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Remove an element to minimize the LCM of the given array

  • Last Updated : 03 Jun, 2021

Given an array arr[] of length N ≥ 2. The task is to remove an element from the given array such that the LCM of the array after removing it is minimized.
Examples: 
 

Input: arr[] = {18, 12, 24} 
Output: 24 
Remove 12: LCM(18, 24) = 72 
Remove 18: LCM(12, 24) = 24 
Remove 24: LCM(12, 18) = 36
Input: arr[] = {5, 15, 9, 36} 
Output: 45 
 

 

Approach: 
 

  • Idea is to find the LCM value of all the sub-sequences of length (N – 1) and removing the element which is not present in the sub-sequence with that LCM. The minimum LCM found would be the answer.
  • To find the LCM of the sub-sequences optimally, maintain a prefixLCM[] and a suffixLCM[] array using single state dynamic programming.
  • The minimum value of LCM(prefixLCM[i – 1], suffixLCM[i + 1]) is the required answer.

Below is the implementation of the above approach: 
 

C++




// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the LCM of two numbers
int lcm(int a, int b)
{
    int GCD = __gcd(a, b);
    return (a * b) / GCD;
}
 
// Function to return the minimum LCM
// after removing a single element
// from the given array
int MinLCM(int a[], int n)
{
 
    // Prefix and Suffix arrays
    int Prefix[n + 2];
    int Suffix[n + 2];
 
    // Single state dynamic programming relation
    // for storing LCM of first i elements
    // from the left in Prefix[i]
    Prefix[1] = a[0];
    for (int i = 2; i <= n; i += 1) {
        Prefix[i] = lcm(Prefix[i - 1], a[i - 1]);
    }
 
    // Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    // Single state dynamic programming relation
    // for storing LCM of all the elements having
    // index greater than or equal to i in Suffix[i]
    for (int i = n - 1; i >= 1; i -= 1) {
        Suffix[i] = lcm(Suffix[i + 1], a[i - 1]);
    }
 
    // If first or last element of
    // the array has to be removed
    int ans = min(Suffix[2], Prefix[n - 1]);
 
    // If any other element is replaced
    for (int i = 2; i < n; i += 1) {
        ans = min(ans, lcm(Prefix[i - 1], Suffix[i + 1]));
    }
 
    // Return the minimum LCM
    return ans;
}
 
// Driver code
int main()
{
    int a[] = { 5, 15, 9, 36 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << MinLCM(a, n);
 
    return 0;
}

Java




// Java implementation of the above approach
class GFG
{
 
// Function to return the LCM of two numbers
static int lcm(int a, int b)
{
    int GCD = __gcd(a, b);
    return (a * b) / GCD;
}
 
// Function to return the minimum LCM
// after removing a single element
// from the given array
static int MinLCM(int a[], int n)
{
 
    // Prefix and Suffix arrays
    int []Prefix = new int[n + 2];
    int []Suffix = new int[n + 2];
 
    // Single state dynamic programming relation
    // for storing LCM of first i elements
    // from the left in Prefix[i]
    Prefix[1] = a[0];
    for (int i = 2; i <= n; i += 1)
    {
        Prefix[i] = lcm(Prefix[i - 1],
                             a[i - 1]);
    }
 
    // Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    // Single state dynamic programming relation
    // for storing LCM of all the elements having
    // index greater than or equal to i in Suffix[i]
    for (int i = n - 1; i >= 1; i -= 1)
    {
        Suffix[i] = lcm(Suffix[i + 1],
                             a[i - 1]);
    }
 
    // If first or last element of
    // the array has to be removed
    int ans = Math.min(Suffix[2],
                       Prefix[n - 1]);
 
    // If any other element is replaced
    for (int i = 2; i < n; i += 1)
    {
        ans = Math.min(ans, lcm(Prefix[i - 1],
                                Suffix[i + 1]));
    }
 
    // Return the minimum LCM
    return ans;
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver code
public static void main(String []args)
{
    int a[] = { 5, 15, 9, 36 };
    int n = a.length;
 
    System.out.println(MinLCM(a, n));
}
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 implementation of
# the above approach
from math import gcd
 
# Function to return the LCM
# of two numbers
def lcm(a, b) :
 
    GCD = gcd(a, b);
    return (a * b) // GCD;
 
# Function to return the minimum LCM
# after removing a single element
# from the given array
def MinLCM(a, n) :
 
    # Prefix and Suffix arrays
    Prefix = [0] * (n + 2);
    Suffix = [0] * (n + 2);
 
    # Single state dynamic programming relation
    # for storing LCM of first i elements
    # from the left in Prefix[i]
    Prefix[1] = a[0];
    for i in range(2, n + 1) :
        Prefix[i] = lcm(Prefix[i - 1],
                             a[i - 1]);
 
    # Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    # Single state dynamic programming relation
    # for storing LCM of all the elements having
    # index greater than or equal to i in Suffix[i]
    for i in range(n - 1, 0, -1) :
        Suffix[i] = lcm(Suffix[i + 1], a[i - 1]);
     
    # If first or last element of
    # the array has to be removed
    ans = min(Suffix[2], Prefix[n - 1]);
 
    # If any other element is replaced
    for i in range(2, n) :
        ans = min(ans, lcm(Prefix[i - 1],
                           Suffix[i + 1]));
     
    # Return the minimum LCM
    return ans;
 
# Driver code
if __name__ == "__main__" :
 
    a = [ 5, 15, 9, 36 ];
    n = len(a);
 
    print(MinLCM(a, n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the above approach
using System;
     
class GFG
{
 
// Function to return the LCM of two numbers
static int lcm(int a, int b)
{
    int GCD = __gcd(a, b);
    return (a * b) / GCD;
}
 
// Function to return the minimum LCM
// after removing a single element
// from the given array
static int MinLCM(int []a, int n)
{
 
    // Prefix and Suffix arrays
    int []Prefix = new int[n + 2];
    int []Suffix = new int[n + 2];
 
    // Single state dynamic programming relation
    // for storing LCM of first i elements
    // from the left in Prefix[i]
    Prefix[1] = a[0];
    for (int i = 2; i <= n; i += 1)
    {
        Prefix[i] = lcm(Prefix[i - 1],
                             a[i - 1]);
    }
 
    // Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    // Single state dynamic programming relation
    // for storing LCM of all the elements having
    // index greater than or equal to i in Suffix[i]
    for (int i = n - 1; i >= 1; i -= 1)
    {
        Suffix[i] = lcm(Suffix[i + 1],
                             a[i - 1]);
    }
 
    // If first or last element of
    // the array has to be removed
    int ans = Math.Min(Suffix[2],
                       Prefix[n - 1]);
 
    // If any other element is replaced
    for (int i = 2; i < n; i += 1)
    {
        ans = Math.Min(ans, lcm(Prefix[i - 1],
                                Suffix[i + 1]));
    }
 
    // Return the minimum LCM
    return ans;
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver code
public static void Main(String []args)
{
    int []a = { 5, 15, 9, 36 };
    int n = a.Length;
 
    Console.WriteLine(MinLCM(a, n));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// JavaScript implementation of the above approach
 
 
// Function to return the LCM of two numbers
function lcm(a, b) {
    let GCD = __gcd(a, b);
    return Math.floor((a * b) / GCD);
}
 
function __gcd(a, b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Function to return the minimum LCM
// after removing a single element
// from the given array
function MinLCM(a, n) {
 
    // Prefix and Suffix arrays
    let Prefix = new Array(n + 2);
    let Suffix = new Array(n + 2);
 
    // Single state dynamic programming relation
    // for storing LCM of first i elements
    // from the left in Prefix[i]
    Prefix[1] = a[0];
    for (let i = 2; i <= n; i += 1) {
        Prefix[i] = lcm(Prefix[i - 1], a[i - 1]);
    }
 
    // Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    // Single state dynamic programming relation
    // for storing LCM of all the elements having
    // index greater than or equal to i in Suffix[i]
    for (let i = n - 1; i >= 1; i -= 1) {
        Suffix[i] = lcm(Suffix[i + 1], a[i - 1]);
    }
 
    // If first or last element of
    // the array has to be removed
    let ans = Math.min(Suffix[2], Prefix[n - 1]);
 
    // If any other element is replaced
    for (let i = 2; i < n; i += 1) {
        ans = Math.min(ans, lcm(Prefix[i - 1], Suffix[i + 1]));
    }
 
    // Return the minimum LCM
    return ans;
}
 
// Driver code
 
let a = [5, 15, 9, 36];
let n = a.length;
 
document.write(MinLCM(a, n));
 
</script>
Output: 
45

 

Time Complexity: O(N * log(M)) where M is the maximum element from the array.
 


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