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Count the numbers that can be reduced to zero or less in a gamein C++
Given an array of positive numbers and two integers A and B. Two players are playing a game in which they will reduce numbers in the array. Player 1 can decrease any element of the array by A and player 2 can increase any element of the array by B. The goal is to find the count of numbers that can be reduced to 0 or less by player 1. The first player makes the first move. The number once reduced to 0 or less can’t be taken into consideration by player 2.
For Example
Input
arr[] = { 1,4,5,2 } A=2, B=3Output
Count of numbers that can be reduced to zero or less in a game are: 1
Explanation
The only number that can be reduced by player 1 is 1 as on first move it will be reduced to −1. Rest all will become greater than A after player 2 increases their value.
Input
arr[] = { 1,4,5,2 } A=4, B=4Output
Count of numbers that can be reduced to zero or less in a game are: 2
Explanation
On first move player 1 reduces 4 to 0. arr[]= [ 1, 0, 5, 2 ] Player 2 will increase 1 to 4. arr[]= [ 5, 0, 5, 2 ] Player 1 will decrease 2 to −2. Arr[] = [ 5, 0, 5, −2 ]. From now onwards all numbers are greater than A so cannot be reduced by player 1 to 0 or less as player 2 is also increasing them simultaneously.
Approach used in the below program is as follows −
In this approach first check if A>B. If yes then in N moves A will reduce all of N elements of arr[] to 0 or less.If A<=B then we will check
all numbers which do not become greater than A even after player 2 adds B to them. Let's say this count as C1.
All numbers that are less than A and become greater than A after player 2 adds B to them. Let's say this count as C2.
Total count will be C= C1+ (C2+1)/2. As in case 2, only half of those will reduce to 0 or less as both players simultaneously increase/decrease them. And player 2 can only increase half of them to more than A. In the meantime, player 1 will reduce the other half to <=0.
Take an integer array containing positive numbers.
Take two variables A and B.
Function reduced_zero(int arr[], int size, int A, int B) will return the count of count the numbers that can be reduced to zero or less in a game
Take the initial count as 0.
Take two variables temp_1 and temp_2 as temporary counts.
If A > B then return the length of the array which is size.
Now traverse array using for loop, for each arr[i] if sum of element and B < A then increment temp_1.
For each element arr[i] <=A, increment temp_2.
Now after the end of a for loop, take count=temp_1+ (temp_2+1)/2. As given in formula.
Return count as result.
Example
#include <bits/stdc++.h>
using namespace std;
int reduced_zero(int arr[], int size, int A, int B){
int count = 0;
int temp_1 = 0, temp_2 = 0;
if (A > B){
return size;
}
for(int i = 0; i < size; i++){
if (A >= arr[i] + B){
temp_1++;
}
else if(A >= arr[i]){
temp_2++;
}
}
int temp = (temp_2 + 1) / 2;
count = temp + temp_1;
return count;
}
int main(){
int arr[] = { 3, 3, 1, 2, 4, 7, 1};
int A = 4, B = 1;
int size = sizeof(arr) / sizeof(arr[0]);
cout<<"Count of numbers that can be reduced to zero or less in a game are: "<<reduced_zero(arr, size, A, B);
return 0;
}Output
If we run the above code it will generate the following output −
Count of numbers that can be reduced to zero or less in a game are: 7
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