STRINGXOR | CONSECUTIVE XOR | July Lunchtime 110 2022 | Problem Solutions | CodeChef

[@_SarveshJoshi]

Whole lotta good explanation brother! Never thought a Carti fan will explain me Competitive Programming Problems

[@JunaidAhmed-jd5kq]

Great Explanation bro😍

[@ka_ka3141]

What's the intuition of alternate strings

[@ashutoshmishra3690]

very well explained thank you

[@dronrahangdale2361]

nice bro like the explanation point to point 😊✌️

[@sarveshsugandhbarnwal9809]

Nice explanation Gg

[@beinghappy9223]

Killer intuition

[@senor_nathak]

Sir java me solution bolo na

[@nikgshort3974]

import java.util.Scanner; class Solution3{ public static void main(String[] args) { Scanner scanner = new Scanner(System.in); int t = scanner.nextInt(); while(t-- > 0){ int n = scanner.nextInt(); String sa = scanner.next(); String sb = scanner.next(); boolean isEqual = true; boolean isBAlter = true; boolean sa0 = false; boolean sa1 = false; boolean sb0 = false; boolean sb1 = false; for(int i = 0 ; i < n ; i++){ if(sa.charAt(i) == '0'){ sa0 = true; }else{ sa1 = true; } if(sb.charAt(i) == '0'){ sb0 = true; } else{ sb1 = true; } if(sa.charAt(i) != sb.charAt(i)){ isEqual = false; } if(i >= 1 && sb.charAt(i) == sb.charAt(i - 1)) isBAlter = false; } if(isEqual) System.out.println("YES"); else{ if((!sa1 && sb1) || isBAlter) System.out.println("NO"); else System.out.println("YES"); } } } }