C Solution:

int marsExploration(char* s) {
    int count=0,i=0;
    char SOS[]="SOS";
    while(*s){
        if(i>=3){
            i=0;
        }
        if(SOS[i]!=*s){
            count++;
        }
        i++;
        s++;
    }
    return count;
}

public static int marsExploration(String str) {
    int count = 0;
    for(int i=1; i<=str.length();i++) {
        char ch = str.charAt(i-1);
        int loc = i%3;
        if((loc == 1 && ch != 'S') || 
            (loc == 2 && ch != 'O') || 
            (loc == 0 && ch != 'S')) 
        count++;
    }
    return count;
}

Python solution with O(n/3) time and O(1) time. I am not sure if there is a more efficient solution. If anyone can find something more efficient please comment:

def marsExploration(s):
    i = 2
    res = 0
    while i < len(s):
        if s[i - 2] != 'S':
            res += 1
        if s[i - 1] != 'O':
            res += 1
        if s[i] != 'S':
            res += 1
            
        i += 3
    
    return res

For C#

return string.Concat(Enumerable.Range(0, s.Length / 3) .Select(s => "SOS")) .Select((c, i) => c != s[i] ? 1 : 0) .Sum();

Here is problem solution in python java c++ c and Javascript - https://programmingoneonone.com/hackerrank-mars-exploration-problem-solution.html