Minimum steps to minimize n as per given condition

Given a number n, count minimum steps to minimize it to 1 according to the following criteria:

If n is divisible by 2 then we may reduce n to n/2.
If n is divisible by 3 then you may reduce n to n/3.
Decrement n by 1.

Examples:

Input : n = 10
Output : 3

Input : 6
Output : 2

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

Greedy Approach (Doesn’t work always) :
As per greedy approach we may choose the step that makes n as low as possible and continue the same, till it reaches 1.

while ( n > 1)
{
    if (n % 3 == 0)
        n /= 3;    
    else if (n % 2 == 0)
        n /= 2;
    else
        n--;
    steps++;
}

If we observe carefully, the greedy strategy doesn’t work here.
Eg: Given n = 10 , Greedy –> 10 /2 = 5 -1 = 4 /2 = 2 /2 = 1 ( 4 steps ).
But the optimal way is –> 10 -1 = 9 /3 = 3 /3 = 1 ( 3 steps ).
So, we must think of dynamic approach for optimal solution.

Dynamic Approach:
For finding minimum steps we have three possibilities for n and they are:

f(n) = 1 + f(n-1)
f(n) = 1 + f(n/2) // if n is divisible by 2
f(n) = 1 + f(n/3) // if n is divisible by 3

Below is memoization based implementation of above recursive formula.

C++

// CPP program to minimize n to 1 by given 
// rule in minimum steps 
#include <bits/stdc++.h> 
using namespace std; 
  
// function to calculate min steps 
int getMinSteps(int n, int *memo) 
{ 
    // base case 
    if (n == 1) 
       return 0; 
    if (memo[n] != -1) 
       return memo[n]; 
  
    // store temp value for n as min( f(n-1), 
    // f(n/2), f(n/3)) +1 
    int res = getMinSteps(n-1, memo); 
  
    if (n%2 == 0) 
        res = min(res, getMinSteps(n/2, memo)); 
    if (n%3 == 0) 
        res = min(res, getMinSteps(n/3, memo)); 
  
    // store memo[n] and return 
    memo[n] = 1 + res; 
    return memo[n]; 
} 
  
// This function mainly initializes memo[] and 
// calls getMinSteps(n, memo) 
int getMinSteps(int n) 
{ 
    int memo[n+1]; 
  
    // initialize memoized array 
    for (int i=0; i<=n; i++) 
        memo[i] = -1; 
  
    return  getMinSteps(n, memo); 
} 
  
// driver program 
int main() 
{ 
    int n = 10; 
    cout << getMinSteps(n); 
    return 0; 
} 

Java

// Java program to minimize n to 1  
// by given rule in minimum steps 
import java.io.*; 
class GFG { 
  
// function to calculate min steps 
static int getMinSteps(int n, int memo[]) 
{ 
    // base case 
    if (n == 1) 
    return 0; 
    if (memo[n] != -1) 
    return memo[n]; 
  
    // store temp value for 
    // n as min( f(n-1), 
    // f(n/2), f(n/3)) +1 
    int res = getMinSteps(n - 1, memo); 
  
    if (n % 2 == 0) 
        res = Math.min(res,  
              getMinSteps(n / 2, memo)); 
    if (n % 3 == 0) 
        res = Math.min(res,  
               getMinSteps(n / 3, memo)); 
  
    // store memo[n] and return 
    memo[n] = 1 + res; 
    return memo[n]; 
} 
  
// This function mainly 
// initializes memo[] and 
// calls getMinSteps(n, memo) 
static int getMinSteps(int n) 
{ 
    int memo[] = new int[n + 1]; 
  
    // initialize memoized array 
    for (int i = 0; i <= n; i++) 
        memo[i] = -1; 
  
    return getMinSteps(n, memo); 
} 
  
    // Driver Code 
    public static void main (String[] args)  
    { 
        int n = 10; 
        System.out.println(getMinSteps(n)); 
    } 
} 
  
// This code is contributed by anuj_67. 

Python3

# Python program to minimize 
# n to 1 by given 
# rule in minimum steps 
  
# function to calculate min steps 
def getMinSteps(n, memo): 
  
    # base case 
    if (n == 1): 
        return 0
    if (memo[n] != -1): 
        return memo[n] 
  
    # store temp value for n as min(f(n-1), 
    # f(n//2), f(n//3)) + 1 
    res = getMinSteps(n-1, memo) 
  
    if (n%2 == 0): 
        res = min(res, getMinSteps(n//2, memo)) 
    if (n%3 == 0): 
        res = min(res, getMinSteps(n//3, memo)) 
  
    # store memo[n] and return 
    memo[n] = 1 + res 
    return memo[n] 
  
# This function mainly 
# initializes memo[] and 
# calls getMinSteps(n, memo) 
def getsMinSteps(n): 
  
    memo = [0 for i in range(n+1)] 
  
    # initialize memoized array 
    for i in range(n+1): 
        memo[i] = -1
  
    return getMinSteps(n, memo) 
  
# driver program 
n = 10
print(getsMinSteps(n)) 
  
# This code is contributed by Soumen Ghosh.    

C#

// C# program to minimize n to 1  
// by given rule in minimum steps 
using System; 
  
class GFG { 
  
    // function to calculate min steps 
    static int getMinSteps(int n, int []memo) 
    { 
        // base case 
        if (n == 1) 
            return 0; 
        if (memo[n] != -1) 
            return memo[n]; 
      
        // store temp value for 
        // n as min( f(n-1), 
        // f(n/2), f(n/3)) +1 
        int res = getMinSteps(n - 1, memo); 
      
        if (n % 2 == 0) 
            res = Math.Min(res,  
                getMinSteps(n / 2, memo)); 
        if (n % 3 == 0) 
            res = Math.Min(res,  
                getMinSteps(n / 3, memo)); 
      
        // store memo[n] and return 
        memo[n] = 1 + res; 
          
        return memo[n]; 
    } 
      
    // This function mainly 
    // initializes memo[] and 
    // calls getMinSteps(n, memo) 
    static int getMinSteps(int n) 
    { 
        int []memo = new int[n + 1]; 
      
        // initialize memoized array 
        for (int i = 0; i <= n; i++) 
            memo[i] = -1; 
      
        return getMinSteps(n, memo); 
    } 
  
    // Driver Code 
    public static void Main ()  
    { 
        int n = 10; 
        Console.WriteLine(getMinSteps(n)); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// PHP program to minimize n to 1 by 
// given rule in minimum steps 
  
// function to calculate min steps 
function getMinSteps( $n, $memo) 
{ 
      
    // base case 
    if ($n == 1) 
        return 0; 
          
    if ($memo[$n] != -1) 
        return $memo[$n]; 
  
    // store temp value for n 
    // as min( f(n-1), 
    // f(n/2), f(n/3)) +1 
    $res = getMinSteps($n - 1, $memo); 
  
    if ($n % 2 == 0) 
        $res = min($res, getMinSteps($n / 2, $memo)); 
    if ($n % 3 == 0) 
        $res = min($res, getMinSteps($n / 3, $memo)); 
  
    // store memo[n] and return 
    $memo[$n] = 1 + $res; 
    return $memo[$n]; 
} 
  
// This function mainly initializes 
// memo[] and calls getMinSteps(n, memo) 
function g_etMinSteps( $n) 
{ 
    $memo= array(); 
  
    // initialize memoized array 
    for($i = 0; $i <= $n; $i++) 
        $memo[$i] = -1; 
  
    return getMinSteps($n, $memo); 
} 
  
    // Driver Code 
    $n = 10; 
    echo g_etMinSteps($n); 
  
// This code is contributed by anuj_67. 
?> 

Output:

Below is a tabulation based solution :

C++

// A tabulation based solution in C++ 
#include <bits/stdc++.h> 
using namespace std; 
   
int getMinSteps(int n) 
{ 
    int table[n+1]; 
    for (int i=0; i<=n; i++) 
        table[i] = n-i; 
    for (int i=n; i>=1; i--) 
    { 
       if (!(i%2)) 
          table[i/2] = min(table[i]+1, table[i/2]); 
       if (!(i%3)) 
          table[i/3] = min(table[i]+1, table[i/3]); 
    } 
    return table[1]; 
} 
   
// driver program 
int main() 
{ 
    int n = 10; 
    cout << getMinSteps(n); 
    return 0; 
} 
// This code is contributed by Jaydeep Rathore 

Java

// A tabulation based  
// solution in Java 
import java.io.*; 
  
class GFG  
{ 
static int getMinSteps(int n) 
{ 
    int table[] = new int[n + 1]; 
    for (int i = 0; i <= n; i++) 
        table[i] = n - i; 
    for (int i = n; i >= 1; i--) 
    { 
    if (!(i % 2 > 0)) 
        table[i / 2] = Math.min(table[i] + 1,  
                                table[i / 2]); 
    if (!(i % 3 > 0)) 
        table[i / 3] = Math.min(table[i] + 1,  
                                table[i / 3]); 
    } 
    return table[1]; 
} 
  
// Driver Code 
public static void main (String[] args) 
{ 
    int n = 10; 
    System.out.print(getMinSteps(n)); 
} 
} 
  
// This code is contributed  
// by anuj_67. 

Python3

# A tabulation based solution in Python3  
  
def getMinSteps(n) : 
      
    table = [0] * (n + 1)  
      
    for i in range(n + 1) : 
        table[i] = n-i 
          
    for i in range(n, 0, -1) : 
          
        if (not(i%2)) : 
            table[i//2] = min(table[i]+1, table[i//2]) 
              
        if (not(i%3)) : 
            table[i//3] = min(table[i]+1, table[i//3]) 
              
    return table[1] 
  
  
# driver program  
if __name__ == "__main__" : 
  
    n = 10 
    print(getMinSteps(n)) 
      
# This code is contributed by Ryuga 

C#

// A tabulation based  
// solution in C# 
using System; 
  
class GFG  
{ 
static int getMinSteps(int n) 
{ 
    int []table = new int[n + 1]; 
    for (int i = 0; i <= n; i++) 
        table[i] = n - i; 
    for (int i = n; i >= 1; i--) 
    { 
    if (!(i % 2 > 0)) 
        table[i / 2] = Math.Min(table[i] + 1,  
                                table[i / 2]); 
    if (!(i % 3 > 0)) 
        table[i / 3] = Math.Min(table[i] + 1,  
                                table[i / 3]); 
    } 
    return table[1]; 
} 
  
// Driver Code 
public static void Main () 
{ 
    int n = 10; 
    Console.WriteLine(getMinSteps(n)); 
} 
} 
  
// This code is contributed  
// by anuj_67. 

PHP

<?php 
// A tabulation based solution in PHP 
  
function getMinSteps( $n) 
{ 
    $table = array(); 
    for ($i = 0; $i <= $n; $i++) 
        $table[$i] = $n - $i; 
    for ($i = $n; $i >= 1; $i--) 
    { 
        if (!($i % 2)) 
            $table[$i / 2] = min($table[$i] +  
                          1, $table[$i / 2]); 
        if (!($i % 3)) 
            $table[$i / 3] = min($table[$i] +  
                          1, $table[$i / 3]); 
    } 
    return $table[1]; 
} 
  
    // Driver Code 
    $n = 10; 
    echo getMinSteps($n); 
  
// This code is contributed by anuj_67. 
?> 

Output:

This article is contributed by Shivam Pradhan (anuj_charm). If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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