# Min operations to reduce N to 1 by multiplying by A or dividing by B

Given a number N and two integers A and B, the task is to check if it is possible to convert the number to 1 by the following two operations:

• Multiply it by A
• Divide it by B

If it is possible to reduce N to 1 then print the minimum number of operations required to achieve it otherwise print “-1”.

Examples:

Input: N = 48, A = 3, B = 12
Output: 3
Explanation:
Below are the 3 operations:
1. Divide 48 by 12 to get 4.
2. Multiply 4 by 3 to get 12.
3.Divide 12 by 12 to get 1.
Hence the total number of operation is 3.

Input: N = 26, A = 3, B = 9
Output: -1
Explanation:
It is not possible to convert 26 to 1.

Approach: The problem can be solved using Greedy Approach. The idea is to check if B is divisible by A or not and on the basis of that we have the below observations:

• If B%A != 0, then it is only possible to convert N to 1 if N is completely divisible by B and it would require N/B steps to do so. whereas if N = 1 then it would require 0 steps, otherwise it’s impossible and prints “-1”.
• If B%A == 0, then consider a variable C whose value is B/A. Divide N by B, using the second operation until it cannot be divided any further, let’s call the number of division as x.
• Again divide the remaining N by C until it cannot be divided any further, let’s call the number of divisions in this operation be y.
• If N does not equal 1 after the above operations then it is impossible to convert N to 1 using the above-mentioned operations and the answer will be “-1”, but if it is equal to 1 then we can use the formula total_steps = x + (2 * y)  to calculate the total minimum steps required.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach  ` `#include   ` `using` `namespace` `std;  ` ` `  `// Function to check if it is possible  ` `// to convert a number N to 1 by a minimum  ` `// use of the two operations  ` `int` `findIfPossible(``int` `n, ``int` `a, ``int` `b)  ` `{  ` `    ``// For the Case b % a != 0  ` `    ``if` `(b % a != 0) {  ` ` `  `        ``// Check if n equal to 1  ` `        ``if` `(n == 1)  ` `            ``return` `0;  ` ` `  `        ``// Check if n is not  ` `        ``// divisible by b  ` `        ``else` `if` `(n % b != 0)  ` `            ``return` `-1;  ` `        ``else` `            ``return` `(``int``)n / b;  ` `    ``}  ` ` `  `    ``// For the Case b % a == 0  ` ` `  `    ``// Initialize a variable 'c'  ` `    ``int` `c = b / a;  ` `    ``int` `x = 0, y = 0;  ` ` `  `    ``// Loop until n is divisible by b  ` `    ``while` `(n % b == 0) {  ` `        ``n = n / b;  ` ` `  `        ``// Count number of divisions  ` `        ``x++;  ` `    ``}  ` ` `  `    ``// Loop until n is divisible by c  ` `    ``while` `(n % c == 0) {  ` `        ``n = n / c;  ` ` `  `        ``// Count number of operations  ` `        ``y++;  ` `    ``}  ` ` `  `    ``// Check if n is reduced to 1  ` `    ``if` `(n == 1) {  ` ` `  `        ``// Count steps  ` `        ``int` `total_steps = x + (2 * y);  ` ` `  `        ``// Return the total number of steps  ` `        ``return` `total_steps;  ` `    ``}  ` `    ``else` `        ``return` `-1;  ` `}  ` ` `  `// Driver Code  ` `int` `main()  ` `{  ` `    ``// Given n, a and b  ` `    ``int` `n = 48;  ` `    ``int` `a = 3, b = 12;  ` ` `  `    ``// Function Call  ` `    ``cout << findIfPossible(n, a, b);  ` `    ``return` `0;  ` `} `

## Java

 `// Java program for the above approach  ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `     `  `// Function to check if it is possible  ` `// to convert a number N to 1 by a minimum  ` `// use of the two operations  ` `static` `int` `findIfPossible(``int` `n, ``int` `a, ``int` `b)  ` `{  ` `     `  `    ``// For the Case b % a != 0  ` `    ``if` `(b % a != ``0``)  ` `    ``{  ` ` `  `        ``// Check if n equal to 1  ` `        ``if` `(n == ``1``)  ` `            ``return` `0``;  ` ` `  `        ``// Check if n is not  ` `        ``// divisible by b  ` `        ``else` `if` `(n % b != ``0``)  ` `            ``return` `-``1``;  ` `        ``else` `            ``return` `(``int``)n / b;  ` `    ``}  ` ` `  `    ``// For the Case b % a == 0  ` ` `  `    ``// Initialize a variable 'c'  ` `    ``int` `c = b / a;  ` `    ``int` `x = ``0``, y = ``0``;  ` ` `  `    ``// Loop until n is divisible by b  ` `    ``while` `(n % b == ``0``)  ` `    ``{  ` `        ``n = n / b;  ` ` `  `        ``// Count number of divisions  ` `        ``x++;  ` `    ``}  ` ` `  `    ``// Loop until n is divisible by c  ` `    ``while` `(n % c == ``0``)  ` `    ``{  ` `        ``n = n / c;  ` ` `  `        ``// Count number of operations  ` `        ``y++;  ` `    ``}  ` ` `  `    ``// Check if n is reduced to 1  ` `    ``if` `(n == ``1``)  ` `    ``{  ` ` `  `        ``// Count steps  ` `        ``int` `total_steps = x + (``2` `* y);  ` ` `  `        ``// Return the total number of steps  ` `        ``return` `total_steps;  ` `    ``}  ` `    ``else` `        ``return` `-``1``;  ` `}  ` ` `  `// Driver Code ` `public` `static` `void` `main(String s[]) ` `{ ` `     `  `    ``// Given n, a and b  ` `    ``int` `n = ``48``;  ` `    ``int` `a = ``3``, b = ``12``;  ` `     `  `    ``// Function Call  ` `    ``System.out.println(findIfPossible(n, a, b)); ` `}  ` `} ` ` `  `// This code is contributed by rutvik_56 `

## Python3

 `# Python3 program for the above approach ` ` `  `# Function to check if it is possible  ` `# to convert a number N to 1 by a minimum  ` `# use of the two operations  ` `def` `FindIfPossible(n, a, b): ` `     `  `    ``# For the Case b % a != 0 ` `    ``if` `(b ``%` `a) !``=` `0``: ` `         `  `    ``# Check if n equal to 1 ` `        ``if` `n ``=``=` `1``: ` `            ``return` `0` `         `  `        ``# Check if n is not  ` `        ``# divisible by b  ` `        ``elif` `(n ``%` `b) !``=` `0``: ` `            ``return` `-``1` `        ``else``: ` `            ``return` `int``(n ``/` `b) ` `     `  `    ``# For the Case b % a == 0  ` `    ``# Initialize a variable 'c'  ` `    ``c ``=` `b ``/` `a ` `    ``x ``=` `0` `    ``y ``=` `0` `     `  `    ``# Loop until n is divisible by b  ` `    ``while` `(n ``%` `b ``=``=` `0``): ` `        ``n ``/``=` `b ` `         `  `    ``# Count number of divisions  ` `        ``x ``+``=` `1` `         `  `    ``# Loop until n is divisible by c ` `    ``while` `(n ``%` `c ``=``=` `0``): ` `        ``n ``/``=` `c ` `         `  `        ``# Count number of operations  ` `        ``y ``+``=` `1` `     `  `    ``# Check if n is reduced to 1  ` `    ``if` `n ``=``=` `1``: ` `         `  `        ``# Count steps ` `        ``total_steps ``=` `x ``+` `2` `*` `y ` `         `  `        ``# Return the total number of steps  ` `        ``return` `total_steps ` `    ``else``: ` `        ``return` `-``1` `         `  `# Driver code ` ` `  `# Given n, a and b  ` `n ``=` `48` `a ``=` `3` `b ``=` `12` ` `  `print``(FindIfPossible(n, a, b)) ` ` `  `# This code is contributed by virusbuddah_ `

## C#

 `// C# program for the above approach  ` `using` `System; ` ` `  `class` `GFG{  ` `     `  `// Function to check if it is possible  ` `// to convert a number N to 1 by a minimum  ` `// use of the two operations  ` `static` `int` `findIfPossible(``int` `n, ``int` `a, ``int` `b)  ` `{  ` `     `  `    ``// For the Case b % a != 0  ` `    ``if` `(b % a != 0)  ` `    ``{  ` ` `  `        ``// Check if n equal to 1  ` `        ``if` `(n == 1)  ` `            ``return` `0;  ` ` `  `        ``// Check if n is not  ` `        ``// divisible by b  ` `        ``else` `if` `(n % b != 0)  ` `            ``return` `-1;  ` `        ``else` `            ``return` `(``int``)n / b;  ` `    ``}  ` ` `  `    ``// For the Case b % a == 0  ` ` `  `    ``// Initialize a variable 'c'  ` `    ``int` `c = b / a;  ` `    ``int` `x = 0, y = 0;  ` ` `  `    ``// Loop until n is divisible by b  ` `    ``while` `(n % b == 0)  ` `    ``{  ` `        ``n = n / b;  ` ` `  `        ``// Count number of divisions  ` `        ``x++;  ` `    ``}  ` ` `  `    ``// Loop until n is divisible by c  ` `    ``while` `(n % c == 0)  ` `    ``{  ` `        ``n = n / c;  ` ` `  `        ``// Count number of operations  ` `        ``y++;  ` `    ``}  ` ` `  `    ``// Check if n is reduced to 1  ` `    ``if` `(n == 1)  ` `    ``{  ` ` `  `        ``// Count steps  ` `        ``int` `total_steps = x + (2 * y);  ` ` `  `        ``// Return the total number of steps  ` `        ``return` `total_steps;  ` `    ``}  ` `    ``else` `        ``return` `-1;  ` `}  ` ` `  `// Driver Code  ` `public` `static` `void` `Main()  ` `{  ` `     `  `    ``// Given n, a and b  ` `    ``int` `n = 48;  ` `    ``int` `a = 3, b = 12;  ` `     `  `    ``// Function call  ` `    ``Console.WriteLine(findIfPossible(n, a, b));  ` `}  ` `}  ` ` `  `// This code is contributed by Stream_Cipher `

Output:

```3
```

Time Complexity: O(log (B/A))
Auxiliary Space: O(1)

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