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:
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 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

Output:

7

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

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