Minimum number of given operations required to reduce a number to 2

Given a positive integer N, the task is to reduce N to 2 by performing the following operations minimum number of times:

• Operation 1: Divide N by 5, if N is exactly divisible by 5.
• Operation 2: Subtract 3 from N.

If it is not possible, print -1.

Examples:

Input: N = 28
Output: 3
Explanation: Operation 1: Subtract 3 from 28. Therefore, N becomes 28 – 3 = 25.
Operation 2: Divide 25 by 5. Therefore, N becomes 25 / 5 = 5.
Operation 3: Subtract 3 from 5. Therefore, N becomes 5 – 3 = 2. 
Hence, the minimum number of operations required is 3.

Input: n=10
Output: 1
Explanation: Operation 1: Divide 10 by 5, so n becomes 10/5=2.
Hence, the minimum operations required is 1.

Naive Approach: The idea is to recursively compute the minimum number of steps required.  

• If the number is not divisible by 5, then subtract 3 from n and recur for the modified value of n, adding 1 to the result.
• Else make two recursive calls, one by subtracting 3 from n and the other by diving n by 5 and return the one with the minimum number of operations, adding 1 to the result.

Time Complexity: O(2ⁿ)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use dynamic programming. Follow these steps to solve this problem.

• Create an array, say dp[n+1] to store minimum operations and initialize all the entries with INT_MAX, where dp[i] stores the minimum number of steps required to reach 2 from i.
• Handle the base case by initializing dp[2] as 0.
• Iterate in the range [2, n] using the variable i
  • If the value of i*5 ? n, then update dp[i*5] to minimum of dp[i*5] and dp[i]+1.
  • If the value of i+3 ? n, then update dp[i+3] to minimum of dp[i+3] and dp[i]+1.
• Print the value of 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 required to reduce n to 2
int findMinOperations(int n)
{
    // Initialize a dp array
    int i, dp[n + 1];

    for (i = 0; i < n + 1; i++) {
        dp[i] = 999999;
    }

    // Handle the base case
    dp[2] = 0;

    // Iterate in the range [2, n]
    for (i = 2; i < n + 1; i++) {

        // Check if i * 5 <= n
        if (i * 5 <= n) {

            dp[i * 5] = min(
dp[i * 5], dp[i] + 1);
        }

        // Check if i + 3 <= n
        if (i + 3 <= n) {

            dp[i + 3] = min(
dp[i + 3], dp[i] + 1);
        }
    }

    // Return the result
    return dp[n];
}

// Driver code
int main()
{
    // Given Input
    int n = 28;

    // Function Call
    int m = findMinOperations(n);

    // Print the result
    if (m != 9999)
        cout << m;
    else
        cout << -1;

    return 0;
}

Java

// Java program for the above approach
public class GFG {

    // Function to find the minimum number
    // of operations required to reduce n to 2
    static int findMinOperations(int n)
    {
        // Initialize a dp array
        int i = 0;
        int dp[] = new int[n + 1];

        for (i = 0; i < n + 1; i++) {
            dp[i] = 999999;
        }

        // Handle the base case
        dp[2] = 0;

        // Iterate in the range [2, n]
        for (i = 2; i < n + 1; i++) {

            // Check if i * 5 <= n
            if (i * 5 <= n) {

                dp[i * 5] = Math.min(dp[i * 5], dp[i] + 1);
            }

            // Check if i + 3 <= n
            if (i + 3 <= n) {

                dp[i + 3] = Math.min(dp[i + 3], dp[i] + 1);
            }
        }

        // Return the result
        return dp[n];
    }

    // Driver code
    public static void main(String[] args)
    {

      // Given Input
        int n = 28;

        // Function Call
        int m = findMinOperations(n);

        // Print the result
        if (m != 9999)
            System.out.println(m);
        else
            System.out.println(-1);
    }
}

// This code is contributed by abhinavjain194

Python3

# Python3 program for the above approach

# Function to find the minimum number
# of operations required to reduce n to 2
def findMinOperations(n):

    # Initialize a dp array
    dp = [0 for i in range(n + 1)]

    for i in range(n + 1):
        dp[i] = 999999

    # Handle the base case
    dp[2] = 0

    # Iterate in the range [2, n]
    for i in range(2, n + 1):

        # Check if i * 5 <= n
        if (i * 5 <= n):
            dp[i * 5] = min(dp[i * 5],
                            dp[i] + 1)

        # Check if i + 3 <= n
        if (i + 3 <= n):
            dp[i + 3] = min(dp[i + 3],
                           dp[i] + 1)

    # Return the result
    return dp[n]

# Driver code
if __name__ == '__main__':

    # Given Input
    n = 28

    # Function Call
    m = findMinOperations(n)

    # Print the result
    if (m != 9999):
        print (m)
    else:
        print (-1)

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;

class GFG{

// Function to find the minimum number
// of operations required to reduce n to 2
static int findMinOperations(int n)
{

    // Initialize a dp array
    int i;
    int []dp = new int[n + 1];

    for(i = 0; i < n + 1; i++)
    {
        dp[i] = 999999;
    }

    // Handle the base case
    dp[2] = 0;

    // Iterate in the range [2, n]
    for(i = 2; i < n + 1; i++)
    {

        // Check if i * 5 <= n
        if (i * 5 <= n)
        {

            dp[i * 5] = Math.Min(dp[i * 5],
                                dp[i] + 1);
        }

        // Check if i + 3 <= n
        if (i + 3 <= n)
        {

            dp[i + 3] = Math.Min(dp[i + 3],
                                dp[i] + 1);
        }
    }

    // Return the result
    return dp[n];
}

// Driver code
public static void Main()
{

    // Given Input
    int n = 28;

    // Function Call
    int m = findMinOperations(n);

    // Print the result
    if (m != 9999)
        Console.Write(m);
    else
        Console.Write(-1);
}
}

// This code is contributed by SURENDRA_GANGWAR

Javascript

<script>

// JavaScript program for the above approach

// Function to find the minimum number
    // of operations required to reduce n to 2
function findMinOperations(n)
{
    // Initialize a dp array
        let i = 0;
        let dp = new Array(n + 1);

        for (i = 0; i < n + 1; i++) {
            dp[i] = 999999;
        }

        // Handle the base case
        dp[2] = 0;

        // Iterate in the range [2, n]
        for (i = 2; i < n + 1; i++) {

            // Check if i * 5 <= n
            if (i * 5 <= n) {

                dp[i * 5] = Math.min(dp[i * 5], dp[i] + 1);
            }

            // Check if i + 3 <= n
            if (i + 3 <= n) {

                dp[i + 3] = Math.min(dp[i + 3], dp[i] + 1);
            }
        }

        // Return the result
        return dp[n];
}

// Driver code
// Given Input
let n = 28;

// Function Call
let m = findMinOperations(n);

// Print the result
if (m != 9999)
    document.write(m);
else
    document.write(-1);


// This code is contributed by unknown2108

</script>

Output:

3

Time Complexity: O(n)
Auxiliary Space: O(n)