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Minimize positive product of two given numbers by at most N decrements
  • Last Updated : 22 Oct, 2020

Given three integers X, Y, and N, the task is to find the minimum possible positive product of X and Y that can be obtained by decreasing either the value of X or Y by 1 at most N times.

Examples:

Input: X = 5, Y= 6, N = 4
Output: 6
Explanation:
Decrease the value of X by 4, X = 5 – 4 = 1 and Y = 6.
Therefore, the minimized product = X * Y = 1 * 6 = 6

Input: X = 49, Y = 4256, N = 10
Output: 165984

Approach: The given problem can be solved based on the following observations:



If X ≤ Y: Reducing X minimizes the product. 
If Y ≤ X: Reducing Y minimizes the product.

Mathematical Proof:
If (X – 2) * Y < (X – 1) * (Y – 1) 
=> X * Y – 2 * Y < X * Y – X – Y + 1 
=> – 2 × Y < -X – Y + 1 
=> Y > X – 1

Follow the steps below to solve the problem:

  • If X ≤ Y: Follow the steps below: 
    • If N < X: Print Y * (X – N) as the answer as reducing X minimizes the product.
    • Otherwise, reduce X to 1 and reduce the remaining N from Y to minimize the product. Therefore, print Y – max(1, N – X + 1)) as the required minimized product.
  • Otherwise, if N < Y, print X * (Y – N) as the minimized product. If N ≥ Y, reduce Y to 1 and print max(X – (N – Y + 1), 1) as the minimized product.

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to minimize
// the product of two numbers
int minProd(int X, int Y,
            int N)
{
    if (X <= Y) {
 
        if (N < X)
 
            // Reducing X, N times,
            // minimizes the product
            return (X - N) * Y;
        else {
 
            // Reduce X to 1 and reduce
            // remaining N from Y
            return max(Y - (N - X + 1), 1);
        }
    }
 
    if (Y >= N)
 
        // Reducing Y, N times,
        // minimizes the product
        return (Y - N) * X;
 
    // Reduce Y to 1 and reduce
    // remaining N from X
    return max(X - (N - Y + 1), 1);
    ;
}
 
// Driver Code
int main()
{
    int X = 47, Y = 42, N = 167;
    cout << minProd(X, Y, N);
}

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Java

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// Java program to implement
// the above approach
import java.io.*;
 
class GFG{
 
// Function to minimize
// the product of two numbers
static int minProd(int X, int Y,
                   int N)
{
    if (X <= Y)
    {
        if (N < X)
   
            // Reducing X, N times,
            // minimizes the product
            return (X - N) * Y;
        else
        {
             
            // Reduce X to 1 and reduce
            // remaining N from Y
            return Math.max(Y - (N - X + 1), 1);
        }
    }
   
    if (Y >= N)
   
        // Reducing Y, N times,
        // minimizes the product
        return (Y - N) * X;
   
    // Reduce Y to 1 and reduce
    // remaining N from X
    return Math.max(X - (N - Y + 1), 1);
}
   
// Driver Code
public static void main (String[] args)
{
    int X = 47, Y = 42, N = 167;
     
    System.out.println(minProd(X, Y, N));
}
}
 
// This code is contributed by code_hunt

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Python3

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# Python3 program to implement
# the above approach
 
# Function to minimize
# the product of two numbers
def minProd(X, Y, N):
    if (X <= Y):
 
        if (N < X):
 
            # Reducing X, N times,
            # minimizes the product
            return (X - N) * Y
        else:
 
            # Reduce X to 1 and reduce
            # remaining N from Y
            return max(Y - (N - X + 1), 1)
 
    if (Y >= N):
 
        # Reducing Y, N times,
        # minimizes the product
        return (Y - N) * X
 
    # Reduce Y to 1 and reduce
    # remaining N from X
    return max(X - (N - Y + 1), 1)
 
# Driver Code
if __name__ == "__main__":
    X = 47
    Y = 42
    N = 167
    print (minProd(X, Y, N))
 
# This code is contributed by Chitranayal

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C#

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// C# program to implement
// the above approach
using System;
class GFG{
 
// Function to minimize
// the product of two numbers
static int minProd(int X,
                   int Y, int N)
{
  if (X <= Y)
  {
    if (N < X)
 
      // Reducing X, N times,
      // minimizes the product
      return (X - N) * Y;
    else
    {
      // Reduce X to 1 and reduce
      // remaining N from Y
      return Math.Max(Y - (N -
                      X + 1), 1);
    }
  }
 
  if (Y >= N)
 
    // Reducing Y, N times,
    // minimizes the product
    return (Y - N) * X;
 
  // Reduce Y to 1 and reduce
  // remaining N from X
  return Math.Max(X - (N -
                  Y + 1), 1);
}
   
// Driver Code
public static void Main(String[] args)
{
  int X = 47, Y = 42, N = 167;
  Console.WriteLine(minProd(X, Y, N));
}
}
 
// This code is contributed by Rajput-Ji

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Output: 

1









 

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

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