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

• Difficulty Level : Basic
• Last Updated : 24 May, 2021

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:
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:

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Approach: Since the game goes on for infinite time, we print N if X > Y. Now we need to solve for X â‰¤ 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 â‰¤ 0 by A.
2. Those which are < X and exceed X on adding Y say count2 only half of which can be reduced to â‰¤ 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[0]);` `    ``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

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## Javascript

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Output:
`2`

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