Dada una array arr[] de N enteros positivos. La tarea es encontrar el LCM mínimo de todos los subarreglos de tamaño mayor que 1.
Ejemplos:
Entrada: arr[] = { 3, 18, 9, 18, 5, 15, 8, 7, 6, 9 }
Salida: 15
Explicación:
LCM del subarreglo {5, 15} es el mínimo, que es 15.Entrada: arr[] = { 4, 8, 12, 16, 20, 24 }
Salida: 8
Explicación:
MCM del subarreglo {4, 8} es el mínimo, que es 8.
Enfoque ingenuo: la idea es generar todos los subarreglos posibles de longitud de al menos 2 y encontrar el LCM de todos los subarreglos formados. Imprime el LCM mínimo entre todos los subarreglos.
Complejidad de Tiempo: O(N 3 )
Espacio Auxiliar: O(1)
Enfoque eficiente: Para optimizar el enfoque anterior, debemos observar que el MCM de dos o más números será menor si y solo si el número de elementos cuyo MCM debe calcularse es mínimo. El valor mínimo posible para el tamaño del subarreglo es 2. Por lo tanto, la idea es encontrar el MCM de todos los pares adyacentes e imprimir el mínimo de ellos.
A continuación se muestra la implementación del enfoque anterior:
C++
// 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
// 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
# 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#
// 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
Javascript
<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
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por divyeshrabadiya07 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA