# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 = a; ` `    ``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, 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); ` ` `  `    ``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 = a; ` `    ``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,  ` `                       ``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 `

Output:

```45
```

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

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