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:
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 operationsint 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 Codeint 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 approachimport java.util.*; class GFG{ // Function to find the minimum cost// to reduce the integer N to 1// by the given operationsstatic 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 codepublic 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 operationsdef 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 Codeif __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 approachusing System; class GFG{ // Function to find the minimum cost// to reduce the integer N to 1// by the given operationsstatic 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 codepublic 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


Output:
7

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

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