# 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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