Question

Add an improved version of the the func4on called

od2ev

that returns

true

if the

number of odd numbers is twice more than the number of even numbers

Function:

bool ev5(int a1[], int l1) {

int evenCount = 0;

for(int i=0;i<l1;i++) {

if(a1[i] % 2 == 0) {

evenCount++;

Answer 1

abc

public class abcexamp {
/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int arr[]={2,12,31,11,99,87};
boolean b=od2ev(arr);
System.out.println("odd numbers is twice greater:"+b);
}
public static boolean od2ev(int arr[]){
int oddcount=0;
int evencount=0;
for(int i=0;i<arr.length;i++){
if(arr[i]%2==0)
evencount++;
else
oddcount++;
}
//System.out.println(evencount+" "+oddcount);
if(evencount*evencount==oddcount)
return true;
else
return false;
}
}

output:
odd numbers is twice greater:true
screenshots:

//as you dint mention any specific programing language ..i implemented in java...executed code is provided..and //also screenshots are provided...thank you