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Minimum Steps to obtain N from 1 by the given operations
  • Last Updated : 13 Jan, 2021

Given an integer N, the task is to find the minimum number of operations needed to obtain the number N starting from 1. Below are the operations: 

  • Add 1 to the current number.
  • Multiply the current number by 2.
  • Multiply the current number by 3.

Print the minimum number of operations required and the corresponding sequence to obtain N.

Examples:  

Input: N = 3 
Output: 

1 3 
Explanation: 
Operation 1: Multiply 1 * 3 = 3. 
Hence, only 1 operation is required.

Input: N = 5 
Output: 

1 2 4 5 
Explanation: 
The minimum required operations are as follows: 
1 * 2 -> 2 
2 * 2 -> 4 
4 + 1 -> 5 



Recursive Approach: Recursively generate every possible combination to reduce N to 1 and calculate the number of operations required. Finally, after exhausting all possible combinations, print the sequence that required the minimum number of operations.

Below is the implementation of the above approach:  

C++

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// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
vector<int> find_sequence(int n)
{
     
    // Base Case
    if (n == 1)
        return {1, -1};
 
    // Recursive Call for n-1
    auto arr = find_sequence(n - 1);
    vector<int> ans = {arr[0] + 1, n - 1};
 
    // Check if n is divisible by 2
    if (n % 2 == 0)
    {
        vector<int> div_by_2 = find_sequence(n / 2);
 
        if (div_by_2[0] < ans[0])
            ans = {div_by_2[0] + 1, n / 2};
    }
 
    // Check if n is divisible by 3
    if (n % 3 == 0)
    {
        vector<int> div_by_3 = find_sequence(n / 3);
 
        if (div_by_3[0] < ans[0])
            vector<int> ans = {div_by_3[0] + 1, n / 3};
    }
 
    // Returns a tuple (a, b), where
    // a: Minimum steps to obtain x from 1
    // b: Previous number
    return ans;
}
 
// Function that find the optimal
// solution
vector<int> find_solution(int n)
{
    auto a = find_sequence(n);
 
    // Print the length
    cout << a[0] << endl;
 
    vector<int> sequence;
    sequence.push_back(n);
 
    //Exit condition for loop = -1
    //when n has reached 1
    while (a[1] != -1)
    {
        sequence.push_back(a[1]);
        auto arr = find_sequence(a[1]);
        a[1] = arr[1];
    }
 
    // Return the sequence
    // in reverse order
    reverse(sequence.begin(),
            sequence.end());
 
    return sequence;
}
 
// Driver Code
int main()
{
     
    // Given N
    int n = 5;
     
    // Function call
    auto i = find_solution(n);
     
    for(int j : i)
        cout << j << " ";
}
 
// This code is contributed by mohit kumar 29

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Java

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// Java program to implement
// the above approach
import java.util.*;
import java.util.Collections;
import java.util.Vector;
 
//Vector<Integer> v = new Vector<Integer>(n);
 
class GFG{
   
static Vector<Integer> find_sequence(int n)
{
    Vector<Integer> temp = new Vector<Integer>();
     
    temp.add(1);
    temp.add(-1);
     
    // Base Case
    if (n == 1)
        return temp;
     
    // Recursive Call for n-1
    Vector<Integer> arr = find_sequence(n - 1);
    Vector<Integer> ans = new Vector<Integer>(n);
    ans.add(arr.get(0) + 1);
    ans.add(n - 1);
     
    // Check if n is divisible by 2
    if (n % 2 == 0)
    {
        Vector<Integer> div_by_2 = find_sequence(n / 2);
         
        if (div_by_2.get(0) < ans.get(0))
        {
            ans.clear();
            ans.add(div_by_2.get(0) + 1);
            ans.add(n / 2);
        }
    }
 
    // Check if n is divisible by 3
    if (n % 3 == 0)
    {
        Vector<Integer> div_by_3 = find_sequence(n / 3);
         
        if (div_by_3.get(0) < ans.get(0))
        {
            ans.clear();
            ans.add(div_by_3.get(0) + 1);
            ans.add(n / 3);
        }
    }
     
    // Returns a tuple (a, b), where
    // a: Minimum steps to obtain x from 1
    // b: Previous number
    return ans;
}
 
// Function that find the optimal
// solution
static Vector<Integer> find_solution(int n)
{
    Vector<Integer> a = find_sequence(n);
     
    // Print the length
    System.out.println(a.get(0));
     
    Vector<Integer> sequence = new Vector<Integer>();
    sequence.add(n);
     
    // Exit condition for loop = -1
    // when n has reached 1
    while (a.get(1) != -1)
    {
        sequence.add(a.get(1));
        Vector<Integer> arr = find_sequence(a.get(1));
        a.set(1, arr.get(1));
    }
     
    // Return the sequence
    // in reverse order
    Collections.reverse(sequence);
     
    return sequence;
}
 
// Driver Code
public static void main(String args[])
{
     
    // Given N
    int n = 5;
     
    // Function call
    Vector<Integer> res = find_solution(n);
     
    for(int i = 0; i < res.size(); i++)
    {
        System.out.print(res.get(i) + " ");
    }
}
}
 
// This code is contributed by Surendra_Gangwar

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Python3

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# Python3 program to implement
# the above approach
 
 
def find_sequence(n):
 
    # Base Case
    if n == 1:
        return 1, -1
  
    # Recursive Call for n-1
    ans = (find_sequence(n - 1)[0] + 1, n - 1)
  
    # Check if n is divisible by 2
    if n % 2 == 0:
        div_by_2 = find_sequence(n // 2)
 
        if div_by_2[0] < ans[0]:
            ans = (div_by_2[0] + 1, n // 2)
  
    # Check if n is divisible by 3
    if n % 3 == 0:
        div_by_3 = find_sequence(n // 3)
 
        if div_by_3[0] < ans[0]:
            ans = (div_by_3[0] + 1, n // 3)
  
    # Returns a tuple (a, b), where
    # a: Minimum steps to obtain x from 1
    # b: Previous number
    return ans
  
# Function that find the optimal
# solution
def find_solution(n):
    a, b = find_sequence(n)
  
    # Print the length
    print(a)
 
    sequence = []
    sequence.append(n)
 
    # Exit condition for loop = -1
    # when n has reached 1
    while b != -1:
        sequence.append(b)
        _, b = find_sequence(b)
  
    # Return the sequence
    # in reverse order
    return sequence[::-1]
 
# Driver Code
 
# Given N
n = 5
 
# Function Call
print(*find_solution(n))

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// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
   
static List<int> find_sequence(int n)
{
  List<int> temp = new List<int>();
 
  temp.Add(1);
  temp.Add(-1);
 
  // Base Case
  if (n == 1)
    return temp;
 
  // Recursive Call for n-1
  List<int> arr = find_sequence(n - 1);
  List<int> ans = new List<int>(n);
  ans.Add(arr[0] + 1);
  ans.Add(n - 1);
 
  // Check if n is divisible by 2
  if (n % 2 == 0)
  {
    List<int> div_by_2 =
         find_sequence(n / 2);
 
    if (div_by_2[0] < ans[0])
    {
      ans.Clear();
      ans.Add(div_by_2[0] + 1);
      ans.Add(n / 2);
    }
  }
 
  // Check if n is divisible
  // by 3
  if (n % 3 == 0)
  {
    List<int> div_by_3 =
         find_sequence(n / 3);
 
    if (div_by_3[0] < ans[0])
    {
      ans.Clear();
      ans.Add(div_by_3[0] + 1);
      ans.Add(n / 3);
    }
  }
 
  // Returns a tuple (a, b), where
  // a: Minimum steps to obtain x
  // from 1 b: Previous number
  return ans;
}
 
// Function that find the optimal
// solution
static List<int> find_solution(int n)
{
  List<int> a = find_sequence(n);
 
  // Print the length
  Console.WriteLine(a[0]);
 
  List<int> sequence =
       new List<int>();
  sequence.Add(n);
 
  // Exit condition for loop = -1
  // when n has reached 1
  while (a[1] != -1)
  {
    sequence.Add(a[1]);
    List<int> arr =
         find_sequence(a[1]);
    a.Insert(1, arr[1]);
  }
 
  // Return the sequence
  // in reverse order
  sequence.Reverse();
  return sequence;
}
 
// Driver Code
public static void Main(String []args)
{   
  // Given N
  int n = 5;
 
  // Function call
  List<int> res = find_solution(n);
 
  for(int i = 0; i < res.Count; i++)
  {
    Console.Write(res[i] + " ");
  }
}
}
 
// This code is contributed by shikhasingrajput

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Output: 

4
1 2 4 5

 

Time Complexity: T(N) = T(N-1) + T(N/2) + T(N/3), where N is given integer. This algorithm results in an exponential time complexity. 

Auxiliary Space: O(1)

Recursion With Memoization Approach: The above approach can be optimized by memoizating the overlapping subproblems.

Below is the implementation of the above approach:  

Python3

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# Python3 program to implement
# the above approach
 
# Function to find the sequence
# with given operations
def find_sequence(n, map):
 
    # Base Case
    if n == 1:
        return 1, -1
 
    # Check if the subproblem
    # is already computed or not
    if n in map:
        return map[n]
 
    # Recursive Call for n-1
    ans = (find_sequence(n - 1, map)[0]\
    + 1, n - 1)
 
    # Check if n is divisible by 2
    if n % 2 == 0:
        div_by_2 = find_sequence(n // 2, map)
 
        if div_by_2[0] < ans[0]:
            ans = (div_by_2[0] + 1, n // 2)
 
    # Check if n is divisible by 3
    if n % 3 == 0:
        div_by_3 = find_sequence(n // 3, map)
 
        if div_by_3[0] < ans[0]:
            ans = (div_by_3[0] + 1, n // 3)
 
    # Memoize
    map[n] = ans
 
    # Returns a tuple (a, b), where
    # a: Minimum steps to obtain x from 1
    # b: Previous state
    return ans
 
# Function to check if a sequence can
# be obtained with given operations
def find_solution(n):
 
    # Stores the computed
    # subproblems
    map = {}
    a, b = find_sequence(n, map)
 
    # Return a sequence in
    # reverse order
    print(a)
    sequence = []
    sequence.append(n)
 
    # If n has reached 1
    while b != -1:
        sequence.append(b)
        _, b = find_sequence(b, map)
 
    # Return sequence in reverse order
    return sequence[::-1]
 
# Driver Code
 
# Given N
n = 5
 
# Function Call
print(*find_solution(n))

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Output: 

4
1 2 4 5

 

Time Complexity: O(N) 
Auxiliary Space: O(N)

Iterative Dynamic Programming Approach: The above approach can be further optimized by using an iterative DP approach. Follow the steps below to solve the problem:  

  1. Create an array dp[] to store the minimum length of sequence that is required to compute 1 to N by the three available operations.
  2. Once the dp[] array is computed, obtain the sequence by traversing the dp[] array from N to 1.

Below is the implementation of the above approach:  

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// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to generate
// the minimum sequence
void find_sequence(int n)
{
 
  // Stores the values for the
  // minimum length of sequences
  int dp[n + 1];
  memset(dp, 0, sizeof(dp));
 
  // Base Case
  dp[1] = 1;
 
  // Loop to build up the dp[]
  // array from 1 to n
  for(int i = 1; i < n + 1; i++)
  {
    if (dp[i] != 0)
    {
 
      // If i + 1 is within limits
      if (i + 1 < n + 1 &&
          (dp[i + 1] == 0 ||
           dp[i + 1] > dp[i] + 1))
      {
 
        // Update the state of i + 1
        // in dp[] array to minimum
        dp[i + 1] = dp[i] + 1;
      }
 
      // If i * 2 is within limits
      if (i * 2 < n + 1 &&
          (dp[i * 2] == 0 ||
           dp[i * 2] > dp[i] + 1))
      {
 
        // Update the state of i * 2
        // in dp[] array to minimum
        dp[i * 2] = dp[i] + 1;
      }
 
      // If i * 3 is within limits
      if (i * 3 < n + 1 &&
          (dp[i * 3] == 0 ||
           dp[i * 3] > dp[i] + 1))
      {
 
        // Update the state of i * 3
        // in dp[] array to minimum
        dp[i * 3] = dp[i] + 1;
      }
    }
  }
 
  // Generate the sequence by
  // traversing the array
  vector<int> sequence;
  while (n >= 1)
  {
 
    // Append n to the sequence
    sequence.push_back(n);
 
    // If the value of n in dp
    // is obtained from n - 1:
    if (dp[n - 1] == dp[n] - 1)
    {
      n--;
    }
 
    // If the value of n in dp[]
    // is obtained from n / 2:
    else if (n % 2 == 0 &&
             dp[(int)floor(n / 2)] == dp[n] - 1)
    {
      n = (int)floor(n / 2);
    }
 
    // If the value of n in dp[]
    // is obtained from n / 3:
    else if (n % 3 == 0 &&
             dp[(int)floor(n / 3)] == dp[n] - 1)
    {
      n = (int)floor(n / 3);
    }
  }
 
  // Print the sequence
  // in reverse order
  reverse(sequence.begin(), sequence.end());
 
  // Print the length of
  // the minimal sequence
  cout << sequence.size() << endl;
  for(int i = 0; i < sequence.size(); i++)
  {
    cout << sequence[i] << " ";
  }
}
 
// Driver code
int main()
{
 
  // Given Number N
  int n = 5;
 
  // Function Call
  find_sequence(n);
 
  return 0;
}
 
// This code is contributed by divyeshrabadiya07

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// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function to generate
// the minimum sequence
public static void find_sequence(int n)
{
     
    // Stores the values for the
    // minimum length of sequences
    int[] dp = new int[n + 1];
    Arrays.fill(dp, 0);
     
    // Base Case
    dp[1] = 1;
     
    // Loop to build up the dp[]
    // array from 1 to n
    for(int i = 1; i < n + 1; i++)
    {
        if (dp[i] != 0)
        {
             
            // If i + 1 is within limits
            if (i + 1 < n + 1 &&
            (dp[i + 1] == 0 ||
             dp[i + 1] > dp[i] + 1))
            {
                 
                // Update the state of i + 1
                // in dp[] array to minimum
                dp[i + 1] = dp[i] + 1;
            }
             
            // If i * 2 is within limits
            if (i * 2 < n + 1 &&
            (dp[i * 2] == 0 ||
             dp[i * 2] > dp[i] + 1))
            {
                 
                // Update the state of i * 2
                // in dp[] array to minimum
                dp[i * 2] = dp[i] + 1;
            }
             
            // If i * 3 is within limits
            if (i * 3 < n + 1 &&
            (dp[i * 3] == 0 ||
             dp[i * 3] > dp[i] + 1))
            {
                 
                // Update the state of i * 3
                // in dp[] array to minimum
                dp[i * 3] = dp[i] + 1;
            }
        }
    }
     
    // Generate the sequence by
    // traversing the array
    List<Integer> sequence = new ArrayList<Integer>();
    while (n >= 1)
    {
         
        // Append n to the sequence
        sequence.add(n);
         
        // If the value of n in dp
        // is obtained from n - 1:
        if (dp[n - 1] == dp[n] - 1)
        {
            n--;
        }
         
        // If the value of n in dp[]
        // is obtained from n / 2:
        else if (n % 2 == 0 &&
         dp[(int)Math.floor(n / 2)] == dp[n] - 1)
        {
            n = (int)Math.floor(n / 2);
        }
         
        // If the value of n in dp[]
        // is obtained from n / 3:
        else if (n % 3 == 0 &&
         dp[(int)Math.floor(n / 3)] == dp[n] - 1)
        {
            n = (int)Math.floor(n / 3);
        }
    }
     
    // Print the sequence
    // in reverse order
    Collections.reverse(sequence);
     
    // Print the length of
    // the minimal sequence
    System.out.println(sequence.size());
    for(int i = 0; i < sequence.size(); i++)
    {
        System.out.print(sequence.get(i) + " ");
    }
}
 
// Driver Code
public static void main (String[] args)
{
     
    // Given Number N
    int n = 5;
     
    // Function Call
    find_sequence(n);
}
}
 
// This code is contributed by avanitrachhadiya2155

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# Python3 program to implement
# the above approach
 
# Function to generate
# the minimum sequence
def find_sequence(n):
 
    # Stores the values for the
    # minimum length of sequences
    dp = [0 for _ in range(n + 1)]
 
    # Base Case
    dp[1] = 1
 
    # Loop to build up the dp[]
    # array from 1 to n
    for i in range(1, n + 1):
 
        if dp[i] != 0:
 
            # If i + 1 is within limits
            if i + 1 < n + 1 and (dp[i + 1] == 0 \
            or dp[i + 1] > dp[i] + 1):
                 
                # Update the state of i + 1
                # in dp[] array to minimum
                dp[i + 1] = dp[i] + 1
 
            # If i * 2 is within limits
            if i * 2 < n + 1 and (dp[i * 2] == 0 \
            or dp[i * 2] > dp[i] + 1):
                 
                # Update the state of i * 2
                # in dp[] array to minimum
                dp[i * 2] = dp[i] + 1
 
            # If i * 3 is within limits
            if i * 3 < n + 1 and (dp[i * 3] == 0 \
            or dp[i * 3] > dp[i] + 1):
                 
                # Update the state of i * 3
                # in dp[] array to minimum
                dp[i * 3] = dp[i] + 1
 
    # Generate the sequence by
    # traversing the array
    sequence = []
    while n >= 1:
 
        # Append n to the sequence
        sequence.append(n)
 
        # If the value of n in dp
        # is obtained from n - 1:
        if dp[n - 1] == dp[n] - 1:
            n = n - 1
 
        # If the value of n in dp[]
        # is obtained from n / 2:
        elif n % 2 == 0 \
        and dp[n // 2] == dp[n] - 1:
            n = n // 2
 
        # If the value of n in dp[]
        # is obtained from n / 3:
        elif n % 3 == 0 \
        and dp[n // 3] == dp[n] - 1:
            n = n // 3
 
    # Return the sequence
    # in reverse order
    return sequence[::-1]
 
# Driver Code
 
# Given Number N
n = 5
 
# Function Call
sequence = find_sequence(n)
 
# Print the length of
# the minimal sequence
print(len(sequence))
 
# Print the sequence
print(*sequence)

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// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
 
  // Function to generate
  // the minimum sequence
  public static void find_sequence(int n)
  {
 
    // Stores the values for the
    // minimum length of sequences
    int[] dp = new int[n + 1];
    Array.Fill(dp, 0);
 
    // Base Case
    dp[1] = 1;
 
    // Loop to build up the dp[]
    // array from 1 to n
    for(int i = 1; i < n + 1; i++)
    {
      if(dp[i] != 0)
      {
 
        // If i + 1 is within limits
        if(i + 1 < n + 1 &&
           (dp[i + 1] == 0 ||
            dp[i + 1] > dp[i] + 1))
        {
 
          // Update the state of i + 1
          // in dp[] array to minimum
          dp[i + 1] = dp[i] + 1;
        }
 
        // If i * 2 is within limits
        if(i * 2 < n + 1 &&
           (dp[i * 2] == 0 ||
            dp[i * 2] > dp[i] + 1))
        {
 
          // Update the state of i * 2
          // in dp[] array to minimum
          dp[i * 2] = dp[i] + 1;
        }
 
        // If i * 3 is within limits
        if(i * 3 < n + 1 &&
           (dp[i * 3] == 0 ||
            dp[i * 3] > dp[i] + 1))
        {
 
          // Update the state of i * 3
          // in dp[] array to minimum
          dp[i * 3] = dp[i] + 1;
        }       
      }
    }
 
    // Generate the sequence by
    // traversing the array
    List<int> sequence = new List<int>();
    while(n >= 1)
    {
 
      // Append n to the sequence
      sequence.Add(n);
 
      // If the value of n in dp
      // is obtained from n - 1:
      if(dp[n - 1] == dp[n] - 1)
      {
        n--;
      }
 
      // If the value of n in dp[]
      // is obtained from n / 2:
      else if(n % 2 == 0 &&
              dp[(int)Math.Floor((decimal)n / 2)] ==
              dp[n] - 1)
      {
        n = (int)Math.Floor((decimal)n / 2);
      }
 
      // If the value of n in dp[]
      // is obtained from n / 3:
      else if(n % 3 == 0 &&
              dp[(int)Math.Floor((decimal)n / 3)] ==
              dp[n] - 1)
      {
        n = (int)Math.Floor((decimal)n / 3);
      }
 
    }
 
    // Print the sequence
    // in reverse order
    sequence.Reverse();
 
    // Print the length of
    // the minimal sequence
    Console.WriteLine(sequence.Count);
    for(int i = 0; i < sequence.Count; i++)
    {
      Console.Write(sequence[i] + " ");
    }
  }
 
  // Driver Code
  static public void Main ()
  {
 
    // Given Number N
    int n = 5;
 
    // Function Call
    find_sequence(n);
  }
}
 
// This code is contributed by rag2127

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Output: 

4
1 3 4 5

 

Time Complexity: O(N) 
Auxiliary Space: O(N)
 

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