# Minimize positive product of two given numbers by at most N decrements

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++ program to implement` `// the above approach` `#include ` `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);` `}`

 `// 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`

 `# 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`

 `// 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`

Output:
```1

```

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

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Improved By : code_hunt, chitranayal, Rajput-Ji

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