Answer:
import java.util.Arrays;
import java.util.Scanner;
public class num1 {
public static void main(String[] args) {
Scanner in = new Scanner (System.in);
int []monthlyRainfall = new int [12];
System.out.println("Enter rainfall for January");
monthlyRainfall[0]= in.nextInt();
for(int i = 1; i<monthlyRainfall.length; i++){
System.out.println("Enter rainfall for the next Month");
monthlyRainfall[i]= in.nextInt();
}
System.out.println("The complete list is:");
System.out.println(Arrays.toString(monthlyRainfall));
//Calculating the total and average rainfall
int totalRainfall =0;
for(int i=0; i<monthlyRainfall.length; i++){
totalRainfall+=monthlyRainfall[i];
}
System.out.println("Total rainfall is "+ totalRainfall);
System.out.println("Average rainfall is "+(totalRainfall/12));
// finding the lowest rainfall
int min = monthlyRainfall[0];
int minMnth =0;
for(int i=0; i<monthlyRainfall.length; i++){
if(min>monthlyRainfall[i]){
min = monthlyRainfall[i];
minMnth =i+1;
}
}
System.out.println("The minimim rainfall is "+min);
System.out.println("The month with the minimum rainfall is "+minMnth);
// finding the highest rainfall
int max = monthlyRainfall[0];
int maxMnth =0;
for(int i=0; i<monthlyRainfall.length; i++){
if(max<monthlyRainfall[i]){
max = monthlyRainfall[i];
maxMnth = i+1;
}
}
System.out.println("The maximum rainfall is "+max);
System.out.println("The month with the maximum rainfall is "+maxMnth);
}
}
Step-by-step explanation:
- Create an array to hold the list of monthly rainfall from january to December int []monthlyRainfall = new int [12];
- Prompt user to enter the values for the monthly rainfall and fill up the array with the values using a for loop
- Print the populated List
- Use a for loop to calculate the Total rainfall by adding elements from index 0-11
- Calculate the average rainfall totalrainfal/12
- Using a for loop, find the maximum and minimum values in the array
- The index of the minimum value plus 1 Gives the month with the lowest rainfall
- The index of the maximum value +1 gives the month with the highest value