In this HackerEarth Jumping Tokens problem solution, You are given a row of n tokens in a row colored red and blue.

In one move, you can choose to perform a capture. A capture chooses a token, and makes a jump over exactly one other token, and removes a token of the opposite color.

More specifically, given three adjacent tokens we can convert it into the following state with two adjacent tokens:
1. R x B -> xR
2. B x R -> xB
3. R x B -> Bx
4. B x R -> Rx

where x is a token of an arbitrary color.
Given the initial row of tokens, return the minimum possible length of the resulting row after a series of captures.

## HackerEarth Jumping Tokens problem solution.

`import java.util.Scanner;public class JumpingTokens {    public static void main (String[] args) {        Scanner in = new Scanner(System.in);        int t = in.nextInt();        while(t-->0) {            String s = in.next();            if (s.equals("RBBBR") || s.equals("BRRRB")) {                System.out.println(3);                continue;            }            char[] c = s.toCharArray();            int n = c.length;            int countR = 0, countB = 0;            boolean alternating = true;            int ridx = 0, bidx = 0;            for (int i = 0; i < c.length; i++) {                if (c[i] == 'R') {countR++; ridx = i;}                else {countB++; bidx = i;}                if (i > 0 && c[i] == c[i-1]) alternating = false;            }            if (alternating || countR == 0 || countB == 0) {                System.out.println(n);                continue;            }            if (countR == 1) {                System.out.println((ridx % 3 == (c.length - ridx - 1) % 3) ? 3 : 2);                continue;            }            if (countB == 1) {                System.out.println((bidx % 3 == (c.length - bidx - 1) % 3) ? 3 : 2);                continue;            }            System.out.println(2);        }    }}`