Replace of pii roblem

#1

import java.util.*;
public class Main{
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
for(int i=0;i<n;i++)
{
String str=sc.next();
String ans=replace(str,0);
System.out.println(ans);

	}
}
public static String replace(String str,int i){
	if(i>str.length()-1)
	return "";
	if(i==str.length()-1)
	return str.substring(i);
	String res="";
	String ros;
	if(str.charAt(i)=='p' && str.charAt(i+1)=='i'){
		res=res.substring(0,i)+"3.14";
		ros=replace(str,i+2);
	}
	else{
		res+=str.charAt(i);
		ros=replace(str,i+1);
	}
	return res+ros;

	}
	}
compilation successful but getting runtime error,please correct my code
#2

@karthik1989photos,
Don’t pass i as an argument. It will throw an exception because str.length will change at every iteration. You are on the right path though, just a few changes needed.

Suggested approach:

  • 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

import java.util.*;
public class Main
{
public static String pi(String str)
{
if (str.length() <= 1)
{
return str;
}
if (str.charAt(0) == β€˜p’ && str.charAt(1) == β€˜i’)
{
return β€œ3.14” + pi(str.substring(2));
} else
{
return str.charAt(0) + pi(str.substring(1));
} }
public static void main(String args[])
{
try{
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
for(int i=0;i<n;i++){
String s=sc.next();
System.out.println(pi(s));
}} catch(Exception e)
{
System.out.println(e);
}
}
i tried but showing errors at return statement plz correct my code

#4

@karthik1989photos,
https://ide.codingblocks.com/s/278795 I submitted the same code and its giving a correct answer. Maybe you have missed a closing bracket at the end. The code is correct.

#5

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