2405. Optimal Partition of String

HashTable

int partitionString(char* s) {
    int start = 0;
    int len = strlen(s);
    int max_len = 0;
    int count = 1;
    int hashtab[256];
    memset(hashtab, -1, sizeof(hashtab));
 
    for (int i = 0; i < len; i++) {
        if (hashtab[s[i]] >= start) {
            start = i;
            count++;
        }
        hashtab[s[i]] = i;
    }
    return count;
}

Bitmask as HashTable

int partitionString(char* s)
{
    int i,n,cnt,found;
    n=(int)strlen(s);
    cnt=1;
    found=0;
    for(i=0;i<n;i++)
    {
        if(found&(1<<(s[i]-'a')))
        {
            found=0;
            cnt++;
        }
        found|=1<<(s[i]-'a');
    }
    return cnt;
}