Mars Exploration


Problem Statement :


A space explorer's ship crashed on Mars! They send a series of SOS messages to Earth for help.

Letters in some of the SOS messages are altered by cosmic radiation during transmission. Given the signal received by Earth as a string, s, determine how many letters of the SOS message have been changed by radiation.

Example

s = 'SOSTOT'

The original message was SOSSOS. Two of the message's characters were changed in transit.


Function Description

Complete the marsExploration function in the editor below.

marsExploration has the following parameter(s):

string s: the string as received on Earth


Returns

int: the number of letters changed during transmission


Input Format

There is one line of input: a single string, s.



Constraints


1  <=   length of s  <=  99
length of s modulo 3 = 0
s will contain only uppercase English letters, ascii[A-Z].


Solution :



title-img


                            Solution in C :

In   C++  :






#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;


int main(){
    string S;
    cin >> S;
    long cnt = 0;
    for (int i = 0; i < S.size() / 3; i++) {
        if (S[3*i+0] != 'S') cnt++;
        if (S[3*i+1] != 'O') cnt++;
        if (S[3*i+2] != 'S') cnt++;
    }
    cout << cnt << endl;
}








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 in = new Scanner(System.in);
        String S = in.next();
        int numChanged = 0;
        
        for(int i = 0; i < S.length(); i++)
        {
            if(i % 3 == 1)
            {
                if(S.charAt(i) != 'O')
                {
                    numChanged++;
                }
            }
            else
            {
                if(S.charAt(i) != 'S')
                {
                    numChanged++;
                }
            }
        }
        
        System.out.println(numChanged);
    }
}









In   C  :








#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
    char* S = (char *)malloc(10240 * sizeof(char));
    scanf("%s",S);
    int i;
    int count=0;
    for(i=0;S[i]!='\0';i+=3){
        if(S[i]!='S'){
            count++;
        }
        if(S[i+1]!='O'){
            count++;
        }
        if(S[i+2]!='S'){
            count++;
        }
    }
    printf("%d",count);
    return 0;
}








In   Python3 :







s = input()
n = "SOS"*(len(s)//3)
cnt = 0
for i in range(len(s)):
    if s[i] != n[i]:
        cnt += 1
print(cnt)
                        




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