Minimum steps to reduce N to 0 by given operations

• Difficulty Level : Medium
• Last Updated : 09 Jun, 2021

Give an integer N, the task is to find the minimum number of moves to reduce N to 0 by one of the following operations:

• Reduce N by 1.
• Reduce N to (N/2), if N is divisible by 2.
• Reduce N to (N/3), if N is divisible by 3.

Examples:

Input: N = 10
Output: 4
Explanation:
Here N = 10
Step 1: Reducing N by 1 i.e., 10 – 1 = 9.
Step 2: Since 9 is divisible by 3, reduce it to N/3 = 9/3 = 3
Step 3: Since again 3 is divisible by 3 again repeating step 2, i.e., 3/3 = 1.
Step 4: 1 can be reduced by the step 1, i.e., 1-1 = 0
Hence, 4 steps are needed to reduce N to 0.

Input: N = 6
Output: 3
Explanation:
Here N = 6
Step 1: Since 6 is divisible by 2, then 6/2 =3
Step 2: since 3 is divisible by 3, then 3/3 = 1.
Step 3: Reduce N to N-1 by 1, 1-1 = 0.
Hence, 3 steps are needed to reduce N to 0.

Naive Approach: The idea is to use recursion for all the possible moves. Below are the steps:

1. Observe that base case for the problem, if N < 2 then for all the cases the answer will be N itself.
2. At every value of N, choose between 2 possible cases:
• Reduce n till n % 2 == 0 and then update n /= 2 with count = 1 + n%2 + f(n/2)
• Reduce n till n % 3 == 0 and then update n /= 3 with count = 1 + n%3 + f(n/3)
3. Both the computation results in the recurrence relation as:

count = 1 + min(n%2 + f(n/2), n%3 + f(n/3))
where, f(n) is the minimum  of moves to reduce N to 0.

Below is the implementation of the above approach:

C++

 // C++ program for the above approach#include using namespace std; // Function to find the minimum// number to steps to reduce N to 0int minDays(int n){         // Base case    if (n < 1)        return n;             // Recursive call to count the    // minimum steps needed    int cnt = 1 + min(n % 2 + minDays(n / 2),                      n % 3 + minDays(n / 3));     // Return the answer    return cnt;} // Driver Codeint main(){         // Given number N    int N = 6;         // Function call    cout << minDays(N);         return 0;} // This code is contributed by 29AjayKumar

Java

 // Java program for the above approachclass GFG{       // Function to find the minimum    // number to steps to reduce N to 0    static int minDays(int n)    {         // Base case        if (n < 1)            return n;         // Recursive Call to count the        // minimum steps needed        int cnt = 1 + Math.min(n % 2 + minDays(n / 2),                               n % 3 + minDays(n / 3));         // Return the answer        return cnt;    }     // Driver Code    public static void main(String[] args)    {        // Given Number N        int N = 6;         // Function Call        System.out.print(minDays(N));    }} // This code is contributed by PrinciRaj1992

Python3

 # Python3 program for the above approach # Function to find the minimum# number to steps to reduce N to 0def minDays(n):     # Base case    if n < 1:        return n           # Recursive Call to count the    # minimum steps needed    cnt = 1 + min(n % 2 + minDays(n // 2),                  n % 3 + minDays(n // 3))     # Return the answer    return cnt # Driver Code # Given Number NN = 6 # Function Callprint(str(minDays(N)))

C#

 // C# program for the above approachusing System;class GFG{ // Function to find the minimum// number to steps to reduce N to 0static int minDays(int n){       // Base case    if (n < 1)        return n;     // Recursive call to count the    // minimum steps needed    int cnt = 1 + Math.Min(n % 2 + minDays(n / 2),                           n % 3 + minDays(n / 3));     // Return the answer    return cnt;} // Driver Codepublic static void Main(String[] args){       // Given number N    int N = 6;     // Function call    Console.Write(minDays(N));}} // This code is contributed by Rajput-Ji

Javascript


Output:
4

Time Complexity: O(2N)
Auxiliary Space: O(1)

Efficient Approach: The idea is to use Dynamic Programming. The above recursive approach results in TLE because of the number of repeated subproblems. To optimize the above method using a dictionary to keep track of values whose recursive call is already performed to reduce the further computation such that value can be accessed faster.

Below is the implementation of the above approach:

C++

 // C++ program for// the above approach#includeusing namespace std; // Function to find the minimum// number to steps to reduce N to 0int count(int n){  // Dictionary for storing  // the precomputed sum  map dp;   // Bases Cases  dp = 0;  dp = 1;   // Check if n is not in dp then  // only call the function so as  // to reduce no of recursive calls  if ((dp.find(n) == dp.end()))    dp[n] = 1 + min(n % 2 +                    count(n / 2), n % 3 +                    count(n / 3));   // Return the answer  return dp[n];} // Driver Codeint main(){     // Given number N  int N = 6;   // Function call  cout << count(N);} // This code is contributed by gauravrajput1

Java

 // Java program for the above approachimport java.util.HashMap; class GFG{     // Function to find the minimum// number to steps to reduce N to 0static int count(int n){         // Dictionary for storing    // the precomputed sum    HashMap dp = new HashMap();     // Bases Cases    dp.put(0, 0);    dp.put(1, 1);     // Check if n is not in dp then    // only call the function so as    // to reduce no of recursive calls    if (!dp.containsKey(n))        dp.put(n, 1 + Math.min(n % 2 +                 count(n / 2), n % 3 +                 count(n / 3)));     // Return the answer    return dp.get(n);} // Driver Codepublic static void main(String[] args){         // Given number N    int N = 6;     // Function call    System.out.println(String.valueOf((count(N))));}} // This code is contributed by Amit Katiyar

Python3

 # Python3 program for the above approach # Function to find the minimum# number to steps to reduce N to 0def count(n):       # Dictionary for storing    # the precomputed sum    dp = dict()     # Bases Cases    dp = 0    dp = 1     # Check if n is not in dp then    # only call the function so as    # to reduce no of recursive calls    if n not in dp:        dp[n] = 1 + min(n % 2 + count(n//2), n % 3 + count(n//3))     # Return the answer    return dp[n]  # Driver Code # Given Number NN = 6 # Function Callprint(str(count(N)))

C#

 // C# program for the above approachusing System;using System.Collections.Generic;public class GFG{     // Function to find the minimum// number to steps to reduce N to 0static int count(int n){       // Dictionary for storing    // the precomputed sum    Dictionary dp = new Dictionary();     // Bases Cases    dp.Add(0, 0);    dp.Add(1, 1);     // Check if n is not in dp then    // only call the function so as    // to reduce no of recursive calls    if (!dp.ContainsKey(n))        dp.Add(n, 1 + Math.Min(n % 2 + count(n / 2),                               n % 3 + count(n / 3)));     // Return the answer    return dp[n];} // Driver Codepublic static void Main(String[] args){       // Given number N    int N = 6;     // Function call    Console.WriteLine(String.Join("",                                  (count(N))));}} // This code is contributed by Rajput-Ji

Javascript


Output:
3

Time Complexity: O(log N)
Auxiliary Space: O(1)

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