# Minimum gcd operations to make all array elements one

Given an array A[] of size N. You can replace any number in the array with the gcd of that element with any of its adjacent elements. Find the minimum number of such operation to make the element of whole array equal to 1. If its not possible print -1.

Examples:

```Input : A[] = {4, 8, 9}
Output : 3
Explanation:
In the first move we choose (8, 9)
gcd(8, 9) = 1. Now the array becomes {4, 1, 9}.
After second move, the array becomes {1, 1, 9}.
After third move the array becomes {1, 1, 1}.

Input : A[] = { 5, 10, 2, 6 }
Output : 5
Explanation:
There is no pair with GCD equal one. We first
consider (5, 10) and replace 10 with 5. Now array
becomes { 5, 5, 2, 6 }. Now we consider pair (5, 2)
and replace 5 with 1, array becomes { 5, 1, 2, 6 }.
We have a 1, so further steps are simple.

Input : A[] = {8, 10, 12}
Output : -1
Explanation:
Its not possible to make all the element equal to 1.

Input : A[] = { 8, 10, 12, 6, 3 }
Output : 7
```

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

• If initially the array contains 1, our answer is the difference between the size of the array and count of ones in the array.
• If the array has no element equal to 1. We need to find the smallest sub array with GCD equal to one. Our result is N + (length of the minimum subarray with GCD 1) – 1. Example cases are { 5, 10, 2, 6 } and { 8, 10, 12, 6, 3 }.

We can find all the sub array in O(N^2) and GCD can be calculated in O(Log N) using Euclidean algorithms.
The overall complexity will be O(N^2 Log N).

Here is the implementation of the above idea.

## C++

 `// CPP program to find minimum GCD operations ` `// to make all array elements one. ` `#include ` `using` `namespace` `std; ` ` `  `// Function to count number of moves. ` `int` `minimumMoves(``int` `A[], ``int` `N) ` `{ ` `   ``// Counting Number of ones. ` `    ``int` `one = 0; ` `    ``for` `(``int` `i = 0; i < N; i++)  ` `        ``if` `(A[i] == 1) ` `            ``one++;  ` ` `  `    ``// If there is a one ` `    ``if` `(one != 0)  ` `        ``return` `N - one;  ` `     `  `    ``// Find smallest subarray with GCD equals ` `    ``// to one. ` `    ``int` `minimum = INT_MAX; ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `        ``int` `g = A[i]; ``// to calculate GCD ` `        ``for` `(``int` `j = i + 1; j < N; j++) { ` `            ``g = __gcd(A[j], g);  ` `            ``if` `(g == 1) { ` `                ``minimum = min(minimum, j - i);  ` `                ``break``; ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``if` `(minimum == INT_MAX) ``// Not Possible ` `        ``return` `-1; ` `    ``else` `        ``return` `N + minimum - 1; ``// Final answer ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `A[] = { 2, 4, 3, 9 }; ` `    ``int` `N = ``sizeof``(A) / ``sizeof``(A); ` `    ``cout << minimumMoves(A, N); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find minimum GCD operations ` `// to make all array elements one. ` `import` `java.util.*; ` ` `  `class` `GFG { ` `     `  `//__gcd function ` `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{ ` `    ``if` `(a == ``0``) ` `    ``return` `b; ` `    ``return` `__gcd(b % a, a); ` `} ` ` `  `// Function to count number of moves. ` `static` `int` `minimumMoves(``int` `A[], ``int` `N)  ` `{ ` `    ``// Counting Number of ones. ` `    ``int` `one = ``0``; ` `    ``for` `(``int` `i = ``0``; i < N; i++) ` `    ``if` `(A[i] == ``1``) ` `        ``one++; ` ` `  `    ``// If there is a one ` `    ``if` `(one != ``0``) ` `    ``return` `N - one; ` ` `  `    ``// Find smallest subarray with  ` `    ``// GCD equals to one. ` `    ``int` `minimum = Integer.MAX_VALUE; ` `    ``for` `(``int` `i = ``0``; i < N; i++) { ` `         `  `    ``// to calculate GCD ` `    ``int` `g = A[i];  ` `    ``for` `(``int` `j = i + ``1``; j < N; j++) { ` `        ``g = __gcd(A[j], g); ` `        ``if` `(g == ``1``) { ` `        ``minimum = Math.min(minimum, j - i); ` `        ``break``; ` `        ``} ` `    ``} ` `    ``} ` ` `  `    ``if` `(minimum == Integer.MAX_VALUE) ``// Not Possible ` `    ``return` `-``1``; ` `    ``else` `    ``return` `N + minimum - ``1``; ``// Final answer ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args)  ` `{ ` `    ``int` `A[] = {``2``, ``4``, ``3``, ``9``}; ` `    ``int` `N = A.length; ` `    ``System.out.print(minimumMoves(A, N)); ` `} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python program to find ` `# minimum GCD operations ` `# to make all array elements one. ` ` `  `#__gcd function ` `def` `__gcd(a,b): ` ` `  `    ``if` `(a ``=``=` `0``): ` `        ``return` `b ` `    ``return` `__gcd(b ``%` `a, a) ` `  `  `# Function to count number of moves. ` `def` `minimumMoves(A,N): ` ` `  `    ``# Counting Number of ones. ` `    ``one ``=` `0` `    ``for` `i ``in` `range``(N):  ` `        ``if` `(A[i] ``=``=` `1``): ` `            ``one``+``=``1` `  `  `    ``# If there is a one ` `    ``if` `(one !``=` `0``):  ` `        ``return` `N ``-` `one  ` `      `  `    ``# Find smallest subarray with GCD equals ` `    ``# to one. ` `    ``minimum ``=` `+``2147483647` `    ``for` `i ``in` `range``(N): ` `        ``g ``=` `A[i] ``# to calculate GCD ` `        ``for` `j ``in` `range``(i ``+` `1``,N): ` `            ``g ``=` `__gcd(A[j], g)  ` `            ``if` `(g ``=``=` `1``): ` `                ``minimum ``=` `min``(minimum, j ``-` `i)  ` `                ``break` `  `  `    ``if` `(minimum ``=``=` `+``2147483647``): ``# Not Possible ` `        ``return` `-``1` `    ``else``: ` `        ``return` `N ``+` `minimum ``-` `1``; ``# Final answer ` ` `  ` `  `# Driver program ` `A ``=` `[ ``2``, ``4``, ``3``, ``9` `] ` `N ``=` `len``(A) ` `print``(minimumMoves(A, N)) ` ` `  `# This code is contributed ` `# by Anant Agarwal. `

## C#

 `// C# program to find minimum GCD operations ` `// to make all array elements one. ` `using` `System; ` ` `  `class` `GFG { ` `     `  `//__gcd function ` `static` `int` `__gcd(``int` `a, ``int` `b)  ` `{ ` `    ``if` `(a == 0) ` `    ``return` `b; ` `    ``return` `__gcd(b % a, a); ` `} ` ` `  `// Function to count number of moves. ` `static` `int` `minimumMoves(``int` `[]A, ``int` `N)  ` `{ ` `    ``// Counting Number of ones. ` `    ``int` `one = 0; ` `    ``for` `(``int` `i = 0; i < N; i++) ` `    ``if` `(A[i] == 1) ` `        ``one++; ` ` `  `    ``// If there is a one ` `    ``if` `(one != 0) ` `    ``return` `N - one; ` ` `  `    ``// Find smallest subarray with  ` `    ``// GCD equals to one. ` `    ``int` `minimum = ``int``.MaxValue; ` `    ``for` `(``int` `i = 0; i < N; i++) { ` `         `  `    ``// to calculate GCD ` `    ``int` `g = A[i];  ` `    ``for` `(``int` `j = i + 1; j < N; j++) { ` `        ``g = __gcd(A[j], g); ` `        ``if` `(g == 1) { ` `        ``minimum = Math.Min(minimum, j - i); ` `        ``break``; ` `        ``} ` `    ``} ` `    ``} ` ` `  `    ``if` `(minimum == ``int``.MaxValue) ``// Not Possible ` `    ``return` `-1; ` `    ``else` `    ``return` `N + minimum - 1; ``// Final answer ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main()  ` `{ ` `    ``int` `[]A = {2, 4, 3, 9}; ` `    ``int` `N = A.Length; ` `    ``Console.WriteLine(minimumMoves(A, N)); ` `} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```4
```

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