In this Mars Exploration HackerRank Problem, 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, , determine how many letters of the SOS message have been changed by radiation.
Example
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, .
Constraints
- will contain only uppercase English letters, ascii[A-Z].
Explanation
Sample 0
= SOSSPSSQSSOR, and signal length . Sami sent SOS messages (i.e.: ).
Expected signal: SOSSOSSOSSOS
Recieved signal: SOSSPSSQSSOR
We print the number of changed letters, which is .
Sample 1
= SOSSOT, and signal length . Sami sent SOS messages (i.e.: ).
Expected Signal: SOSSOS
Received Signal: SOSSOT
We print the number of changed letters, which is .
Mars Exploration Hacker Rank Problem Solutions
I will Provide solution in Multiple programming languages for you. If you are not able to find the code in required language then please share in comments so that our team can help you.
Problem Solution in JavaScript
function marsExploration(s) {
let numberOfSos = s.length / 3;
let correctSignal = 'SOS'.repeat(numberOfSos);
let numberOfChanges = 0;
for (let i=0; i< s.length; i++){
if(s[i] !== correctSignal[i]){
numberOfChanges +=1;
}
}
return numberOfChanges;
}Problem Solution in Python
def marsExploration(s):
length = len(s)
numner_messages = length/3
expected_signal = 'SOS'*int(numner_messages)
count = 0
for char1, char2 in zip(s,expected_signal):
if char1 != char2:
count +=1
return(count)Problem Solution in Java
public static int marsExploration(String s) {
int count = 0;
for(int i = 0; i<s.length(); i=i+3){
String sub = s.substring(i,i+3);
if(!sub.equals("SOS")){
if(sub.charAt(0)!='S'){
count++;
}
if(sub.charAt(1)!='O'){
count++;
}
if(sub.charAt(2)!='S'){
count++;
}
}
}
return count;
}