Cost of Ballons Problem Solution
Problem Statement:
You are conducting a contest .... while purchasing balloons.
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| Cost of Ballons problem |
Detailed Problem Statement: Cost of Ballons
Solution Code in Java :
import java.io.BufferedReader;
import java.io.InputStreamReader;
class TestClass {
public static void main(String args[]) throws Exception {
//testCases t
BufferedReader br = new BufferedReader(new InputStreamReader(System. in ));
int t = Integer.parseInt(br.readLine());
if (t >= 1 && t <= 10) {
for (int q = 0; q < t; q++) {
int x1 = 0,
x2 = 0,
n = 0,
counterCol0 = 0,
counterCol1 = 0,
result = 0;
int[] a = new int[2];
int[][] arr = new int[10][2];
String lines = br.readLine();
String[] strs = lines.trim().split("\\s+");
for (int i = 0; i < strs.length; i++) {
a[i] = Integer.parseInt(strs[i]);
}
x1 = a[0]; //ballon numbers
x2 = a[1]; // ballons numbers
//n number of participants
n = Integer.parseInt(br.readLine());
if (n >= 1 && n <= 10) {
for (int i = 0; i < n; i++) {
String line2 = br.readLine();
String[] str2 = line2.trim().split("\\s+");
for (int y = 0; y < 2; y++) {
arr[i][y] = Integer.parseInt(str2[y]);
//storing input as 1 1
// 1 0
//now logic comes
}
}
}
//here reading the participant answer from arr error
for (int i = 0; i < n; i++) {
for (int y = 0; y < 2; y++) {
if (y == 0) {
if (arr[i][y] == 1) counterCol0++;
//System.out.println(counterCol0 + ""+arr[i][y]);
}
if (y == 1) {
if (arr[i][y] == 1) counterCol1++;
// System.out.println(counterCol1 + ""+arr[i][y]);
}
}
}
//
if (counterCol0 >= counterCol1) {
//col0 is max
if (x1 <= x2) {
result = (x2 * counterCol1) + (x1 * counterCol0);
}
else {
//x1 is max
result = (x1 * counterCol1) + (x2 * counterCol0);
}
}
else if (counterCol0 < counterCol1) {
if (x1 <= x2) {
//x2 is max
result = (x2 * counterCol0) + (x1 * counterCol1);
}
else {
//x1 is max
result = (x1 * counterCol0) + (x2 * counterCol1);
}
}
System.out.println(result);
}
}
}
}
Hello Everyone i hope your Code works,i had test it on my PC using open JDK 1.8 and on HackerEarth open JDK 1.7 version. Comment down for any query, errors and new Problems.
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