = 1; i < N ; i++) { if (min > A[i]) { min = A[i]; } } return min; } // Function to return the count partitions // possible from the given array such that // the minimum element of any partition // divides all the other elements // of that partition static int countPartitions(int []A, int N) { // Initialize the count variable int count = 0; int i, j; for (i = 0; i < N; i++) { // Find the minimum element int min_elem = min_element(A, N); // Break if no minimum element present if (min_elem == INT_MAX) break; // Increment the count if // minimum element present count++; // Replace all the element // divisible by min_elem for (j = 0; j < N; j++) { if (A[j] % min_elem == 0) A[j] = INT_MAX; } } return count; } // Driver code public static void Main() { int []arr = { 7, 6, 5, 4, 3, 2, 2, 3 }; int N = arr.Length; Console.WriteLine(countPartitions(arr, N)); } } // This code is contributed by AnkitRai01 |
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Output:
4
Time Complexity: O(N2)
