Minimum number of operations required to reduce N to 1

• Difficulty Level : Easy
• Last Updated : 23 Dec, 2021

Given an integer element ‘N’, the task is to find the minimum number of operations that need to be performed to make ‘N’ equal to 1.
The allowed operations to be performed are:

1. Decrement N by 1.
2. Increment N by 1.
3. If N is a multiple of 3, you can divide N by 3.

Examples:

Input: N = 4
Output:
4 – 1 = 3
3 / 3 = 1
The minimum number of operations required is 2.

Input: N = 8
Output:
8 + 1 = 9
9 / 3 = 3
3 / 3 = 1
The minimum number of operations required is 3.

Approach:

• If the number is a multiple of 3, divide it by 3.
• If the number modulo 3 is 1, decrement it by 1.
• If the number modulo 3 is 2, increment it by 1.
• There is an exception when the number is equal to 2, in this case the number should be decremented by 1.
• Repeat the above steps until the number is greater than 1 and print the count of operations performed in the end.

Below is the implementation of the above approach:

C++

 // C++ implementation of above approach#includeusing namespace std; // Function that returns the minimum// number of operations to be performed// to reduce the number to 1int count_minimum_operations(long long n){     // To stores the total number of    // operations to be performed    int count = 0;    while (n > 1) {         // if n is divisible by 3        // then reduce it to n / 3        if (n % 3 == 0)            n /= 3;         // if n modulo 3 is 1        // decrement it by 1        else if (n % 3 == 1)            n--;        else {            if (n == 2)                n--;                         // if n modulo 3 is 2            // then increment it by 1            else                n++;        }         // update the counter        count++;    }    return count;} // Driver codeint main(){     long long n = 4;    long long ans = count_minimum_operations(n);    cout<

Java

 // Java implementation of above approach class GFG {     // Function that returns the minimum    // number of operations to be performed    // to reduce the number to 1    static int count_minimum_operations(long n)    {         // To stores the total number of        // operations to be performed        int count = 0;        while (n > 1) {             // if n is divisible by 3            // then reduce it to n / 3            if (n % 3 == 0)                n /= 3;             // if n modulo 3 is 1            // decrement it by 1            else if (n % 3 == 1)                n--;            else {                if (n == 2)                    n--;                 // if n modulo 3 is 2                // then increment it by 1                else                    n++;            }             // update the counter            count++;        }        return count;    }     // Driver code    public static void main(String[] args)    {         long n = 4;        long ans = count_minimum_operations(n);        System.out.println(ans);    }}

Python3

 # Python3 implementation of above approach # Function that returns the minimum# number of operations to be performed# to reduce the number to 1def count_minimum_operations(n):     # To stores the total number of    # operations to be performed    count = 0    while (n > 1) :         # if n is divisible by 3        # then reduce it to n / 3        if (n % 3 == 0):            n //= 3         # if n modulo 3 is 1        # decrement it by 1        elif (n % 3 == 1):            n -= 1        else :            if (n == 2):                n -= 1                         # if n modulo 3 is 2            # then increment it by 1            else:                n += 1                 # update the counter        count += 1         return count # Driver codeif __name__ =="__main__":    n = 4    ans = count_minimum_operations(n)    print (ans) # This code is contributed# by ChitraNayal

C#

 // C# implementation of above approachusing System; public class GFG{             // Function that returns the minimum    // number of operations to be performed    // to reduce the number to 1    static int count_minimum_operations(long n)    {         // To stores the total number of        // operations to be performed        int count = 0;        while (n > 1) {             // if n is divisible by 3            // then reduce it to n / 3            if (n % 3 == 0)                n /= 3;             // if n modulo 3 is 1            // decrement it by 1            else if (n % 3 == 1)                n--;            else {                if (n == 2)                    n--;                 // if n modulo 3 is 2                // then increment it by 1                else                    n++;            }             // update the counter            count++;        }        return count;    }     // Driver code    static public void Main (){             long n = 4;        long ans = count_minimum_operations(n);        Console.WriteLine(ans);    }}

PHP

 1)    {         // if n is divisible by 3        // then reduce it to n / 3        if (\$n % 3 == 0)            \$n /= 3;         // if n modulo 3 is 1        // decrement it by 1        else if (\$n % 3 == 1)            \$n--;        else        {            if (\$n == 2)                \$n--;                         // if n modulo 3 is 2            // then increment it by 1            else                \$n++;        }         // update the counter        \$count++;    }    return \$count;} // Driver code\$n = 4; \$ans = count_minimum_operations(\$n);echo \$ans, "\n"; // This code is contributed by akt_mit?>

Javascript


Output
2

Recursive Approach: The recursive approach is similar to the approach used above.

Below is the implementation:

C++

 // C++ implementation of above approach#include using namespace std; // Function that returns the minimum// number of operations to be performed// to reduce the number to 1int count_minimum_operations(long long n){     // Base cases    if (n == 2) {        return 1;    }    else if (n == 1) {        return 0;    }    if (n % 3 == 0) {        return 1 + count_minimum_operations(n / 3);    }    else if (n % 3 == 1) {        return 1 + count_minimum_operations(n - 1);    }    else {        return 1 + count_minimum_operations(n + 1);    }} // Driver codeint main(){     long long n = 4;    long long ans = count_minimum_operations(n);    cout << ans << endl;    return 0;} // This code is contributed by koulick_sadhu

Java

 // Java implementation of above approachimport java.util.*; class GFG{     // Function that returns the minimum// number of operations to be performed// to reduce the number to 1public static int count_minimum_operations(int n){         // Base cases    if (n == 2)    {        return 1;    }    else if (n == 1)    {        return 0;    }    if (n % 3 == 0)    {        return 1 + count_minimum_operations(n / 3);    }    else if (n % 3 == 1)    {        return 1 + count_minimum_operations(n - 1);    }    else    {        return 1 + count_minimum_operations(n + 1);    }} // Driver codepublic static void main(String []args){    int n = 4;    int ans = count_minimum_operations(n);         System.out.println(ans);}} // This code is contributed by avanitrachhadiya2155

Python3

 # Python3 implementation of above approach # Function that returns the minimum# number of operations to be performed# to reduce the number to 1def count_minimum_operations(n):         # Base cases    if (n == 2):        return 1    elif (n == 1):        return 0    if (n % 3 == 0):        return 1 + count_minimum_operations(n / 3)    elif (n % 3 == 1):        return 1 + count_minimum_operations(n - 1)    else:        return 1 + count_minimum_operations(n + 1) # Driver Coden = 4ans = count_minimum_operations(n) print(ans) # This code is contributed by divyesh072019

C#

 // C# implementation of above approachusing System;class GFG {         // Function that returns the minimum    // number of operations to be performed    // to reduce the number to 1    static int count_minimum_operations(int n)    {              // Base cases        if (n == 2) {            return 1;        }        else if (n == 1) {            return 0;        }        if (n % 3 == 0) {            return 1 + count_minimum_operations(n / 3);        }        else if (n % 3 == 1) {            return 1 + count_minimum_operations(n - 1);        }        else {            return 1 + count_minimum_operations(n + 1);        }    }     // Driver code  static void Main() {    int n = 4;    int ans = count_minimum_operations(n);    Console.WriteLine(ans);  }} // This code is contributed by divyeshrabadiya07

Javascript


Output
2

Another Method (Efficient):

DP using memoization(Top down approach)

We can avoid the repeated work done by storing the operations performed calculated so far. We just need to store all the values in an array.

C++

 // C++ implementation of above approach#include using namespace std; int static dp; // Function that returns the minimum// number of operations to be performed// to reduce the number to 1int count_minimum_operations(long long n){    // Base cases    if (n == 2) {        return 1;    }    if (n == 1) {        return 0;    }    if(dp[n] != -1)    {        return dp[n];    }    if (n % 3 == 0) {        dp[n] = 1 + count_minimum_operations(n / 3);    }    else if (n % 3 == 1) {        dp[n] = 1 + count_minimum_operations(n - 1);    }    else {        dp[n] = 1 + count_minimum_operations(n + 1);    }    return dp[n];} // Driver codeint main(){    long long n = 4;      memset(dp, -1, sizeof(dp));    long long ans = count_minimum_operations(n);    cout << ans << endl;    return 0;} // This code is contributed by Samim Hossain Mondal

Java

 // Java implementation of above approachimport java.util.*;public class GFG{     static int []dp = new int;     // Function that returns the minimum// number of operations to be performed// to reduce the number to 1public static int count_minimum_operations(int n){         // Base cases    if (n == 2)    {        return 1;    }    else if (n == 1)    {        return 0;    }    if(dp[n] != -1)    {        return dp[n];    }    if (n % 3 == 0) {        dp[n] = 1 + count_minimum_operations(n / 3);    }    else if (n % 3 == 1) {        dp[n] = 1 + count_minimum_operations(n - 1);    }    else {        dp[n] = 1 + count_minimum_operations(n + 1);    }    return dp[n];} // Driver codepublic static void main(String []args){    int n = 4;         for(int i = 0; i < 1001; i++) {        dp[i] = -1;    }         int ans = count_minimum_operations(n);    System.out.println(ans);}} // This code is contributed by Samim Hossain Mondal.

Python

 # Python3 implementation of above approach # Function that returns the minimum# number of operations to be performed# to reduce the number to 1dp = [-1 for i in range(1001)] def count_minimum_operations(n):         # Base cases    if (n == 2):        return 1    elif (n == 1):        return 0    if(dp[n] != -1):        return dp[n]             elif (n % 3 == 0):        dp[n] = 1 + count_minimum_operations(n / 3)             elif (n % 3 == 1):        dp[n] = 1 + count_minimum_operations(n - 1)     else:        dp[n] = 1 + count_minimum_operations(n + 1)     return dp[n] # Driver Coden = 4ans = count_minimum_operations(n) print(ans) # This code is contributed by Samim Hossain Mondal

C#

 // C# implementation of above approachusing System;class GFG{   static int []dp = new int;   // Function that returns the minimum  // number of operations to be performed  // to reduce the number to 1  public static int count_minimum_operations(int n)  {     // Base cases    if (n == 2)    {      return 1;    }    else if (n == 1)    {      return 0;    }    if(dp[n] != -1)    {      return dp[n];    }    if (n % 3 == 0) {      dp[n] = 1 + count_minimum_operations(n / 3);    }    else if (n % 3 == 1) {      dp[n] = 1 + count_minimum_operations(n - 1);    }    else {      dp[n] = 1 + count_minimum_operations(n + 1);    }    return dp[n];  }   // Driver code  public static void Main()  {    int n = 4;     for(int i = 0; i < 1001; i++) {      dp[i] = -1;    }     int ans = count_minimum_operations(n);    Console.Write(ans);  }} // This code is contributed by Samim Hossain Mondal.

Javascript


Output
2

Time Complexity: O(n)

Auxiliary Space: O(n)

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