Searching for the operation with less Effort

    import java.util.*;
public class A17 {

    static String cappyturnaround(String[]positions,int i, int j) 
    {   for
            ( i=0; i<positions.length;i++) 
    {
        for
            ( j=i+1; j< positions.length;j++) 
    {
        if
            (positions[i]!="H"||positions[j]!="H"||positions[j]!="V"||positions[i]!="V") 
    { 
                throw new IllegalArgumentException(positions[i|j]);
    }
    }

            }



for
( i=0; i<positions.length; i++) 
{
for
    ( j=i+1; j< positions.length;j++) 
{
if
(positions[i]=="H"||positions[j] == "H") 
{  positions[i]=positions[i].replace("H", "V");

}
}


}
return positions[i|j];
        }

    static String cappyturnaround2(String[]positions,int i, int j) 
    {  
        return "x";}






}

I am searching for a method to count the H's and V's to turn the Element which appears Less in to the other one e.g. 4 H's 2V's --> 6 H's. 4H's 6V's-> 10 V's

So far i just solved the problem in general so that I just can replace one Element without reviewing the effort. it would also be possible to solve the problem just for chars.