Answer is correct

#1

import java.util.*;
public class Main {
public static void main(String args[]) {
Scanner sc= new Scanner (System.in);
int test= sc.nextInt();
for(int i=0;i<test;i++){
String s= sc.nextLine();
System.out.println(pi(s));
}

}

public static String pi(String s){
	if(s.length()==1||s.length()==0){
		return s;
	}
	if(s.charAt(0)=='p'&&s.charAt(1)=='i'){
		s="3.14"+pi(s.substring(2,s.length()));
	}else {
		s=s.charAt(0)+pi(s.substring(1,s.length()));
	}
	return s;
}

}

#2

Hi @Hardik-Bhardwaj-392610454600906
For input
3
xpix
xabpixx3.15x
xpipippixx
Your code is giving output :-

x3.14x
xab3.14xx3.15x
But Actual output that should come is :-
x3.14x
xab3.14xx3.15x
x3.143.14p3.14xx
Use this approach in your you will be able to solve this problem

  • Traverse the string. At each instance check whether substring starting from index 0 till index 2 is same as β€œpi”.
  • If so , obtain the result for remaining string from index 2 i.e. s.substring(2) recursively . Store this result in say β€˜ros’ . Now return β€œ3.14” + ros.
  • If however , it was not so , obtain the result for the remaining string starting from index 1 and add s[0] to it i.e. return s[0] + ros.
#3

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#4

Hi @Hardik-Bhardwaj-392610454600906