Minimum LCM of all subarrays of length at least 2

Given an array arr[] of N positive integers. The task is to find the minimum LCM of all subarrays of size greater than 1.

Examples: 

Input: arr[] = { 3, 18, 9, 18, 5, 15, 8, 7, 6, 9 } 
Output: 15 
Explanation: 
LCM of subarray {5, 15} is minimum which is 15.

Input: arr[] = { 4, 8, 12, 16, 20, 24 } 
Output:
Explanation: 
LCM of subarray {4, 8} is minimum which is 8.

 

Naive Approach: The idea is to generate all possible subarrays of length at least 2 and find the LCM of all the subarrays formed. Print the minimum LCM among all the subarrays.



Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach we have to observe that the LCM of two or more numbers will be less if and only if the number of elements whose LCM has to be calculated is minimum. The minimum possible value for subarray size is 2. Therefore the idea is to find the LCM of all the adjacent pairs and print the minimum of them.

Below is the implementation of the above approach:

C++

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// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to find LCM pf two numbers
int LCM(int a, int b)
{
    // Initialise lcm value
    int lcm = a > b ? a : b;
  
    while (true) {
  
        // Check for divisibility
        // of a and b by the lcm
        if (lcm % a == 0 && lcm % b == 0)
            break;
        else
            lcm++;
    }
    return lcm;
}
  
// Function to find the Minimum LCM of
// all subarrays of length greater than 1
void findMinLCM(int arr[], int n)
{
  
    // Store the minimum LCM
    int minLCM = INT_MAX;
  
    // Traverse the array
    for (int i = 0; i < n - 1; i++) {
  
        // Find LCM of consecutive element
        int val = LCM(arr[i], arr[i + 1]);
  
        // Check if the calculated LCM is
        // less than the minLCM then update it
        if (val < minLCM) {
            minLCM = val;
        }
    }
  
    // Print the minimum LCM
    cout << minLCM << endl;
}
  
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 4, 8, 12, 16, 20, 24 };
  
    // Size of the array
    int n = sizeof(arr) / sizeof(arr[0]);
  
    // Function Call
    findMinLCM(arr, n);
    return 0;
}

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Java

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// Java program for the above approach 
import java.util.*;
  
class GFG{ 
  
// Function to find LCM pf two numbers 
static int LCM(int a, int b) 
      
    // Initialise lcm value 
    int lcm = a > b ? a : b; 
  
    while (true)
    
          
        // Check for divisibility 
        // of a and b by the lcm 
        if (lcm % a == 0 && lcm % b == 0
            break
        else
            lcm++; 
    
    return lcm; 
  
// Function to find the Minimum LCM of 
// all subarrays of length greater than 1 
static void findMinLCM(int arr[], int n) 
  
    // Store the minimum LCM 
    int minLCM = Integer.MAX_VALUE; 
  
    // Traverse the array 
    for(int i = 0; i < n - 1; i++) 
    
          
        // Find LCM of consecutive element 
        int val = LCM(arr[i], arr[i + 1]); 
  
        // Check if the calculated LCM is 
        // less than the minLCM then update it 
        if (val < minLCM)
        
            minLCM = val; 
        
    
      
    // Print the minimum LCM 
    System.out.print(minLCM + "\n"); 
  
// Driver Code 
public static void main(String[] args) 
      
    // Given array arr[] 
    int arr[] = { 4, 8, 12, 16, 20, 24 }; 
  
    // Size of the array 
    int n = arr.length; 
  
    // Function call 
    findMinLCM(arr, n); 
}
}
  
// This code is contributed by amal kumar choubey

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Python3

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# Python3 program for the above approach
import sys
  
# Function to find LCM pf two numbers 
def LCM(a, b):
  
    # Initialise lcm value 
    lcm = a if a > b else
  
    while (True): 
  
        # Check for divisibility 
        # of a and b by the lcm 
        if (lcm % a == 0 and lcm % b == 0):
            break
        else:
            lcm += 1
      
    return lcm 
  
# Function to find the Minimum LCM of 
# all subarrays of length greater than 1 
def findMinLCM(arr, n): 
  
    # Store the minimum LCM 
    minLCM = sys.maxsize
  
    # Traverse the array 
    for i in range(n - 1): 
  
        # Find LCM of consecutive element 
        val = LCM(arr[i], arr[i + 1]) 
  
        # Check if the calculated LCM is 
        # less than the minLCM then update it 
        if (val < minLCM): 
            minLCM = val 
          
    # Print the minimum LCM 
    print(minLCM) 
  
# Driver Code 
  
# Given array arr[] 
arr = [ 4, 8, 12, 16, 20, 24
  
# Size of the array 
n = len(arr)
  
# Function call 
findMinLCM(arr, n) 
  
# This code is contributed by sanjoy_62

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C#

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// C# program for the above approach 
using System;
  
class GFG{ 
  
// Function to find LCM pf two numbers 
static int LCM(int a, int b) 
      
    // Initialise lcm value 
    int lcm = a > b ? a : b; 
  
    while (true)
    
          
        // Check for divisibility 
        // of a and b by the lcm 
        if (lcm % a == 0 && lcm % b == 0) 
            break
        else
            lcm++; 
    
    return lcm; 
  
// Function to find the Minimum LCM of 
// all subarrays of length greater than 1 
static void findMinLCM(int []arr, int n) 
  
    // Store the minimum LCM 
    int minLCM = int.MaxValue; 
  
    // Traverse the array 
    for(int i = 0; i < n - 1; i++) 
    
          
        // Find LCM of consecutive element 
        int val = LCM(arr[i], arr[i + 1]); 
  
        // Check if the calculated LCM is 
        // less than the minLCM then update it 
        if (val < minLCM)
        
            minLCM = val; 
        
    
      
    // Print the minimum LCM 
    Console.Write(minLCM + "\n"); 
  
// Driver Code 
public static void Main(String[] args) 
      
    // Given array []arr 
    int []arr = { 4, 8, 12, 16, 20, 24 }; 
  
    // Size of the array 
    int n = arr.Length; 
  
    // Function call 
    findMinLCM(arr, n); 
}
}
  
// This code is contributed by PrinciRaj1992

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Output: 

8

Time Complexity: O(N)
Auxiliary Space: O(1)

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