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: 8
Explanation:
LCM of subarray {4, 8} is minimum which is 8.
Brute Force Approach:
- initialize a variable min_lcm to INT_MAX.
- Next, we use nested loops to generate all possible subarrays of a length greater than 1.
- For each subarray, we calculate its LCM by iterating over its elements and calculating the LCM of each pair of elements using the LCM function defined earlier.
- We compare this LCM with the current minimum LCM and update it if the LCM is smaller.
- Finally, we print the minimum LCM obtained as the output.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to find the LCM of 2 numbers int LCM( int a, int b)
{ return (a * b) / __gcd(a, b);
} // Function to find the Minimum LCM of // all subarrays of length greater than 1 void findMinLCM( int arr[], int n)
{ int min_lcm = INT_MAX;
// Generate all possible subarrays
for ( int i = 0; i < n; i++)
{
for ( int j = i+1; j < n; j++)
{
int lcm = arr[i];
for ( int k = i+1; k <= j; k++)
{
lcm = LCM(lcm, arr[k]);
}
if (lcm < min_lcm)
{
min_lcm = lcm;
}
}
}
cout << min_lcm << 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;
} |
import java.util.*;
public class Main {
// Function to find the LCM of 2 numbers
public static int LCM( int a, int b) {
return (a * b) / gcd(a, b);
}
// Function to find the GCD of 2 numbers
public static int gcd( int a, int b) {
if (b == 0 ) {
return a;
} else {
return gcd(b, a % b);
}
}
// Function to find the Minimum LCM of
// all subarrays of length greater than 1
public static void findMinLCM( int [] arr, int n) {
int min_lcm = Integer.MAX_VALUE;
// Generate all possible subarrays
for ( int i = 0 ; i < n; i++) {
for ( int j = i+ 1 ; j < n; j++) {
int lcm = arr[i];
for ( int k = i+ 1 ; k <= j; k++) {
lcm = LCM(lcm, arr[k]);
}
if (lcm < min_lcm) {
min_lcm = lcm;
}
}
}
System.out.println(min_lcm);
}
// 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);
}
} |
# Python program for the above approach # Function to find the LCM of 2 numbers import math
def LCM(a, b):
return (a * b) / / math.gcd(a, b)
# Function to find the Minimum LCM of # all subarrays of length greater than 1 def findMinLCM(arr, n):
min_lcm = float ( 'inf' )
# Generate all possible subarrays
for i in range (n):
for j in range (i + 1 , n):
lcm = arr[i]
for k in range (i + 1 , j + 1 ):
lcm = LCM(lcm, arr[k])
if lcm < min_lcm:
min_lcm = lcm
print (min_lcm)
# Driver Code if __name__ = = '__main__' :
# Given array arr[]
arr = [ 4 , 8 , 12 , 16 , 20 , 24 ]
# Size of the array
n = len (arr)
# Function Call
findMinLCM(arr, n)
|
using System;
public class Program
{ // Function to find the LCM of 2 numbers public static int LCM( int a, int b)
{ return (a * b) / GCD(a, b);
} // Function to find the GCD of 2 numbers
public static int GCD( int a, int b)
{ if (a == 0) return b;
return GCD(b % a, a);
} // Function to find the Minimum LCM of // all subarrays of length greater than 1 public static void FindMinLCM( int [] arr, int n)
{ int min_lcm = int .MaxValue;
// Generate all possible subarrays
for ( int i = 0; i < n; i++)
{
for ( int j = i + 1; j < n; j++)
{
int lcm = arr[i];
for ( int k = i + 1; k <= j; k++)
{
lcm = LCM(lcm, arr[k]);
}
if (lcm < min_lcm)
{
min_lcm = lcm;
}
}
}
Console.WriteLine(min_lcm);
} public static void Main()
{ // Given array arr[]
int [] arr = { 4, 8, 12, 16, 20, 24 };
// Size of the array
int n = arr.Length;
// Function Call
FindMinLCM(arr, n);
} } |
// Function to find the LCM of 2 numbers function LCM(a, b) {
return (a * b) / gcd(a, b);
} // Function to find the GCD of 2 numbers function gcd(a, b) {
if (b === 0) {
return a;
} else {
return gcd(b, a % b);
}
} // Function to find the Minimum LCM of // all subarrays of length greater than 1 function findMinLCM(arr) {
let min_lcm = Number.MAX_VALUE;
// Generate all possible subarrays
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
let lcm = arr[i];
for (let k = i + 1; k <= j; k++) {
lcm = LCM(lcm, arr[k]);
}
if (lcm < min_lcm) {
min_lcm = lcm;
}
}
}
console.log(min_lcm);
} // Driver Code // Given array arr[] const arr = [4, 8, 12, 16, 20, 24]; // Function Call findMinLCM(arr); |
8
Time Complexity: O(N^3)
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++ 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;
} |
// 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 |
# 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 b
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 |
// 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 |
<script> // Javascript program for the above approach // Function to find LCM of two numbers function LCM(a, b)
{ // Initialise lcm value
let 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 function findMinLCM(arr, n)
{ // Store the minimum LCM
let minLCM = Number.MAX_VALUE;
// Traverse the array
for (let i = 0; i < n - 1; i++) {
// Find LCM of consecutive element
let 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
document.write(minLCM + "<br>" );
} // Driver Code // Given array arr[]
let arr = [ 4, 8, 12, 16, 20, 24 ];
// Size of the array
let n = arr.length;
// Function Call
findMinLCM(arr, n);
// This code is contributed by Mayank Tyagi </script> |
8
Time Complexity: O(N)
Auxiliary Space: O(1)