# Count the numbers that can be reduced to zero or less in a game

Given two integers X and Y and an array of N integers. Player A can decrease any element of the array by X and Player B can increase any element of the array by Y. The task is to count the number of elements that A can reduce to 0 or less. They both play optimally for infinite time with A making the first move.
Note: A number once reduced to zero or less cannot be increased.

Examples:

Input: a[] = {1, 2, 4, 2, 3}, X = 3, Y = 3
Output: 2
A reduces 2 to -1
B increases 1 to 4
A reduces 2 to -1
B increases 4 to 7 and the game goes on.

Input: a[] = {1, 2, 4, 2, 3}, X = 3, Y = 2
Output: 5

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

Approach: Since the game goes on for infinite time, we print N if X > Y. Now we need to solve for X ≤ Y. The numbers can be of two types:

1. Those which do not exceed X on adding Y say count1 which can be reduced to ≤ 0 by A.
2. Those which are < X and exceed X on adding Y say count2 only half of which can be reduced to ≤ 0 by A as they are playing optimally and B will try to increase any one of those number so that it becomes > X in each one of his turns.

So, the answer will be count1 + ((count2 + 1) / 2).

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of numbers ` `int` `countNumbers(``int` `a[], ``int` `n, ``int` `x, ``int` `y) ` `{ ` ` `  `    ``// Base case ` `    ``if` `(y < x) ` `        ``return` `n; ` ` `  `    ``// Count the numbers ` `    ``int` `count1 = 0, count2 = 0; ` `    ``for` `(``int` `i = 0; i < n; i++) { ` ` `  `        ``if` `(a[i] + y <= x) ` `            ``count1++; ` `        ``else` `if` `(a[i] <= x) ` `            ``count2++; ` `    ``} ` ` `  `    ``int` `number = (count2 + 1) / 2 + count1; ` ` `  `    ``return` `number; ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `a[] = { 1, 2, 4, 2, 3 }; ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a); ` ` `  `    ``int` `x = 3, y = 3; ` `    ``cout << countNumbers(a, n, x, y); ` `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` ` `  `class` `GFG ` `{ ` `    ``// Function to return the count of numbers ` `    ``static` `int` `countNumbers(``int` `a[], ``int` `n, ``int` `x, ``int` `y) ` `    ``{ ` `     `  `        ``// Base case ` `        ``if` `(y < x) ` `            ``return` `n; ` `     `  `        ``// Count the numbers ` `        ``int` `count1 = ``0``, count2 = ``0``; ` `        ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``{ ` `            ``if` `(a[i] + y <= x) ` `                ``count1++; ` `            ``else` `if` `(a[i] <= x) ` `                ``count2++; ` `        ``} ` `        ``int` `number = (count2 + ``1``) / ``2` `+ count1; ` `        ``return` `number; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String []args) ` `    ``{ ` `        ``int` `a[] = { ``1``, ``2``, ``4``, ``2``, ``3` `}; ` `        ``int` `n = a.length; ` `        ``int` `x = ``3``, y = ``3``; ` `        ``System.out.println(countNumbers(a, n, x, y));     ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

## Python3

 `# Python3 implementation of the approach ` ` `  `# Function to return the count of numbers ` `def` `countNumbers( a,  n, x, y): ` ` `  ` `  `    ``# Base case ` `    ``if` `(y < x): ` `        ``return` `n ` ` `  `    ``# Count the numbers ` `    ``count1 ``=` `0` `    ``count2 ``=` `0` `    ``for` `i ``in` `range` `( ``0``, n): ` ` `  `        ``if` `(a[i] ``+` `y <``=` `x): ` `            ``count1 ``=` `count1 ``+` `1` `        ``elif` `(a[i] <``=` `x): ` `            ``count2 ``=` `count2 ``+` `1` `     `  ` `  `    ``number ``=` `(count2 ``+` `1``) ``/``/` `2` `+` `count1 ` ` `  `    ``return` `number ` ` `  ` `  `# Driver Code ` `a ``=` `[ ``1``, ``2``, ``4``, ``2``, ``3` `] ` `n ``=` `len``(a) ` ` `  `x ``=` `3` `y ``=` `3` `print``(countNumbers(a, n, x, y)) ` ` `  `# This code is contributed by ihritik `

## C#

 `// C# implementation of the approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// Function to return the count of numbers ` `    ``static` `int` `countNumbers(``int` `[]a, ``int` `n, ``int` `x, ``int` `y) ` `    ``{ ` `     `  `        ``// Base case ` `        ``if` `(y < x) ` `            ``return` `n; ` `     `  `        ``// Count the numbers ` `        ``int` `count1 = 0, count2 = 0; ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `        ``{ ` `            ``if` `(a[i] + y <= x) ` `                ``count1++; ` `            ``else` `if` `(a[i] <= x) ` `                ``count2++; ` `        ``} ` `        ``int` `number = (count2 + 1) / 2 + count1; ` `        ``return` `number; ` `    ``} ` `     `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int` `[] a = { 1, 2, 4, 2, 3 }; ` `        ``int` `n = a.Length; ` `        ``int` `x = 3, y = 3; ` `        ``Console.WriteLine(countNumbers(a, n, x, y)); ` `    ``} ` `} ` ` `  `// This code is contributed by ihritik `

## PHP

 ` `

Output:

```2
```

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