CountDistinctSlices

// 100% solution

int solution(int M, vector<int> &A) {  
    int n = A.size();
    long result = 0;
    int front = 0;
    int back = 0;
    
    vector<bool> seen(M, false);
    
    while (front < n) {
        if (back < n && !seen[A[back]]) {
            seen[A[back]] = true;
            ++back;
        }
        else {
            result += (back - front);
            if (result > 1000000000) {
                return 1000000000;
            }
            seen[A[front]] = false;
            front++;
        }
    }
    
    return result;
}

// 70% solution
#include <set>

int solution(int M, vector<int> &A) {
    int result = 0;    
    set<int> cat;
    for (size_t i = 0; i < A.size(); ++i) {
        for (size_t j = i; j < A.size(); ++j) {
            if (cat.insert(A[j]).second) {
                if (++result >= 1000000000) {
                    return 1000000000;
                };
                continue;
            }  
            break;
        }
        cat.clear();
    }    
    return result;
}