Funny String


Problem Statement :


In this challenge, you will determine whether a string is funny or not. To determine whether a string is funny, create a copy of the string in reverse e.g. abc  ->  cab. Iterating through each string, compare the absolute difference in the ascii values of the characters at positions 0 and 1, 1 and 2 and so on to the end. If the list of absolute differences is the same for both strings, they are funny.

Determine whether a give string is funny. If it is, return Funny, otherwise return Not Funny.


Function Description

Complete the funnyString function in the editor below.

funnyString has the following parameter(s):

string s: a string to test


Returns

string: either Funny or Not Funny


Input Format

The first line contains an integer s, the number of queries.
The next q lines each contain a string, s.

Constraints


1  <=  q  <=  10
2  <=  length of  s  <=  10000



Solution :



title-img


                            Solution in C :

In   C++  :







#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include<cstring>
using namespace std;


int main() {
  
    int t;
    cin>>t;
    while(t--)
        {
        char arr[10001],brr[10001];
        cin>>arr;
        int len=strlen(arr);
        int j=len-2;
        int i;
        for(i=1;i<len;i++,j--)
            {
            if(abs(arr[i]-arr[i-1])!=abs(arr[j]-arr[j+1]))
                {
                cout<<"Not Funny\n";
                break;
            }
        }
        if(i==len)
        cout<<"Funny\n";
        
    }
    return 0;
}








In   Java  :







import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner sc=new Scanner(System.in);
        int t=sc.nextInt();
        for(int j=0;j<t;j++)
        {
        String s=sc.next();
        int len=s.length();
        int stop=(len/2);
        int flag=0;
        int i=0;
        for(i=0;i<stop;i++)
            {
            int diff_fwd=(int)(s.charAt(i))-(int)(s.charAt(i+1));
            int diff_back=(int)(s.charAt(len-1-i))-(int)(s.charAt(len-1-(i+1)));
            if(Math.abs(diff_fwd)!=Math.abs(diff_back))
                {
                    flag=1;
                    break;
                }
        }
        if(flag==1)
            System.out.println("Not Funny");
        else
            System.out.println("Funny");
    }
}
}








In  C :






#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
    int T, N, funny, i, r;
    char S[10001];

    scanf("%d\n", &T);
    while (T--) {
        scanf("%s\n", S);
        N = strlen(S);
        funny = 1;
        for(i=1, r=N-2; i<N; i++, r--) {
            if (fabs(S[i]-S[i-1]) != fabs(S[r]-S[r+1])) {
                funny = 0;
                break;
            }
        }
        printf("%sFunny\n", funny ? "" : "Not ");
    }
    return 0;
}








In   Python3  :







def isFunny(s):
    for i in range(1, len(s)):
        if abs(s[i] - s[i-1]) != abs(s[-i-1] - s[-i]): return False
    return True
    

if __name__ == "__main__":
    num_cases = int(input());
    for case in range(num_cases):
        s = [ord(c) for c in input()];
        if isFunny(s): print("Funny")
        else: print("Not Funny");
                        








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