# Competitive Coding : Cost of balloons October 16, 2020 by shekhar

### Problem Statement:

You are conducting a contest at your college. This contest consists of two problems and n participants. You know the problem that a candidate will solve during the contest.
You provide a balloon to a participant after he or she solves a problem. There are only green and purple-colored balloons available in a market. Each problem must have a balloon associated with it as a prize for solving that specific problem. You can distribute balloons to each participant by performing the following operation:
1. Use green-colored balloons for the first problem and purple-colored balloons for the second problem
2. Use purple-colored balloons for the first problem and green-colored balloons for the second problem
You are given the cost of each balloon and problems that each participant solve. Your task is to print the minimum price that you have to pay while purchasing balloons.
Input format
• First line: T that denotes the number of test cases (1≤T≤10)
• For each test case:
o First line: Cost of green and purple-colored balloons
o Second line: n that denotes the number of participants (1≤n≤10)
• Next n lines: Contain the status of users. For example, if the value of the jth integer in the ith row is 0, then it depicts that the ith participant has not solved the jth problem. Similarly, if the value of the jth integer in the ith row is 1, then it depicts that the ith participant has solved the jth problem.
Output format
For each test case, print the minimum cost that you have to pay to purchase balloons.

If you don’t need the explanation, just scroll down to the end of this page to get the full code for this question. If you have any doubts, you can ask me in comment section.

### Let’s Code it:

###### Step1 : Take the Inputs(explanation for each lines of code is written as comments) ###### Step2 : Define a method which takes the result array, cost of two types of balloon and number of participants as input and return the total cost as output. The logic is just to iterate the array ,multiply the first column with first price and 2nd column with Second price and finally find the sum of all costs and return the value(total cost). Use static keyword in the method if you don’t want to create instance of the class to call the method. If you want to create an object remove the static keyword and call the method by creating an object of the class. >

###### Step3 : Now the only thing left is getting the output and printing it. We have to find the minimum cost. The selection of prizes can be done in 2 possible ways i.e Green coloured balloons for solving 1st question and purple coloured balloons for solving second question.Purple coloured balloons for solving 1st question and green coloured balloons for solving second question.We can do that by calling the method 2 times by alternating the prices of 2 balloons(2nd and 3rd parameters of the method) and then by comparing the two values we can get the minimum cost and print that value. ###### Full code (Open in Laptop or use Desktop view in Smartphone) :

``````import java.util.Scanner;
class TestClass {
public static void main(String args[] ) throws Exception {
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
for(int k = 0; k < T ; k++ ){
int green_cost = sc.nextInt();
int purple_cost = sc.nextInt();
int n = sc.nextInt();
int[][] solve = new int[n];
for(int i = 0; i < n; i++){
for(int j = 0; j < 2; j++){
solve[i][j] = sc.nextInt();
}
}
int amt1 = get_amt(solve,green_cost,purple_cost,n);
int amt2 = get_amt(solve,purple_cost,green_cost,n);
if(amt1 < amt2){
System.out.println(amt1);
}
else{
System.out.println(amt2);
}
}
}
public static int get_amt(int[][] arr, int p1, int p2,int n){
int amt = 0;
for(int i = 0; i < n; i++){
for(int j = 0; j < 2; j++){
if(j == 0){
amt += (arr[i][j] * p1);
}
else{
amt += (arr[i][j] * p2);
}
}
}
return amt;
}
}``````