# 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 = 3Output: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 = 2Output:5

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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:

- Those which do not exceed
**X**on adding**Y**say**count1**which can be reduced to**â‰¤ 0**by**A**. - 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 <bits/stdc++.h>` `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

`<?php` `// PHP implementation of the approach` `// Function to return the count of numbers` `function` `countNumbers(` `$a` `, ` `$n` `, ` `$x` `, ` `$y` `)` `{` ` ` `// Base case` ` ` `if` `(` `$y` `< ` `$x` `)` ` ` `return` `$n` `;` ` ` `// Count the numbers` ` ` `$count1` `= 0 ;` ` ` `$count2` `= 0 ;` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `{` ` ` `if` `(` `$a` `[` `$i` `] + ` `$y` `<= ` `$x` `)` ` ` `$count1` `++;` ` ` `else` `if` `(` `$a` `[` `$i` `] <= ` `$x` `)` ` ` `$count2` `++;` ` ` `}` ` ` `$number` `= ` `floor` `((` `$count2` `+ 1) / 2) + ` `$count1` `;` ` ` `return` `$number` `;` `}` `// Driver Code` `$a` `= ` `array` `( 1, 2, 4, 2, 3 );` `$n` `= sizeof(` `$a` `);` `$x` `= 3;` `$y` `= 3;` `echo` `countNumbers(` `$a` `, ` `$n` `, ` `$x` `, ` `$y` `);` `// This code is contributed by Ryuga` `?>` |

## Javascript

`<script>` `// javascript implementation of the approach` `// Function to return the count of numbers` `function` `countNumbers(a, n, x, y)` `{` ` ` `// Base case` ` ` `if` `(y < x)` ` ` `return` `n;` ` ` ` ` `var` `i;` ` ` `// Count the numbers` ` ` `var` `count1 = 0, count2 = 0;` ` ` `for` `(i = 0; i < n; i++) {` ` ` `if` `(a[i] + y <= x)` ` ` `count1++;` ` ` `else` `if` `(a[i] <= x)` ` ` `count2++;` ` ` `}` ` ` `var` `number = (count2 + 1) / 2 + count1;` ` ` `return` `number;` `}` `// Driver Code` ` ` `var` `a = [1, 2, 4, 2, 3];` ` ` `var` `n = a.length;` ` ` `var` `x = 3, y = 3;` ` ` `document.write(parseInt(countNumbers(a, n, x, y)));` `// This code is contributed by ipg2016107.` `</script>` |

**Output:**

2