Longest Duplicate Substring

#define p 10000007
//#define lli long long int

vector<int> roll;

class Solution {
    bool match(string s,int len,int size,string& ans)
    {
        unordered_map<int,vector<int>> m;   //Key->hashValue...Value->starting index of substring
        int hash=0;   //curr window hash
        for(int i=0;i<size;++i)
            hash = (hash*26 + (s[i]-'a'))%p;
        
        m[hash].push_back(0);
        //Rolling hash (sliding window)
        for(int j=size;j<len;++j)
        {
            hash = ((hash-roll[size-1]*(s[j-size]-'a'))%p + p)%p;
            hash = (hash*26 + (s[j]-'a'))%p;
            if(m.find(hash)!=m.end())
            {
                for(auto start: m[hash])
                {
                    string temp = s.substr(start,size);
                    string curr = s.substr(j-size+1,size);
                    if(temp.compare(curr)==0)
                    {
                        ans = temp;
                        return true;
                    }
                }
            }           
            m[hash].push_back(j-size+1);
        }
        return false;
    }

public:
    string longestDupSubstring(string S) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        
        int len = S.size();
        if(len==0)
            return "";
        
        //Store all rolling hash values
        roll.resize(len);  //Allocating fixed space to array
        roll[0] = 1;    //Since 26^0 = 1
        for(int i=1;i<len;++i)
            roll[i] = (26*roll[i-1])%p;
        
        int low=1,high=len,mid;
        string ans = "",temp;
        while(low<=high)
        {
            temp="";
            mid = low+(high-low)/2;
            bool flag = match(S,len,mid,temp);
            if(flag)
            {
                if(temp.size()>ans.size())
                    ans=temp;
                low = mid+1;
            }
            else
                high = mid-1;
        }
        return ans;
    }
};

Why +p is there on line 18
hash = ((hash-roll[size-1]*(s[j-size]-'a'))%p + p)%p;

because it might possible that after deleting from the hash value becomes negative so we generally adds the modulo value to make it positive and then take simple modulo you can take some examples and you will figure out

Why +p is there on line 18
hash = ((hash-roll[size-1]*(s[j-size]-'a'))%p + p)%p;

because it might possible that after deleting from the hash value becomes negative so we generally adds the modulo value to make it positive and then take simple modulo you can take some examples and you will figure out

Ya I need to study modulo operations.