Remove an element to minimize the LCM of the given array

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

45

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



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Improved By : AnkitRai01, princiraj1992