Minimum cost to reduce the integer N to 1 as per given conditions

Given four integers N, X, P, and Q, the task is to find the minimum cost to make N to 1 by the following two operations:

• Subtract 1 from N with cost as P.
• Divide N by X(if N is divisible by X), with cost Q.

Examples:

Input: N = 5, X = 2, P = 2, Q = 3 
Output: 7 
Explanation:
Operation 1: 5 – 1 -> cost = 2 
Operation 2: 4 / 2 -> cost = 3 
Operation 3: 2 – 1 -> cost = 2 
Minimum total cost is 2 + 3 + 2 = 7.

Input: N = 6, X = 6, P = 2, Q = 1 
Output: 1
Explanation:
Operation 1:  6 / 6 with cost = 1, hence that would be the minimum.

Approach: This problem can be solved using Greedy Approach. Below are the observations:

1. If x = 1, then the answer is (N – 1) * P.
2. Otherwise, if N is less than X, then it is only possible to decrease the number by 1, so the answer is (N – 1) * P.
3. Otherwise, take the first operation until N is not divisible by X.
4. Choose the second operation optimally by comparing the first and second operations i.e., if we can perform the first operation such that the cost of reducing N to 1 is minimum, else choose the second operation.

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 cost
// to reduce the integer N to 1
// by the given operations
int min_cost(int n, int x, int p, int q)
{
    // Check if x is 1
    if (x == 1) {

        // Print the answer
        cout << (n - 1) * p << endl;
        return 0;
    }

    // Prestore the answer
    int ans = (n - 1) * p;
    int pre = 0;

    // Iterate till n exists
    while (n > 1) {

        // Divide N by x
        int tmp = n / x;

        if (tmp < 0)
            break;

        pre += (n - tmp * x) * p;

        // Reduce n by x
        n /= x;

        // Add the cost
        pre += q;

        // Update the answer
        ans = min(ans,
                  pre + (n - 1) * p);
    }

    // Return the answer
    return ans;
}

// Driver Code
int main()
{
    // Initialize the variables
    int n = 5, x = 2, p = 2, q = 3;

    // Function call
    cout << min_cost(n, x, p, q);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;

class GFG{

// Function to find the minimum cost
// to reduce the integer N to 1
// by the given operations
static int min_cost(int n, int x,
                    int p, int q)
{

    // Check if x is 1
    if (x == 1)
    {

        // Print the answer
        System.out.println((n - 1) * p);
        return 0;
    }

    // Prestore the answer
    int ans = (n - 1) * p;
    int pre = 0;

    // Iterate till n exists
    while (n > 1)
    {

        // Divide N by x
        int tmp = n / x;

        if (tmp < 0)
            break;

        pre += (n - tmp * x) * p;

        // Reduce n by x
        n /= x;

        // Add the cost
        pre += q;

        // Update the answer
        ans = Math.min(ans,
                       pre + (n - 1) * p);
    }

    // Return the answer
    return ans;
}

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

    // Initialize the variables
    int n = 5, x = 2, p = 2, q = 3;

    // Function call
    System.out.println(min_cost(n, x, p, q));
}
}

// This code is contributed by offbeat

Python3

# Python3 program for the above approach

# Function to find the minimum cost
# to reduce the integer N to 1
# by the given operations
def min_cost(n, x, p, q):

  # Check if x is 1
  if (x == 1):

    # Print the answer
    print((n - 1) * p)
    return 0

  # Prestore the answer
  ans = (n - 1) * p
  pre = 0

  # Iterate till n exists
  while (n > 1):

    # Divide N by x
    tmp = n // x

    if (tmp < 0):
      break

    pre += (n - tmp * x) * p

    # Reduce n by x
    n //= x

    # Add the cost
    pre += q

    # Update the answer
    ans = min(ans, pre + (n - 1) * p)

  # Return the answer
  return ans

# Driver Code
if __name__ == '__main__':

  # Initialize the variables
  n = 5; x = 2;
  p = 2; q = 3;

  # Function call
  print(min_cost(n, x, p, q))

# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the minimum cost
// to reduce the integer N to 1
// by the given operations
static int min_cost(int n, int x,
                    int p, int q)
{

    // Check if x is 1
    if (x == 1)
    {

        // Print the answer
        Console.WriteLine((n - 1) * p);
        return 0;
    }

    // Prestore the answer
    int ans = (n - 1) * p;
    int pre = 0;

    // Iterate till n exists
    while (n > 1)
    {

        // Divide N by x
        int tmp = n / x;

        if (tmp < 0)
            break;

        pre += (n - tmp * x) * p;

        // Reduce n by x
        n /= x;

        // Add the cost
        pre += q;

        // Update the answer
        ans = Math.Min(ans,
                       pre + (n - 1) * p);
    }

    // Return the answer
    return ans;
}

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

    // Initialize the variables
    int n = 5, x = 2, p = 2, q = 3;

    // Function call
    Console.WriteLine(min_cost(n, x, p, q));
}
}

// This code is contributed by princiraj1992

Javascript

<script>
        // JavaScript Program to implement
        // the above approach

        // Function to find the minimum cost
        // to reduce the integer N to 1
        // by the given operations
        function min_cost(n, x, p, q) {
            // Check if x is 1
            if (x == 1) {

                // Print the answer
                document.write((n - 1) * p + "<br>");
                return 0;
            }

            // Prestore the answer
            let ans = (n - 1) * p;
            let pre = 0;

            // Iterate till n exists
            while (n > 1) {

                // Divide N by x
                let tmp = Math.floor(n / x);

                if (tmp < 0)
                    break;

                pre += (n - tmp * x) * p;

                // Reduce n by x
                n = Math.floor(n / x)

                // Add the cost
                pre += q;

                // Update the answer
                ans = Math.min(ans,
                    pre + (n - 1) * p);
            }

            // Return the answer
            return ans;
        }

        // Driver Code

        // Initialize the variables
        let n = 5, x = 2, p = 2, q = 3;

        // Function call
        document.write(min_cost(n, x, p, q));

// This code is contributed by Potta Lokesh
    </script>

Output:

7

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