# Minimum steps to reach N from 1 by multiplying each step by 2, 3, 4 or 5

Given an integer N, the task is to find the minimum number of steps to reach the number N from 1 by multiplying each step by 2, 3, 4 or 5. If it is not possible to reach N, print -1.
Examples:

Input: N = 10
Output:
Explanation:
Initial number = 1
Step 1: Multiply it by 2, Current Number = 2
Step 2: Multiply it by 5, Current Number = 10
Therefore, Minimum 2 steps required to reach 10.
Input: N = 13
Output: -1
Explanation:
There is no way reach 13 using any given operations

Approach: The idea is to use Greedy Algorithm to choose the operation that should be performed at each step and perform the operations in the reverse manner that is instead of going from 1 to N, find the operations required to reach N to 1. Below is the illustration of the steps:

• Apply the operations below until N is greater than 1.
• Check if N is divisible by 5, Then increase steps by 1 and reduce N to N/5
• Else, check if N is divisible by 4, Then increase steps by 1 and reduce N to N/4
• Else, check if N is divisible by 3, Then increase steps by 1 and reduce N to N/3
• Else, check if N is divisible by 2, Then increase steps by 1, and reduce N to N/2
• If at any step no operation can be applied then there is no possible set of operations to reach N from 1. Therefore, return -1.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find``// minimum number of steps``// to reach N from 1` `#include ` `using` `namespace` `std;` `// Function to find a minimum number``// of steps to reach N from 1``int` `Minsteps(``int` `n)``{``    ``int` `ans = 0;` `    ``// Check until N is greater``    ``// than 1 and operations``    ``// can be applied``    ``while` `(n > 1) {` `        ``// Condition to choose the``        ``// operations greedily``        ``if` `(n % 5 == 0) {` `            ``ans++;``            ``n = n / 5;``            ``continue``;``        ``}``        ``else` `if` `(n % 4 == 0) {``            ``ans++;``            ``n = n / 4;``            ``continue``;``        ``}``        ``else` `if` `(n % 3 == 0) {``            ``ans++;``            ``n = n / 3;``            ``continue``;``        ``}``        ``else` `if` `(n % 2 == 0) {``            ``ans++;``            ``n = n / 2;``            ``continue``;``        ``}``        ``return` `-1;``    ``}``    ``return` `ans;``}` `// Driver code``int` `main()``{``    ``int` `n = 10;``    ``cout << Minsteps(n);``    ``return` `0;``}`

## Java

 `// Java implementation to find``// minimum number of steps``// to reach N from 1` `import` `java.util.*;` `class` `GFG{` `// Function to find a minimum number``// of steps to reach N from 1``static` `int` `Minsteps(``int` `n)``{``    ``int` `ans = ``0``;` `    ``// Check until N is greater``    ``// than 1 and operations``    ``// can be applied``    ``while` `(n > ``1``)``    ``{``        ` `        ``// Condition to choose the``        ``// operations greedily``        ``if` `(n % ``5` `== ``0``)``        ``{``            ``ans++;``            ``n = n / ``5``;``            ``continue``;``        ``}``        ``else` `if` `(n % ``4` `== ``0``)``        ``{``            ``ans++;``            ``n = n / ``4``;``            ``continue``;``        ``}``        ``else` `if` `(n % ``3` `== ``0``) ``        ``{``            ``ans++;``            ``n = n / ``3``;``            ``continue``;``        ``}``        ``else` `if` `(n % ``2` `== ``0``) ``        ``{``            ``ans++;``            ``n = n / ``2``;``            ``continue``;``        ``}``        ``return` `-``1``;``    ``}``    ``return` `ans;``}` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `n = ``10``;``    ``System.out.print(Minsteps(n));``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 implementation to find ``# minimum number of steps ``# to reach N from 1 ` `# Function to find a minimum number ``# of steps to reach N from 1 ``def` `Minsteps(n): ` `    ``ans ``=` `0` `    ``# Check until N is greater ``    ``# than 1 and operations ``    ``# can be applied ``    ``while` `(n > ``1``):` `        ``# Condition to choose the ``        ``# operations greedily ``        ``if` `(n ``%` `5` `=``=` `0``): ``            ``ans ``=` `ans ``+` `1``            ``n ``=` `n ``/` `5``            ``continue` `        ``elif` `(n ``%` `4` `=``=` `0``): ``            ``ans ``=` `ans ``+` `1``            ``n ``=` `n ``/` `4``            ``continue` `        ``elif` `(n ``%` `3` `=``=` `0``):``            ``ans ``=` `ans ``+` `1``            ``n ``=` `n ``/` `3``            ``continue` `        ``elif` `(n ``%` `2` `=``=` `0``):``            ``ans ``=` `ans ``+` `1``            ``n ``=` `n ``/` `2``            ``continue` `        ``return` `-``1` `    ``return` `ans` `# Driver code ``n ``=` `10``print``(Minsteps(n))` `# This code is contributed by Pratik`

## C#

 `// C# implementation to find``// minimum number of steps``// to reach N from 1``using` `System;` `class` `GFG{` `// Function to find a minimum number``// of steps to reach N from 1``static` `int` `Minsteps(``int` `n)``{``    ``int` `ans = 0;` `    ``// Check until N is greater``    ``// than 1 and operations``    ``// can be applied``    ``while` `(n > 1)``    ``{``        ` `        ``// Condition to choose the``        ``// operations greedily``        ``if` `(n % 5 == 0)``        ``{``            ``ans++;``            ``n = n / 5;``            ``continue``;``        ``}``        ``else` `if` `(n % 4 == 0)``        ``{``            ``ans++;``            ``n = n / 4;``            ``continue``;``        ``}``        ``else` `if` `(n % 3 == 0) ``        ``{``            ``ans++;``            ``n = n / 3;``            ``continue``;``        ``}``        ``else` `if` `(n % 2 == 0) ``        ``{``            ``ans++;``            ``n = n / 2;``            ``continue``;``        ``}``        ``return` `-1;``    ``}``    ``return` `ans;``}` `// Driver code``public` `static` `void` `Main()``{``    ``int` `n = 10;``    ``Console.Write(Minsteps(n));``}``}` `// This code is contributed by rutvik_56`

## Javascript

 ``

Output:
`2`

Time Complexity: O(log n)

Auxiliary Space: O(1)

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