int marsExploration(char* s) { char* a = malloc(strlen(s)*sizeof(char)); strcpy(a,s); int count=0; for(int i=0;a[i]!='\0';i++){ if(a[i]!='S' && a[i]!='O') count++; } return count; } Why it is showing only some test cases are fail?

Here is my c++ solution

int marsExploration(string s) { int sz = s.size(); string strSos("SOS");

int no_sos = sz/ 3;

for(int i = 0; i < no_sos-1; i++)
    strSos.append("SOS");

int cnt = 0;
for( int i = 0; i < sz; i++)
{
    if( strSos[i] != s[i] )
        cnt++;
}
return cnt;

}

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