# Minimum number of operations required to reduce N to 1

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

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// CPP implementation of above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns the minimum ` `// number of operations to be performed ` `// to reduce the number to 1 ` `int` `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 code ` `int` `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 1 ` `def` `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 code ` `if` `__name__ ``=``=``"__main__"``: ` `    ``n ``=` `4` `    ``ans ``=` `count_minimum_operations(n) ` `    ``print` `(ans) ` ` `  `# This code is contributed  ` `# by ChitraNayal `

## C#

 `// C# implementation of above approach ` `using` `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 ` `?> `

Output:

```2
```

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