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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 ``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

 ``
Output:
`45`

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

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