HackerEarth Jumping Tokens problem solution YASH PAL, 31 July 202413 February 2026 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:R x B -> xRB x R -> xBR x B -> BxB 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); } }} coding problems solutions HackerEarth HackerEarth