First day we won the game we are in lead.
Now
Today problems learning Outcomes:
- now we know how to take input as given in question (from 1st question).
- We know how to store the taken input in arrays.
- We know how to use Integer.parseInt() function.
- We learned how to apply logic like adding elements and storing in new array.
we use Integer.parseInt(value); here value can be string or float or long any datatype.
we use Stirng[] arr= str.split(” “) function to split a string.
we use int[] arr=new int[]{31,2,31,23,12}; to create new arry of int.
or we use int[] arr=new int[size]; to declare array of size size;
Problem 1
Given an integer array A of size N, find sum of elements in it. Input: First line contains an integer denoting the test cases 'T'. T testcases follow. Each testcase contains two lines of input. First line contains N the size of the array A. The second line contains the elements of the array. Output: For each testcase, print the sum of all elements of the array in separate line. Input Format Input: 2 3 3 2 1 4 1 2 3 4 Constraints Constraints:4 1 <= T <= 100 1<= N<= 100 1 <= Ai <= 100 Output Format Output: 6 10
Solution
import java.util.Scanner;
public class validate {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner iv=new Scanner(System.in);
int k=Integer.parseInt(iv.nextLine());
int result[]=new int[k];
for(int i=0;i<k;i++) {
int temp=Integer.parseInt(iv.nextLine());
String[] arr=iv.nextLine().split(” “);
int total=0;
for(int j=0;j<temp;j++) {
total+=Integer.parseInt(arr[j]);
}
result[i]=total;
}
for(int i=0;i<k;i++) {
System.out.println(result[i]);
}
}
Problem 2: Triangles can be formed or not from a input array.
int[] canMakeTrinagle(int[] arr){
int[] finalop=new int[arr.length-2];
for(int i=0;i<arr.length-2;i++){
if(arr[i]+arr[i+1]>arr[i+2] && arr[i+2]+arr[i+1]>arr[i] && arr[i+2]+arr[i]>arr[i+1]){
finalop[i]=1;
}
else{
finalop[i]=0;
}
}
return finalop;
}
Developer's & Smart Students Secret 