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.

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