Maximize the value of x + y + z such that ax + by + cz = n

Given integers n, a, b and c, the task is to find the maximum value of x + y + z such that ax + by + cz = n.

Examples:

Input: n = 10, a = 5, b = 3, c = 4
Output: 3
x = 0, y = 2 and z = 1

Input: n = 50, a = 8, b = 10, c = 2
Output: 25
x = 0, y = 0 and z = 25

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: Fix the values of x and y then the value of z can be calculated as z = (n – (ax + by)) / c. If current value of z is an integer then update the maximum value of x + y + z found so far.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach #include using namespace std;    // Function to return the maximum value of (x + y + z) // such that (ax + by + cz = n) int maxResult(int n, int a, int b, int c) {     int maxVal = 0;        // i represents possible values of a * x     for (int i = 0; i <= n; i += a)            // j represents possible values of b * y         for (int j = 0; j <= n - i; j += b) {             float z = (n - (i + j)) / c;                // If z is an integer             if (floor(z) == ceil(z)) {                 int x = i / a;                 int y = j / b;                 maxVal = max(maxVal, x + y + (int)z);             }         }        return maxVal; }    // Driver code int main() {     int n = 10, a = 5, b = 3, c = 4;     cout << maxResult(n, a, b, c);        return 0; }

Java

 // Java implementation of the approach import java.util.*;    class GFG {    // Function to return the maximum value of (x + y + z) // such that (ax + by + cz = n) static int maxResult(int n, int a, int b, int c) {     int maxVal = 0;        // i represents possible values of a * x     for (int i = 0; i <= n; i += a)            // j represents possible values of b * y         for (int j = 0; j <= n - i; j += b)         {             float z = (n - (i + j)) / c;                // If z is an integer             if (Math.floor(z) == Math.ceil(z))             {                 int x = i / a;                 int y = j / b;                 maxVal = Math.max(maxVal, x + y + (int)z);             }         }        return maxVal; }    // Driver code public static void main(String args[]) {     int n = 10, a = 5, b = 3, c = 4;     System.out.println(maxResult(n, a, b, c)); } }    // This code is contributed by // Surendra_Gangwar

Python 3

 # Python3 implementation of the approach  from math import *;    # Function to return the maximum value  # of (x + y + z) such that (ax + by + cz = n)  def maxResult(n, a, b, c) :      maxVal = 0;         # i represents possible values of a * x      for i in range(0, n + 1, a) :            # j represents possible values of b * y          for j in range(0, n - i + 1, b) :              z = (n - (i + j)) / c;                 # If z is an integer              if (floor(z) == ceil(z)) :                  x = i // a;                  y = j // b;                  maxVal = max(maxVal, x + y + int(z));        return maxVal;     # Driver code  if __name__ == "__main__" :             n = 10     a = 5     b = 3     c = 4            print(maxResult(n, a, b, c));        # This code is contributed by Ryuga

C#

 // C# implementation of the approach  using System;    class GFG  {     // Function to return the maximum value of (x + y + z)  // such that (ax + by + cz = n)  static int maxResult(int n, int a, int b, int c)  {      int maxVal = 0;         // i represents possible values of a * x      for (int i = 0; i <= n; i += a)             // j represents possible values of b * y          for (int j = 0; j <= n - i; j += b)          {              float z = (n - (i + j)) / c;                 // If z is an integer              if (Math.Floor(z) == Math.Ceiling(z))              {                  int x = i / a;                  int y = j / b;                  maxVal = Math.Max(maxVal, x + y + (int)z);              }          }      return maxVal;  }     // Driver code  public static void Main(String []args)  {      int n = 10, a = 5, b = 3, c = 4;      Console.WriteLine(maxResult(n, a, b, c));  }  }     // This code has been contributed by 29AjayKumar

PHP



Output:

3

Time Complexity: O(N2)

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