# Minimum LCM of all subarrays of length at least 2

Given an array arr[] of N positive integers. The task is to find the minimum LCM of all subarrays of size greater than 1.

Examples:

Input: arr[] = { 3, 18, 9, 18, 5, 15, 8, 7, 6, 9 }
Output: 15
Explanation:
LCM of subarray {5, 15} is minimum which is 15.

Input: arr[] = { 4, 8, 12, 16, 20, 24 }
Output:
Explanation:
LCM of subarray {4, 8} is minimum which is 8.

Naive Approach: The idea is to generate all possible subarrays of length at least 2 and find the LCM of all the subarrays formed. Print the minimum LCM among all the subarrays.

Time Complexity: O(N3)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach we have to observe that the LCM of two or more numbers will be less if and only if the number of elements whose LCM has to be calculated is minimum. The minimum possible value for subarray size is 2. Therefore the idea is to find the LCM of all the adjacent pairs and print the minimum of them.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` `#include ` `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); ` ` `  `    ``// 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 `

Output:

```8
```

Time Complexity: O(N)
Auxiliary Space: O(1)

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