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:



  1. Find the minimum element in the array which is not equal to INT_MAX.
  2. Remove all the elements (replace by INT_MAX) from the array divisible by the minimum element.
  3. 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++

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// 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

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// 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

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# 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#

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// 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)

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Improved By : AnkitRai01