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Wordle | codechef solution

Rajnish Tripathi 06:00

 

Problem

Chef invented a modified wordle.

There is a hidden word � and a guess word �, both of length 5.

Chef defines a string � to determine the correctness of the guess word. For the ��ℎ index:

  • If the guess at the ��ℎ index is correct, the ��ℎ character of � is G.
  • If the guess at the ��ℎ index is wrong, the ��ℎ character of � is B.

Given the hidden word � and guess �, determine string �.

Input Format

  • First line will contain �, number of test cases. Then the test cases follow.
  • Each test case contains of two lines of input.
  • First line contains the string � - the hidden word.
  • Second line contains the string � - the guess word.

Output Format

For each test case, print the value of string �.

You may print each character of the string in uppercase or lowercase (for example, the strings BgBgB, BGBGB, bgbGB and bgbgb will all be treated as identical).

Constraints

  • 1≤�≤1000
  • ∣�∣=∣�∣=5
  • �,� contain uppercase english alphabets only.

Sample 1:

Input
Output
3
ABCDE
EDCBA
ROUND
RINGS
START
STUNT

BBGBB
GBBBB
GGBBG

Explanation:

Test Case 1: Given string �=ABCDE and �=EDCBA. The string � is:

  • Comparing the first indices, A≠E, thus, �[1]=B.
  • Comparing the second indices, B≠D, thus, �[2]=B.
  • Comparing the third indices, C=C, thus, �[3]=G.
  • Comparing the fourth indices, D≠B, thus, �[4]=B.
  • Comparing the fifth indices, E≠A, thus, �[5]=B.
    Thus, �=BBGBB.

Test Case 2: Given string �=ROUND and �=RINGS. The string � is:

  • Comparing the first indices, R=R, thus, �[1]=G.
  • Comparing the second indices, O≠I, thus, �[2]=B.
  • Comparing the third indices, U≠N, thus, �[3]=B.
  • Comparing the fourth indices, N≠G, thus, �[4]=B.
  • Comparing the fifth indices, D≠S, thus, �[5]=B.
    Thus, �=GBBBB.
Code(c++):-
#include <iostream>
using namespace std;

int main() {
    // your code goes here
    int t;
    cin>>t;
    while(t--)
    {
     string a,b,c="";
     cin>>a>>b;
     for(int i=0;a[i];i++)
     {
     if(a[i]==b[i])
     c+="G";
     else
     c+="B";
     }
     cout<<c<<endl;
    }
    return 0;
}


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