# Minimum number of operations required to reduce N to 1

• Difficulty Level : Easy
• Last Updated : 29 Oct, 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:

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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``#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``?>`

## 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 1``int` `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 code``int` `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 approach``import` `java.util.*;` `class` `GFG{``    ` `// 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` `(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``public` `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 1``def` `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 Code``n ``=` `4``ans ``=` `count_minimum_operations(n)` `print``(ans)` `# This code is contributed by divyesh072019`

## C#

 `// C# implementation of above approach``using` `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 1``int` `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 code``int` `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`
Output
`2`

Time Complexity: O(n)

Auxiliary Space: O(n)

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