Given an array **arr[] **of size** N**, the task is to split the entire array into a minimum number of subarrays such that for each subarray, the GCD of the first and last element of the subarray is greater than 1.

**Examples:**

Input:arr[] = {2, 3, 4, 4, 4, 3}Output:2Explanation:

Split the given array into [2, 3, 4, 4, 4] and [3].

The first subarray has gcd(2, 4) = 2 which is more than 1 and

The second subarray has gcd(3, 3) = 3 which is also more than 1.

Input:arr[] = {1, 2, 3}Output:0Explanation:

There is no possible splitting of the given array into subarrays in which the GCD of first and last element of the subarray is more than 1.

**Naive Approach:** The simplest approach to solve the problem is to perform all possible splits in the given array and for each split, check if all the subarrays have GCD of its first and last element greater than 1 or not. For all subarrays for which it is found to be true, store the count of subarrays. Finally, print the minimum count obtained among those splits.**Time Complexity:** O(2^{N}) **Auxiliary Space:** O(1)

**Efficient Approach:** The approach is based on the idea that the first and the last element of the original array will always be used. The first element will be used in the first subarray split, the last element will be used in the last subarray split. To minimize the count of valid subarrays, follow the steps below:

- Fix a
**right**pointer to the last element of the original array**arr[]**, and find the leftmost element in the original array such that**GCD(left, right) > 1**. If such an element is not found, there is no valid answer. - If such an element is found, that means we have found a valid subarray. Then change the
**right**pointer to**(left – 1)**, and again start searching for a valid subarray. - Repeat the above step until
**right**is more than**0**and keep on increasing the count of subarrays found till now. - Print the value of count after all the above steps.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the` `// minimum number of subarrays` `int` `minSubarrays(` `int` `arr[], ` `int` `n)` `{` ` ` `// Right pointer` ` ` `int` `right = n - 1;` ` ` `// Left pointer` ` ` `int` `left = 0;` ` ` `// Count of subarrays` ` ` `int` `subarrays = 0;` ` ` `while` `(right >= 0) {` ` ` `for` `(left = 0; left <= right;` ` ` `left += 1) {` ` ` `// Find GCD(left, right)` ` ` `if` `(__gcd(arr[left],` ` ` `arr[right])` ` ` `> 1) {` ` ` `// Found a valid large` ` ` `// subarray between` ` ` `// arr[left, right]` ` ` `subarrays += 1;` ` ` `right = left - 1;` ` ` `break` `;` ` ` `}` ` ` `// Searched the arr[0..right]` ` ` `// and found no subarray` ` ` `// having size >1 and having` ` ` `// gcd(left, right) > 1` ` ` `if` `(left == right` ` ` `&& __gcd(arr[left],` ` ` `arr[right])` ` ` `== 1) {` ` ` `return` `0;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `subarrays;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 6;` ` ` `int` `arr[] = { 2, 3, 4, 4, 4, 3 };` ` ` `// Function call` ` ` `cout << minSubarrays(arr, N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to find the` `// minimum number of subarrays` `static` `int` `minSubarrays(` `int` `arr[], ` `int` `n)` `{` ` ` `// Right pointer` ` ` `int` `right = n - ` `1` `;` ` ` `// Left pointer` ` ` `int` `left = ` `0` `;` ` ` `// Count of subarrays` ` ` `int` `subarrays = ` `0` `;` ` ` `while` `(right >= ` `0` `)` ` ` `{` ` ` `for` `(left = ` `0` `; left <= right; left += ` `1` `)` ` ` `{` ` ` `// Find GCD(left, right)` ` ` `if` `(__gcd(arr[left],` ` ` `arr[right]) > ` `1` `)` ` ` `{` ` ` `// Found a valid large` ` ` `// subarray between` ` ` `// arr[left, right]` ` ` `subarrays += ` `1` `;` ` ` `right = left - ` `1` `;` ` ` `break` `;` ` ` `}` ` ` `// Searched the arr[0..right]` ` ` `// and found no subarray` ` ` `// having size >1 and having` ` ` `// gcd(left, right) > 1` ` ` `if` `(left == right &&` ` ` `__gcd(arr[left],` ` ` `arr[right]) == ` `1` `)` ` ` `{` ` ` `return` `0` `;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `subarrays;` `}` `static` `int` `__gcd(` `int` `a, ` `int` `b) ` `{ ` ` ` `return` `b == ` `0` `? a : __gcd(b, a % b); ` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `N = ` `6` `;` ` ` `int` `arr[] = {` `2` `, ` `3` `, ` `4` `, ` `4` `, ` `4` `, ` `3` `};` ` ` `// Function call` ` ` `System.out.print(minSubarrays(arr, N));` `}` `}` `// This code is contributed by Rajput-Ji` |

## Python3

`# Python3 program for the above approach` `from` `math ` `import` `gcd` `# Function to find the` `# minimum number of subarrays` `def` `minSubarrays(arr, n):` ` ` ` ` `# Right pointer` ` ` `right ` `=` `n ` `-` `1` ` ` `# Left pointer` ` ` `left ` `=` `0` ` ` `# Count of subarrays` ` ` `subarrays ` `=` `0` ` ` `while` `(right >` `=` `0` `):` ` ` `for` `left ` `in` `range` `(right ` `+` `1` `):` ` ` `# Find GCD(left, right)` ` ` `if` `(gcd(arr[left], arr[right]) > ` `1` `):` ` ` `# Found a valid large` ` ` `# subarray between` ` ` `# arr[left, right]` ` ` `subarrays ` `+` `=` `1` ` ` `right ` `=` `left ` `-` `1` ` ` `break` ` ` `# Searched the arr[0..right]` ` ` `# and found no subarray` ` ` `# having size >1 and having` ` ` `# gcd(left, right) > 1` ` ` `if` `(left ` `=` `=` `right ` `and` ` ` `__gcd(arr[left],` ` ` `arr[right]) ` `=` `=` `1` `):` ` ` `return` `0` ` ` `return` `subarrays` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` ` ` `N ` `=` `6` ` ` `arr ` `=` `[ ` `2` `, ` `3` `, ` `4` `, ` `4` `, ` `4` `, ` `3` `]` ` ` `# Function call` ` ` `print` `(minSubarrays(arr, N))` ` ` `# This code is contributed by mohit kumar 29` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG{` `// Function to find the` `// minimum number of subarrays` `static` `int` `minSubarrays(` `int` `[] arr, ` `int` `n)` `{` ` ` `// Right pointer` ` ` `int` `right = n - 1;` ` ` `// Left pointer` ` ` `int` `left = 0;` ` ` `// Count of subarrays` ` ` `int` `subarrays = 0;` ` ` `while` `(right >= 0)` ` ` `{` ` ` `for` `(left = 0; left <= right; left += 1)` ` ` `{` ` ` `// Find GCD(left, right)` ` ` `if` `(__gcd(arr[left],` ` ` `arr[right]) > 1)` ` ` `{` ` ` `// Found a valid large` ` ` `// subarray between` ` ` `// arr[left, right]` ` ` `subarrays += 1;` ` ` `right = left - 1;` ` ` `break` `;` ` ` `}` ` ` `// Searched the arr[0..right]` ` ` `// and found no subarray` ` ` `// having size >1 and having` ` ` `// gcd(left, right) > 1` ` ` `if` `(left == right &&` ` ` `__gcd(arr[left],` ` ` `arr[right]) == 1)` ` ` `{` ` ` `return` `0;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `subarrays;` `}` ` ` `static` `int` `__gcd(` `int` `a, ` `int` `b) ` `{ ` ` ` `return` `b == 0 ? a : __gcd(b, a % b); ` `}` ` ` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` `int` `N = 6;` ` ` `int` `[] arr = {2, 3, 4, 4, 4, 3};` ` ` `// Function call` ` ` `Console.Write(minSubarrays(arr, N));` `}` `}` `// This code is contributed by Chitranayal` |

## Javascript

`<script>` `// javascript program for the above approach` ` ` `// Function to find the` ` ` `// minimum number of subarrays` ` ` `function` `minSubarrays(arr , n) {` ` ` `// Right pointer` ` ` `var` `right = n - 1;` ` ` `// Left pointer` ` ` `var` `left = 0;` ` ` `// Count of subarrays` ` ` `var` `subarrays = 0;` ` ` `while` `(right >= 0) {` ` ` `for` `(left = 0; left <= right; left += 1) {` ` ` `// Find GCD(left, right)` ` ` `if` `(__gcd(arr[left], arr[right]) > 1) {` ` ` `// Found a valid large` ` ` `// subarray between` ` ` `// arr[left, right]` ` ` `subarrays += 1;` ` ` `right = left - 1;` ` ` `break` `;` ` ` `}` ` ` `// Searched the arr[0..right]` ` ` `// and found no subarray` ` ` `// having size >1 and having` ` ` `// gcd(left, right) > 1` ` ` `if` `(left == right && __gcd(arr[left], arr[right]) == 1) {` ` ` `return` `0;` ` ` `}` ` ` `}` ` ` `}` ` ` `return` `subarrays;` ` ` `}` ` ` `function` `__gcd(a , b) {` ` ` `return` `b == 0 ? a : __gcd(b, a % b);` ` ` `}` ` ` `// Driver Code` ` ` ` ` `var` `N = 6;` ` ` `var` `arr = [ 2, 3, 4, 4, 4, 3 ];` ` ` `// Function call` ` ` `document.write(minSubarrays(arr, N));` `// This code contributed by umadevi9616` `</script>` |

**Output:**

2

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

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