# Maximum length subarray with LCM equal to product

Given an arr[], the task is to find the maximum length of the sub-array such that LCM of the sub-array is equal to the product of numbers in the sub-array. If no valid sub-array exists then print -1.
Note: The length of the sub-array must be ≥ 2.

Examples:

Input: arr[] = { 6, 10, 21 }
Output: 2
The sub-array { 10, 21 } satisfies the condition.

Input: arr[] = { 2, 2, 4 }
Output: -1
No sub-array satisfies the condition. Hence the output is -1.

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

Naive Approach: Run nested loops to check the condition for every possible sub-array of length ≥ 2. If the sub-array satisfies the condition, then update ans = max(ans, length(sub-array)). Print the ans in the end.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `#define ll long long ` ` `  `// Function to calculate gcd(a, b) ` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == 0) ` `        ``return` `a; ` `    ``return` `gcd(b, a % b); ` `} ` ` `  `// Function to return max length of subarray ` `// that satisfies the condition ` `int` `maxLengthSubArray(``const` `int``* arr, ``int` `n) ` `{ ` `    ``int` `maxLen = -1; ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` `            ``ll lcm = 1LL * arr[i]; ` `            ``ll product = 1LL * arr[i]; ` ` `  `            ``// Update LCM and product of the ` `            ``// sub-array ` `            ``for` `(``int` `k = i + 1; k <= j; k++) { ` `                ``lcm = (((arr[k] * lcm)) /  ` `                          ``(gcd(arr[k], lcm))); ` `                ``product = product * arr[k]; ` `            ``} ` ` `  `            ``// If the current sub-array satisfies  ` `            ``// the condition ` `            ``if` `(lcm == product) { ` ` `  `                ``// Choose the maximum ` `                ``maxLen = max(maxLen, j - i + 1); ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``return` `maxLen; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 6, 10, 21 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``cout << maxLengthSubArray(arr, n); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` `import` `java.util.*; ` ` `  `class` `GFG ` `{ ` ` `  `// Function to calculate gcd(a, b) ` `static` `int` `gcd(``int` `a, ``int` `b) ` `{ ` `    ``if` `(b == ``0``) ` `        ``return` `a; ` `    ``return` `gcd(b, a % b); ` `} ` ` `  `// Function to return max length of subarray ` `// that satisfies the condition ` `static` `int` `maxLengthSubArray(``int` `arr[], ``int` `n) ` `{ ` `    ``int` `maxLen = -``1``; ` `    ``for` `(``int` `i = ``0``; i < n - ``1``; i++)  ` `    ``{ ` `        ``for` `(``int` `j = i + ``1``; j < n; j++)  ` `        ``{ ` `            ``int` `lcm = ``1` `* arr[i]; ` `            ``int` `product = ``1` `* arr[i]; ` ` `  `            ``// Update LCM and product of the ` `            ``// sub-array ` `            ``for` `(``int` `k = i + ``1``; k <= j; k++)  ` `             `  `            ``{ ` `                ``lcm = (((arr[k] * lcm)) /  ` `                        ``(gcd(arr[k], lcm))); ` `                ``product = product * arr[k]; ` `            ``} ` ` `  `            ``// If the current sub-array satisfies  ` `            ``// the condition ` `            ``if` `(lcm == product) ` `            ``{ ` ` `  `                ``// Choose the maximum ` `                ``maxLen = Math.max(maxLen, j - i + ``1``); ` `            ``} ` `        ``} ` `    ``} ` `    ``return` `maxLen; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String args[]) ` `{ ` `    ``int` `arr[] = { ``6``, ``10``, ``21` `}; ` `    ``int` `n = arr.length; ` `    ``System.out.println(maxLengthSubArray(arr, n)); ` `} ` `} ` ` `  `// This code is contributed by ` `// Shashank_Sharma `

## Python3

 `# Python3 implementation of the  ` `# above approach ` ` `  `# Function to calculate gcd(a, b) ` `def` `gcd(a, b): ` `    ``if` `(b ``=``=` `0``): ` `        ``return` `a ` `    ``return` `gcd(b, a ``%` `b) ` ` `  `# Function to return max length of  ` `# subarray that satisfies the condition ` `def` `maxLengthSubArray(arr, n): ` ` `  `    ``maxLen ``=` `-``1` `    ``for` `i ``in` `range``(n ``-` `1``): ` `        ``for` `j ``in` `range``(n): ` `            ``lcm ``=` `arr[i] ` `            ``product ``=` `arr[i] ` ` `  `            ``# Update LCM and product of the ` `            ``# sub-array ` `            ``for` `k ``in` `range``(i ``+` `1``, j ``+` `1``): ` `                ``lcm ``=` `(((arr[k] ``*` `lcm)) ``/``/`  `                    ``(gcd(arr[k], lcm))) ` `                ``product ``=` `product ``*` `arr[k] ` `             `  `            ``# If the current sub-array satisfies  ` `            ``# the condition ` `            ``if` `(lcm ``=``=` `product):  ` ` `  `                ``# Choose the maximum ` `                ``maxLen ``=` `max``(maxLen, j ``-` `i ``+` `1``) ` `    ``return` `maxLen ` ` `  `# Driver code ` `arr ``=` `[``6``, ``10``, ``21` `] ` `n ``=` `len``(arr) ` `print``(maxLengthSubArray(arr, n)) ` ` `  `# This code is contributed by  ` `# mohit kumar 29 `

## C#

// C# implementation of the above approach
using System;

class GFG
{

// Function to calculate gcd(a, b)
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}

// Function to return max length of subarray
// that satisfies the condition
static int maxLengthSubArray(int[] arr, int n)
{
int maxLen = -1;
for (int i = 0; i < n - 1; i++) { for (int j = i + 1; j < n; j++) { int lcm = 1 * arr[i]; int product = 1 * arr[i]; // Update LCM and product of the // sub-array for (int k = i + 1; k <= j; k++) { lcm = (((arr[k] * lcm)) / (gcd(arr[k], lcm))); product = product * arr[k]; } // If the current sub-array satisfies // the condition if (lcm == product) { // Choose the maximum maxLen = Math.Max(maxLen, j - i + 1); } } } return maxLen; } // Driver code public static void Main() { int[] arr = { 6, 10, 21 }; int n = arr.Length; Console.Write(maxLengthSubArray(arr, n)); } } // This code is contributed by ita_c [tabby title="PHP"]

 ` `

Output:

```2
```

Efficient Approach: A sub-array will have its LCM equal to its product when no two elements in the sub-array have any common factor.
For example:

arr[] = { 6, 10, 21 }

Prime factorization yields:
arr[] = { 2 * 3, 2 * 5, 3 * 7 }

[6, 10] has 2 as a common factor.
[6, 10, 21] has 2 as a common factor between 6 and 10.
Sub-array [10, 21] has no common factor between any 2 elements. Therefore, answer = 2.

Firstly, prime factorization of numbers is done to deal with factors. To calculate the sub-array in which no 2 elements have a common factor, we use the two pointer technique.
Two pointers run, both from the right and they represent the current sub-array. We add elements in the sub-array from the right. Now there are two scenarios:

1. An element is added in the current sub-array if it has no factor in common to the current elements in the sub-array. If a common factor is found, then starting from the left, elements are subsequently eliminated until no common factor is found with the newly added element.
2. If there are no common factors between newly added element and existing elements, then update ans = max(ans, length of sub-array)

Below is the implementation of the above approach:

 `// C++ implementation of the above approach ` `#include ` ` `  `#define pb push_back ` `#define N 100005 ` `#define MAX 1000002 ` `#define mem(a, b) memset(a, b, sizeof(a)) ` ` `  `using` `namespace` `std; ` ` `  `int` `prime[MAX]; ` ` `  `// Stores array of primes for every element ` `vector<``int``> v[N]; ` ` `  `// Stores the position of a prime in the subarray ` `// in two pointer technique ` `int` `f[MAX]; ` ` `  `// Function to store smallest prime factor of numbers ` `void` `sieve() ` `{ ` `    ``prime[0] = prime[1] = 1; ` `    ``for` `(``int` `i = 2; i < MAX; i++) { ` `        ``if` `(!prime[i]) { ` `            ``for` `(``int` `j = i * 2; j < MAX; j += i) { ` `                ``if` `(!prime[j]) ` `                    ``prime[j] = i; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``for` `(``int` `i = 2; i < MAX; i++) { ` ` `  `        ``// If number is prime, ` `        ``// then smallest prime factor is the ` `        ``// number itself ` `        ``if` `(!prime[i]) ` `            ``prime[i] = i; ` `    ``} ` `} ` ` `  `// Function to return maximum length of subarray ` `// with LCM = product ` `int` `maxLengthSubArray(``int``* arr, ``int` `n) ` `{ ` `    ``// Initialize f with -1 ` `    ``mem(f, -1); ` ` `  `    ``for` `(``int` `i = 0; i < n; ++i) { ` ` `  `        ``// Prime factorization of numbers ` `        ``// Store primes in a vector for every element ` `        ``while` `(arr[i] > 1) { ` `            ``int` `p = prime[arr[i]]; ` `            ``arr[i] /= p; ` `            ``v[i].pb(p); ` `        ``} ` `    ``} ` ` `  `    ``// Two pointers l and r  ` `    ``// denoting left and right of subarray ` `    ``int` `l = 0, r = 1, ans = -1; ` ` `  `    ``// f is a mapping. ` `    ``// prime -> index in the current subarray ` `    ``// With the help of f, ` `    ``// we can detect whether a prime has ` `    ``// already occurred in the subarray ` `    ``for` `(``int` `i : v[0]) { ` `        ``f[i] = 0; ` `    ``} ` ` `  `    ``while` `(l <= r && r < n) { ` `        ``int` `flag = 0; ` ` `  `        ``for` `(``int` `i = 0; i < v[r].size(); i++) { ` ` `  `            ``// Map the prime to the index ` `            ``if` `(f[v[r][i]] == -1 || f[v[r][i]] == r) { ` `                ``f[v[r][i]] = r; ` `            ``} ` ` `  `            ``// If already occurred then, ` `            ``// start removing elements from the left ` `            ``else` `{ ` `                ``flag = 1; ` `                ``break``; ` `            ``} ` `        ``} ` ` `  `        ``// Remove elements if flag = 1 ` `        ``if` `(flag) { ` ` `  `            ``// Nullify entries of element at index 'l' ` `            ``for` `(``int` `i : v[l]) { ` `                ``f[i] = -1; ` `            ``} ` ` `  `            ``// Increment 'l' ` `            ``l++; ` `        ``} ` `        ``else` `{ ` ` `  `            ``// Maximize the answer when ` `            ``// no common factor is found ` `            ``ans = max(ans, r - l + 1); ` `            ``r++; ` `        ``} ` `    ``} ` ` `  `    ``// One length subarray is discarded ` `    ``if` `(ans == 1) ` `        ``ans = -1; ` ` `  `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``sieve(); ` `    ``int` `arr[] = { 6, 10, 21 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``cout << maxLengthSubArray(arr, n); ` `    ``return` `0; ` `} `

Output:

```2
```

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