Skip to content
Related Articles

Related Articles

Table of Contents

Improve Article
Save Article
Like Article

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

  • Last Updated : 27 Oct, 2021

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)

My Personal Notes
arrow_drop_up
Next

Largest odd divisor Game to check which player wins
Recommended Articles
Page :
Count of Binary Strings possible as per given conditions
30, Jul 20
Maximize the last Array element as per the given conditions
12, Aug 20
Maximum modified Array sum possible by choosing elements as per the given conditions
19, Aug 20
Maximum sum obtained by dividing Array into several subarrays as per given conditions
24, Jun 21
Minimum cost to complete given tasks if cost of 1, 7 and 30 days are given
09, Jul 21
Minimum cost to reduce A and B to 0 using square root or divide by 2
12, Oct 21
Maximum items that can be bought from the cost Array based on given conditions
16, May 20
Minimum cost to empty Array where cost of removing an element is 2^(removed_count) * arr[i]
20, Jul 20
Maximize cost of segment having weight at most K from given weight and cost of N items
09, Dec 21
Minimize cost to travel from source to destination in a Matrix based on given row change and column change cost
27, Dec 21
Minimize cost to reach bottom right from top left corner of Matrix with given separate cost for each move
11, Jan 22
Maximize Profit by trading stocks based on given rate per day
02, Jul 20
Arrange the numbers in the Array as per given inequalities
23, Jul 20
Nth angle of a Polygon whose initial angle and per angle increment is given
25, Jul 20
Count minimum character replacements required such that given string satisfies the given conditions
29, Oct 20
Minimum time required to print given string from a circular container based on given conditions
25, Aug 21
Minimize cost to split an array into K subsets such that the cost of each element is its product with its position in the subset
03, Jan 21
Minimize cost to connect the graph by connecting any pairs of vertices having cost at least 0
27, Sep 21
Minimum initial vertices to traverse whole matrix with given conditions
29, Nov 17
Minimum increment/decrement operations required on Array to satisfy given conditions
19, Aug 20
Minimum steps to determine the subsequence with max 1s based on given conditions
18, Nov 21
Minimum number of candies required to distribute among children based on given conditions
11, Jun 21
Minimum number of given operations required to reduce the array to 0 element
07, May 19
Minimum replacement of pairs by their LCM required to reduce given array to its LCM
08, Sep 20
Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
rutvik_56
@rutvik_56
Vote for difficulty





Improved By :
Article Tags :
Practice Tags :
Report Issue

We use cookies to ensure you have the best browsing experience on our website. By using our site, you
acknowledge that you have read and understood our
Cookie Policy &
        Privacy Policy

Lightbox

Start Your Coding Journey Now!