Longest Duplicate Substring in C++

Suppose we have a string S, consider all duplicated contiguous substrings that occur 2 or more times. (The occurrences may overlap.), We have to find the duplicated substring that has the longest possible length. If there is no such substrings, then return a blank string. As the answer may very large, so return in mod 10^9 + 7.

So, if the input is like "ababbaba", then the output will be "bab"

To solve this, we will follow these steps −

• m := 1e9 + 7

• Define a function add(), this will take a, b,

• return ((a mod m) + (b mod m)) mod m

• Define a function sub(), this will take a, b,

• return ((a mod m) - (b mod m) + m) mod m

• Define a function mul(), this will take a, b,

• return ((a mod m) * (b mod m)) mod m

• Define an array power

• Define a function ok(), this will take x, s,

• if x is same as 0, then −

  • return empty string

• Define one map called hash

• current := 0

• for initialize i := 0, when i < x, update (increase i by 1), do −

  • current := add(mul(current, 26), s[i] - 'a')

• hash[current] := Define an array (1, 0)

• n := size of s

• for initialize i := x, when i < n, update (increase i by 1), do −

  • current := sub(current, mul(power[x - 1], s[i - x] - 'a'))

  • current := add(mul(current, 26), s[i] - 'a')

  • if count is member of hash, then −

    • for all it in hash[current] −

      • if substring of s from it to x - 1 is same as substring of s from i - x + 1 to x - 1, then −

        • return substring of s from it to x - 1

  • Otherwise

    • insert i - x + 1 at the end of hash[current]

• return empty string

• From the main method, do the following −

• ret := empty string

• n := size of S

• power := Define an array of size n and fill this with 1

• for initialize i := 1, when i < n, update (increase i by 1), do −

  • power[i] := mul(power[i - 1], 26)

• low := 0, high := n - 1

• while low <= high, do −

  • mid := low + (high - low) /2

  • temp := ok(mid, S)

  • if size of temp is same as 0, then −

    • high := mid - 1

  • Otherwise

    • if size of temp > size of ret, then −

      • ret := temp

    • low := mid + 1

• return ret

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
   public:
   int m = 1e9 + 7;
   int add(lli a, lli b){
      return ((a % m) + (b % m)) % m;
   }
   int sub(lli a, lli b){
      return ((a % m) - (b % m) + m) % m;
   }
   int mul(lli a, lli b){
      return ((a % m) * (b % m)) % m;
   }
   vector<int> power;
   string ok(int x, string s){
      if (x == 0)
      return "";
      unordered_map<int, vector<int> > hash;
      lli current = 0;
      for (int i = 0; i < x; i++) {
         current = add(mul(current, 26), s[i] - 'a');
      }
      hash[current] = vector<int>(1, 0);
      int n = s.size();
      for (int i = x; i < n; i++) {
         current = sub(current, mul(power[x - 1], s[i - x] -
         'a'));
         current = add(mul(current, 26), s[i] - 'a');
         if (hash.count(current)) {
            for (auto& it : hash[current]) {
               if (s.substr(it, x) == s.substr(i - x + 1, x)) {
                  return s.substr(it, x);
               }
            }
         } else {
            hash[current].push_back(i - x + 1);
         }
      }
      return "";
   }
   string longestDupSubstring(string S){
      string ret = "";
      int n = S.size();
      power = vector<int>(n, 1);
      for (int i = 1; i < n; i++) {
         power[i] = mul(power[i - 1], 26);
      }
      int low = 0;
      int high = n - 1;
      while (low <= high) {
         int mid = low + (high - low) / 2;
         string temp = ok(mid, S);
         if (temp.size() == 0) {
            high = mid - 1;
         } else {
            if (temp.size() > ret.size())
            ret = temp;
            low = mid + 1;
         }
      }
      return ret;
   }
};
main(){
   Solution ob;
   cout << (ob.longestDupSubstring("ababbaba"));
}

Input

"ababbaba"

Output

bab