# Minimize operations to reduce N to 2 by repeatedly reducing by 3 or dividing by 5

Given a positive integer **N**, the task is to find the minimum number of operations needed to convert** N** to **2** either by decrementing **N** by **3** or dividing **N** by **5** if **N** is divisible by **5**. If it is not possible to reduce **N** to **2**, then print **“-1”**.

**Examples:**

Input:N =10Output:1Explanation:

Following are the operations performed to reduce N to 2:

- Dividing N by 5, reduces N to 10/5 = 2.
After the above operations, N is reduced to 2. Therefore, the minimum number of operations required is 1.

Input:N = 25Output:2

**Approach:** The given problem can be solved by using **Dynamic Programming**, the idea is to start iterating from** 2** and perform both the operations in a reverse manner i.e., instead of subtracting, perform addition of **3** and instead of dividing, perform multiplication with **5** at every state and store the minimum number of operations for every possible value of **N** in the array **dp[]**.

If the value of **N** is reached, then print the value of** dp[N]** as the minimum number of operations. Otherwise, print **-1**. Follow the steps below to solve the problem:

- Initialize an auxiliary array, say
**dp[]**of size**(N + 1)**, and initialize all array elements with**INT_MAX**. - Set the value of
**dp[2]**equal to**0**. - Iterate over the range
**[0, N]**, and update the value of**dp[i]**as:- dp[i * 5] = min(dp[i * 5], dp[i] + 1).
- dp[i + 3] = min(dp[i + 3], dp[i] + 1).

- If the value of
**dp[N]**is**INT_MAX**, then print**-1**. Otherwise, print**dp[N]**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 operations to reduce N to 2 by` `// dividing N by 5 or decrementing by 3` `int` `minimumOperations(` `int` `N)` `{` ` ` `// Initialize the dp array` ` ` `int` `dp[N + 1];` ` ` `int` `i;` ` ` `// Initialize the array dp[]` ` ` `for` `(` `int` `i = 0; i <= N; i++) {` ` ` `dp[i] = 1e9;` ` ` `}` ` ` `// For N = 2 number of operations` ` ` `// needed is zero` ` ` `dp[2] = 0;` ` ` `// Iterating over the range [1, N]` ` ` `for` `(i = 2; i <= N; i++) {` ` ` `// If it's not possible to` ` ` `// create current N` ` ` `if` `(dp[i] == 1e9)` ` ` `continue` `;` ` ` `// Multiply with 5` ` ` `if` `(i * 5 <= N) {` ` ` `dp[i * 5] = min(dp[i * 5],` ` ` `dp[i] + 1);` ` ` `}` ` ` `// Adding the value 3` ` ` `if` `(i + 3 <= N) {` ` ` `dp[i + 3] = min(dp[i + 3],` ` ` `dp[i] + 1);` ` ` `}` ` ` `}` ` ` `// Checking if not possible to` ` ` `// make the number as 2` ` ` `if` `(dp[N] == 1e9)` ` ` `return` `-1;` ` ` `// Return the minimum number` ` ` `// of operations` ` ` `return` `dp[N];` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `N = 25;` ` ` `cout << minimumOperations(N);` ` ` `return` `0;` `}` |

## Java

`// Java program for the above approach` `import` `java.io.*;` `class` `GFG {` ` ` `// Function to find the minimum number` ` ` `// of operations to reduce N to 2 by` ` ` `// dividing N by 5 or decrementing by 3` ` ` `static` `int` `minimumOperations(` `int` `N)` ` ` `{` ` ` ` ` `// Initialize the dp array` ` ` `int` `[] dp = ` `new` `int` `[N + ` `1` `];` ` ` `int` `i;` ` ` `// Initialize the array dp[]` ` ` `for` `(i = ` `0` `; i <= N; i++) {` ` ` `dp[i] = (` `int` `)1e9;` ` ` `}` ` ` `// For N = 2 number of operations` ` ` `// needed is zero` ` ` `dp[` `2` `] = ` `0` `;` ` ` `// Iterating over the range [1, N]` ` ` `for` `(i = ` `2` `; i <= N; i++) {` ` ` `// If it's not possible to` ` ` `// create current N` ` ` `if` `(dp[i] == (` `int` `)1e9)` ` ` `continue` `;` ` ` `// Multiply with 5` ` ` `if` `(i * ` `5` `<= N) {` ` ` `dp[i * ` `5` `] = Math.min(dp[i * ` `5` `], dp[i] + ` `1` `);` ` ` `}` ` ` `// Adding the value 3` ` ` `if` `(i + ` `3` `<= N) {` ` ` `dp[i + ` `3` `] = Math.min(dp[i + ` `3` `], dp[i] + ` `1` `);` ` ` `}` ` ` `}` ` ` `// Checking if not possible to` ` ` `// make the number as 2` ` ` `if` `(dp[N] == 1e9)` ` ` `return` `-` `1` `;` ` ` `// Return the minimum number` ` ` `// of operations` ` ` `return` `dp[N];` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `N = ` `25` `;` ` ` `System.out.println(minimumOperations(N));` ` ` `}` `}` `// This code is contributed by Potta Lokesh` |

## C#

`// C# program for above approach` `using` `System;` `class` `GFG{` ` ` `// Function to find the minimum number` ` ` `// of operations to reduce N to 2 by` ` ` `// dividing N by 5 or decrementing by 3` ` ` `static` `int` `minimumOperations(` `int` `N)` ` ` `{` ` ` ` ` `// Initialize the dp array` ` ` `int` `[] dp = ` `new` `int` `[N + 1];` ` ` `int` `i;` ` ` `// Initialize the array dp[]` ` ` `for` `(i = 0; i <= N; i++) {` ` ` `dp[i] = (` `int` `)1e9;` ` ` `}` ` ` `// For N = 2 number of operations` ` ` `// needed is zero` ` ` `dp[2] = 0;` ` ` `// Iterating over the range [1, N]` ` ` `for` `(i = 2; i <= N; i++) {` ` ` `// If it's not possible to` ` ` `// create current N` ` ` `if` `(dp[i] == (` `int` `)1e9)` ` ` `continue` `;` ` ` `// Multiply with 5` ` ` `if` `(i * 5 <= N) {` ` ` `dp[i * 5] = Math.Min(dp[i * 5], dp[i] + 1);` ` ` `}` ` ` `// Adding the value 3` ` ` `if` `(i + 3 <= N) {` ` ` `dp[i + 3] = Math.Min(dp[i + 3], dp[i] + 1);` ` ` `}` ` ` `}` ` ` `// Checking if not possible to` ` ` `// make the number as 2` ` ` `if` `(dp[N] == 1e9)` ` ` `return` `-1;` ` ` `// Return the minimum number` ` ` `// of operations` ` ` `return` `dp[N];` ` ` `}` `// Driver Code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `N = 25;` ` ` `Console.Write(minimumOperations(N));` `}` `}` `// This code is contributed by sanjoy_62.` |

## Javascript

`<script>` ` ` `// JavaScript Program for the above approach` ` ` `// Function to find the minimum number` ` ` `// of operations to reduce N to 2 by` ` ` `// dividing N by 5 or decrementing by 3` ` ` `function` `minimumOperations(N)` ` ` `{` ` ` ` ` `// Initialize the dp array` ` ` `let dp = ` `new` `Array(N + 1);` ` ` `let i;` ` ` `// Initialize the array dp[]` ` ` `for` `(i = 0; i <= N; i++) {` ` ` `dp[i] = 1e9;` ` ` `}` ` ` `// For N = 2 number of operations` ` ` `// needed is zero` ` ` `dp[2] = 0;` ` ` `// Iterating over the range [1, N]` ` ` `for` `(i = 2; i <= N; i++) {` ` ` `// If it's not possible to` ` ` `// create current N` ` ` `if` `(dp[i] == 1e9)` ` ` `continue` `;` ` ` `// Multiply with 5` ` ` `if` `(i * 5 <= N) {` ` ` `dp[i * 5] = Math.min(dp[i * 5], dp[i] + 1);` ` ` `}` ` ` `// Adding the value 3` ` ` `if` `(i + 3 <= N) {` ` ` `dp[i + 3] = Math.min(dp[i + 3], dp[i] + 1);` ` ` `}` ` ` `}` ` ` `// Checking if not possible to` ` ` `// make the number as 2` ` ` `if` `(dp[N] == 1e9)` ` ` `return` `-1;` ` ` `// Return the minimum number` ` ` `// of operations` ` ` `return` `dp[N];` ` ` `}` ` ` `// Driver Code` ` ` `let N = 25;` ` ` `document.write(minimumOperations(N));` `// This code is contributed by sanjoy_62.` `</script>` |

## Python3

`# Python 3 program for the above approach` `# Function to find the minimum number` `# of operations to reduce N to 2 by` `# dividing N by 5 or decrementing by 3` `def` `minimumOperations(N):` ` ` `# Initialize the dp array` ` ` `dp ` `=` `[` `0` `for` `i ` `in` `range` `(N ` `+` `1` `)]` ` ` `# Initialize the array dp[]` ` ` `for` `i ` `in` `range` `(N` `+` `1` `):` ` ` `dp[i] ` `=` `1000000000` ` ` `# For N = 2 number of operations` ` ` `# needed is zero` ` ` `dp[` `2` `] ` `=` `0` ` ` `# Iterating over the range [1, N]` ` ` `for` `i ` `in` `range` `(` `2` `,N` `+` `1` `,` `1` `):` ` ` `# If it's not possible to` ` ` `# create current N` ` ` `if` `(dp[i] ` `=` `=` `1000000000` `):` ` ` `continue` ` ` `# Multiply with 5` ` ` `if` `(i ` `*` `5` `<` `=` `N):` ` ` `dp[i ` `*` `5` `] ` `=` `min` `(dp[i ` `*` `5` `], dp[i] ` `+` `1` `)` ` ` `# Adding the value 3` ` ` `if` `(i ` `+` `3` `<` `=` `N):` ` ` `dp[i ` `+` `3` `] ` `=` `min` `(dp[i ` `+` `3` `], dp[i] ` `+` `1` `)` ` ` `# Checking if not possible to` ` ` `# make the number as 2` ` ` `if` `(dp[N] ` `=` `=` `1000000000` `):` ` ` `return` `-` `1` ` ` `# Return the minimum number` ` ` `# of operations` ` ` `return` `dp[N]` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `N ` `=` `25` ` ` `print` `(minimumOperations(N))` |

**Output:**

2

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

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