# Minimum decrements or division by a proper divisor required to reduce N to 1

Given a positive integer **N**, the task is to find the minimum number of operations required to reduce **N** to **1** by repeatedly dividing **N** by its proper divisors or by decreasing **N** by **1**.

**Examples:**

Input:N = 9Output:3Explanation:

The proper divisors of N(= 9) are {1, 3}. Following operations are performed to reduced N to 1:Operation 1:Divide N(= 9) by 3(which is a proper divisor of N(= 9) modifies the value of N to 9/3 = 1.Operation 2:Decrementing the value of N(= 3) by 1 modifies the value of N to 3 – 1 = 2.Operation 3:Decrementing the value of N(= 2) by 1 modifies the value of N to 2 – 1 = 1.

Therefore, the total number of operations required is 3.

Input:N = 4Output:2

**Approach:** The given problem can be solved based on the following observations:

- If the value of
**N**is even, then it can be reduced to value**2**by dividing**N**by**N / 2**followed by decrementing**2**to**1**. Therefore, the minimum number of steps required is**2**. - Otherwise, the value of
**N**can be made even by decrementing it and can be reduced to**1**using the above steps.

Follow the steps given below to solve the problem

- Initialize a variable, say
**cnt**as**0**, to store the minimum number of steps required to reduce**N**to**1**. - Iterate a loop until
**N**reduces to**1**and perform the following steps:- If the value of
**N**is equal to**2**or N is odd, then update the value of**N = N – 1**and increment**cnt**by**1**. - Otherwise, update the value of
**N = N / (N / 2)**and increment**cnt**by 1.

- If the value of
- After completing the above steps, print the value of
**cnt**as the result.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to find the minimum number` `// of steps required to reduce N to 1` `int` `reduceToOne(` `long` `long` `int` `N)` `{` ` ` `// Stores the number` ` ` `// of steps required` ` ` `int` `cnt = 0;` ` ` `while` `(N != 1) {` ` ` `// If the value of N` ` ` `// is equal to 2 or N is odd` ` ` `if` `(N == 2 or (N % 2 == 1)) {` ` ` `// Decrement N by 1` ` ` `N = N - 1;` ` ` `// Increment cnt by 1` ` ` `cnt++;` ` ` `}` ` ` `// If N is even` ` ` `else` `if` `(N % 2 == 0) {` ` ` `// Update N` ` ` `N = N / (N / 2);` ` ` `// Increment cnt by 1` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` `// Return the number` ` ` `// of steps obtained` ` ` `return` `cnt;` `}` `// Driver Code` `int` `main()` `{` ` ` `long` `long` `int` `N = 35;` ` ` `cout << reduceToOne(N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `class` `GFG{` ` ` `// Function to find the minimum number` `// of steps required to reduce N to 1` `static` `int` `reduceToOne(` `long` `N)` `{` ` ` ` ` `// Stores the number` ` ` `// of steps required` ` ` `int` `cnt = ` `0` `;` ` ` `while` `(N != ` `1` `)` ` ` `{` ` ` ` ` `// If the value of N` ` ` `// is equal to 2 or N is odd` ` ` `if` `(N == ` `2` `|| (N % ` `2` `== ` `1` `))` ` ` `{` ` ` ` ` `// Decrement N by 1` ` ` `N = N - ` `1` `;` ` ` `// Increment cnt by 1` ` ` `cnt++;` ` ` `}` ` ` `// If N is even` ` ` `else` `if` `(N % ` `2` `== ` `0` `)` ` ` `{` ` ` ` ` `// Update N` ` ` `N = N / (N / ` `2` `);` ` ` `// Increment cnt by 1` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` `// Return the number` ` ` `// of steps obtained` ` ` `return` `cnt;` `}` `// Driver Code` `public` `static` `void` `main(String[] args)` `{` ` ` `long` `N = ` `35` `;` ` ` ` ` `System.out.println(reduceToOne(N));` `}` `}` `// This code is contributed by Dharanendra L V.` |

## Python3

`# python program for the above approach` `# Function to find the minimum number` `# of steps required to reduce N to 1` `def` `reduceToOne(N):` ` ` ` ` `# Stores the number` ` ` `# of steps required` ` ` `cnt ` `=` `0` ` ` `while` `(N !` `=` `1` `):` ` ` `# If the value of N` ` ` `# is equal to 2 or N is odd` ` ` `if` `(N ` `=` `=` `2` `or` `(N ` `%` `2` `=` `=` `1` `)):` ` ` `# Decrement N by 1` ` ` `N ` `=` `N ` `-` `1` ` ` `# Increment cnt by 1` ` ` `cnt ` `+` `=` `1` ` ` `# If N is even` ` ` `elif` `(N ` `%` `2` `=` `=` `0` `):` ` ` `# Update N` ` ` `N ` `=` `N ` `/` `(N ` `/` `2` `)` ` ` `# Increment cnt by 1` ` ` `cnt ` `+` `=` `1` ` ` `# Return the number` ` ` `# of steps obtained` ` ` `return` `cnt` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `N ` `=` `35` ` ` `print` `(reduceToOne(N))` `# This code is contributed by mohit kumar 29.` |

## C#

`// C# program for the above approach` `using` `System;` ` ` `class` `GFG` `{` `// Function to find the minimum number` `// of steps required to reduce N to 1` `static` `int` `reduceToOne(` `long` `N)` `{` ` ` ` ` `// Stores the number` ` ` `// of steps required` ` ` `int` `cnt = 0;` ` ` `while` `(N != 1)` ` ` `{` ` ` ` ` `// If the value of N` ` ` `// is equal to 2 or N is odd` ` ` `if` `(N == 2 || (N % 2 == 1))` ` ` `{` ` ` ` ` `// Decrement N by 1` ` ` `N = N - 1;` ` ` `// Increment cnt by 1` ` ` `cnt++;` ` ` `}` ` ` `// If N is even` ` ` `else` `if` `(N % 2 == 0)` ` ` `{` ` ` ` ` `// Update N` ` ` `N = N / (N / 2);` ` ` `// Increment cnt by 1` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` `// Return the number` ` ` `// of steps obtained` ` ` `return` `cnt;` `}` ` ` `// Driver Code` `public` `static` `void` `Main()` `{` ` ` `long` `N = 35;` ` ` ` ` `Console.WriteLine(reduceToOne(N));` `}` `}` `// This code s contributed by code_hunt.` |

## Javascript

`<script>` `// Javascript program for the above approach` `// Function to find the minimum number` `// of steps required to reduce N to 1` `function` `reduceToOne( N)` `{` ` ` `// Stores the number` ` ` `// of steps required` ` ` `let cnt = 0;` ` ` `while` `(N != 1) {` ` ` `// If the value of N` ` ` `// is equal to 2 or N is odd` ` ` `if` `(N == 2 || (N % 2 == 1)) {` ` ` `// Decrement N by 1` ` ` `N = N - 1;` ` ` `// Increment cnt by 1` ` ` `cnt++;` ` ` `}` ` ` `// If N is even` ` ` `else` `if` `(N % 2 == 0) {` ` ` `// Update N` ` ` `N = Math.floor(N / Math.floor(N / 2));` ` ` `// Increment cnt by 1` ` ` `cnt++;` ` ` `}` ` ` `}` ` ` `// Return the number` ` ` `// of steps obtained` ` ` `return` `cnt;` `}` `// Driver Code` `let N = 35;` `document.write(reduceToOne(N));` `// This code is contributed by jana_sayantan.` `</script>` |

**Output:**

3

**Time Complexity:** O(1)**Auxiliary Space:** O(1)

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