Minimum gcd operations to make all array elements one

16 Jun, 2022
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 
 
 
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++
Java
Python3
C#
PHP
Javascript
// CPP program to find minimum GCD operations// to make all array elements one.#include <bits/stdc++.h>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 codeint main(){    int A[] = { 2, 4, 3, 9 };    int N = sizeof(A) / sizeof(A[0]);    cout << minimumMoves(A, N);    return 0;}
// Java program to find minimum GCD operations// to make all array elements one.import java.util.*; class GFG {     //__gcd functionstatic 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 codepublic 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.
# Python program to find# minimum GCD operations# to make all array elements one. #__gcd functiondef __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 programA = [ 2, 4, 3, 9 ]N = len(A)print(minimumMoves(A, N)) # This code is contributed# by Anant Agarwal.
// C# program to find minimum GCD operations// to make all array elements one.using System; class GFG {     //__gcd functionstatic 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 codepublic 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// PHP program to find minimum// GCD operations to make all// array elements one. // __gcd functionfunction __gcd($a, $b){    if ($a == 0)    return $b;    return __gcd($b % $a, $a);} // Function to count// number of moves.function minimumMoves($A, $N){    // Counting Number of ones.    $one = 0;    for ($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.    $minimum = PHP_INT_MAX;    for ($i = 0; $i < $N; $i++)    {             // to calculate GCD    $g = $A[$i];    for ($j = $i + 1;         $j < $N; $j++)    {        $g = __gcd($A[$j], $g);        if ($g == 1)        {            $minimum = min($minimum,                           $j - $i);            break;        }    }    }     if ($minimum == PHP_INT_MAX) // Not Possible    return -1;    else    return $N + $minimum - 1; // Final answer} // Driver code$A = array(2, 4, 3, 9);$N = sizeof($A);echo (minimumMoves($A, $N)); // This code is contributed// by akt_mit.?>
<script> // JavaScript program to find minimum// GCD operations to make all// array elements one. // __gcd functionfunction __gcd(a, b){    if (a == 0)    return b;    return __gcd(b % a, a);} // Function to count// number of moves.function minimumMoves(A, N){    // Counting Number of ones.    let one = 0;    for (let 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.    let minimum = Number.MAX_SAFE_INTEGER;    for (let i = 0; i < N; i++)    {             // to calculate GCD    let g = A[i];    for (let j = i + 1;         j < N; j++)    {        g = __gcd(A[j], g);        if (g == 1)        {            minimum = Math.min(minimum,                           j - i);            break;        }    }    }     if (minimum == Number.MAX_SAFE_INTEGER) // Not Possible    return -1;    else    return N + minimum - 1; // Final answer} // Driver codelet A = [2, 4, 3, 9];let N = A.length;document.write(minimumMoves(A, N)); // This code is contributed by _saurabh_jaiswal. </script>
Output: 
4
Time Complexity: O(n2*log(n))
Auxiliary Space: O(1) 
vt_m
jit_t
_saurabh_jaiswal
ajaymakvana
GCD-LCM
Arrays
Mathematical
Arrays
Mathematical
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