Rolling Hash - Question 2

1316. Distinct Echo Substrings

Return the number of distinct non-empty substrings of text that can be written as the concatenation of some string with itself (i.e. it can be written as a + a where a is some string).

Constraints:

1 <= text.length <= 2000

text has only lowercase English letters.

Analysis:

There are O(N^2) substring in total, and if we check each substring one by one, the time complexity for checking is O(N). So the overall time complexity is O(N^3).

The longest test case may have N as 2000, so N^3 ~ 10^10, which is too large.

We need to find some other ways faster.

One way is to use a rolling hashing method. For a fixed length, we just scan the string from left to right once with double pointers.

The first pointer is the ending of the first half; the second one is the ending of the second half. If they are the same, or give the same hash value, then we consider they are the same.

To avoid repeating counting, we can use a set.

See the code below,

class Solution {
public:
    int distinctEchoSubstrings(string text) {
        unordered_set<int> st;
        int n = text.size();
        for(int i=1; i<=n/2; ++i) {
            long h1 = 0, h2 = 0, p = 199, mod = 1e9+7, r = 1;
            for(int j=0, k=i; k<n; ++j, ++k) {
                if(j<i) r = r*p%mod;
                h1 = (h1*p%mod + text[j]-'a'+1)%mod;
                h2 = (h2*p%mod + text[k]-'a'+1)%mod;
                if(j>=i) {
                    h1 = (h1 + mod - (text[j-i]-'a'+1)*r%mod)%mod;
                    h2 = (h2 + mod - (text[k-i]-'a'+1)*r%mod)%mod;
                }
                if(j>=i-1 && h1 == h2) {
                    // cout<<text.substr(j-i+1, i*2)<<endl;
                    st.insert(h1);
                }
            }
        }
        return st.size();
    }
};

If do not want to use the rolling hash method, we can just use a counting to record the same number of elements in order. 

If the counting is the same as the length, we find one valid answer; when moving onto the next element, we just need to decrease the counting by 1;

If the two poninters point to two different elements, we need to reset the value of counting to be 0;

If the two pointers point to elements with the same value, the counting will be increased by 1. 

 See the code below,

class Solution {
public:
    int distinctEchoSubstrings(string text) {
        unordered_set<string> st;
        int n = text.size();
        for(int i=1; i<=n/2; ++i) {
            long ct = 0;
            for(int j=0, k=i; k<n; ++j, ++k) {
                if(text[j] == text[k]) ++ct;
                else ct = 0;
                if(ct == i) {
                    st.insert(text.substr(j-i+1, i));
                    --ct;
                }
            }
        }
        return st.size();
    }
};

Upper Layer