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
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:
- Those which do not exceed X on adding Y say count1 which can be reduced to ≤ 0 by A.
- 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 <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 = 3y = 3print(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 ?> |
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