Skip to main content

Alternate Characters

Problem Statement
Shashank likes strings in which consecutive characters are different. For example, he likesABABA, while he doesn't like ABAA. Given a string containing characters A and B only, he wants to change it into a string he likes. To do this, he is allowed to delete the characters in the string.
Your task is to find the minimum number of required deletions.
Input Format
The first line contains an integer T, i.e. the number of test cases. 
The next T lines contain a string each.
Output Format
For each test case, print the minimum number of deletions required.
Constraints
1≤T≤10 
1≤ length of string ≤105
Sample Input
5
AAAA
BBBBB
ABABABAB
BABABA
AAABBB
Sample Output
3
4
0
0
4
Explanation
AAAA ⟹ A, 3 deletions
BBBBB ⟹ B, 4 deletions
ABABABAB ⟹ ABABABAB, 0 deletions
BABABA ⟹ BABABA, 0 deletions
AAABBB ⟹ AB, 4 deletions because to convert it to AB we need to delete 2 A's and 2 B's

import java.util.Scanner;

public class Solution {
    static int numOfDels(String inpStr){
                                char[] chrArray=inpStr.toCharArray();
                                int len=chrArray.length;
                                int count=0;
                                for(int i=0;i<len-1;i++){
                                                if(chrArray[i]==chrArray[i+1]){
                                                                count++;
                                                }
                                }
                                return count;
                }
    public static void main(String[] args) {
        /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
        Scanner scan=new Scanner(System.in);
        int n=scan.nextInt();
        while(n>0){
                String strInp=scan.next();
                System.out.println(numOfDels(strInp));
        }
        //System.out.println(numOfDels(scan.nex());
    }
}

Score: 30.00
Test Case #0:  0.09s
Test Case #1:  0.09s
Test Case #2:  0.09s
Test Case #3:  0.09s
Test Case #4:  0.14s
Test Case #5:  0.15s
Test Case #6:  0.14s
Test Case #7:  0.24s
Test Case #8:  0.23s
Test Case #9:  0.3s
Test Case #10:  0.3s
Test Case #11:  0.29s
Test Case #12:  0.28s


Comments

Post a Comment

Popular posts from this blog

Java Interface

Problem Statement A Java interface can only contain method signatures and fields. Interface can be used to achieve polymorphism. In this problem you will practice your knowledge on interfaces. You are given an interface   AdvancedArithmetic   which contains a method signature   public abstract int divisorSum(int n) . You need to write a class called MyCalculator which implements the interface. divisorSum   function just takes an integer as input and return the sum of all its divisors. For example divisors of 6 are 1,2,3 and 6, so   divisorSum   should return 12. Value of n will be at most 1000. Read the partially completed code in the editor and complete it. You just need to write the MyCalculator class only.   Your class shouldn't be public. Sample Input 6 Sample Output I implemented: AdvancedArithmetic 12 Explanation Divisors of 6 are 1,2,3 and 6. 1+2+3+6=12. import java.util.*; interface AdvancedArithmetic{   p...
Post a Comment
Read more

change directory (cd) function for an abstract file system ( Java Implementation )

Write a function that provides change directory (cd) function for an abstract file system. Notes: Root path is '/'. Path separator is '/'. Parent directory is addressable as "..". Directory names consist only of English alphabet letters (A-Z and a-z). For example, new Path("/a/b/c/d").cd("../x").getPath() should return "/a/b/c/x". Note: The evaluation environment uses '\' as the path separator. public class Path {     private String path;     public Path(String path) {         this.path = path;     }     public String getPath() {         return path;     }     public Path cd(String newPath) {         //throw new UnsupportedOperationException("Waiting to be implemented."); String[] newP=newPath.split("/");     String[] oldP=path.split("/");     int lnCount=0;     for(String str:newP){     if(st...
6 comments
Read more

Java Abstract Class

Problem Statement A Java abstract class is a class that can't be instantiated, that means you cannot create new instances of an abstract class. It works as a base for subclasses. You should learn about Java Inheritence before attempting this challenge. Following is an example of abstact class: abstract class Book {     String title;     abstract void setTitle(String s);     String getTitle()     {         return title;     } } If you try to create an instance of this class like the following line you will get an error: Book new_novel=new Book(); You have to create another class that extends the abstract class. Then you can create instance of the new class. Notice that   setTitle   method is abstract too and has no body. That means you must implement the body of the that method in the child class. In the editor we have provided ...
Post a Comment
Read more