Unable to think how i will write its code

#1

how i will make mask and how to include and not include elements in it , and what will be the base case and what will be s in starting ( is it OR of all the values in string) ??

#2

@aryan_007
Yeah the question doesn’t use bitmasking but instead has a very interesting approach.
The dp part is precalculation of the prefix sums for each digit at every index i.

    F(i,0,n-1)
    {
        cnt[i][x[i]-'0']++;
        if(i>=1)
        {
            F(j,0,9)
                cnt[i][j] += cnt[i-1][j];
        }
    }

and hass is the required amount

    F(i,0,m-1)
    {
        ll y;
        cin>>y;
        hass[y]++;
    }

Now the idea to solve is that ,for every index i ,find the closest index j,(j>=i), such that this range contains all the digits present in hass atleast once. So this can be done using prefix sums we precalculated.

            F(j,0,9)
            {
                if(i>=1)
                {
                    if(hass[j] > cnt[mid][j]-cnt[i-1][j])
                        check = 0;
                }
                else
                {
                    if(hass[j] > cnt[mid][j])
                        check = 0;
                }
            }

Now if you are familiar with how binary search works, we can find this j in log(n) time using binary search.
Once you find this j you are supposed to add the number of substrings starting from i which follow the property, so that would be simply

        if(id!=-1)
        {
            ans += (n-id);
        }

So the complexity of the solution is O(Nlog(N)*10)