Longest Non-Overlapping Repeated Substring using Suffix Tree/Array (Algorithm Only)

Question

I need to find the longest non-overlapping repeated substring in a String. I have the suffix tree and suffix array of the string available.

When overlapping is allowed,

Answer 1

Generate suffix array and sort in O(nlogn).ps: There is more effective algorithm like DC3 and Ukkonen algorithm. example:

String : ababc Suffix array: start-index of substring | substring
0 - ababc
2 - abc
1 - babc
3 - bc
4 - c

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compare each two consecutive substring and get common prefix with following constraint:
Say i1 is index of substring "ababc": 0
say i2 is index of substring "abc":2
common prefix is "ab" , length of common prefix is len

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abs(i1-i2) >len //avoid overlap

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go through suffix array with solution, and you will get the result of "ababc", which is "ab";

The whole solution will run O(nlogn)

However, there will be a special case: "aaaaa" that this solution can't solve thoroughly.
Welcome to discuss and come up to a solution of O(nlogn) but not O(n^2)

Answer 2

Complete code:

#include <bits/stdc++.h>
using namespace std;
int cplen(string a,string b){
    int i,to=min(a.length(),b.length());
    int ret=0;
    for(i=0;i<to;i++){
        if(a[i]==b[i])ret++;
        else {
            return ret;}
        }
    return ret;
    }
int main(){
    {
        int len,i;
        string str;
        cin>>str;
        len=str.length();
        vector<pair<string,int> >vv;
        map<char,int>hatbc;
        string pp="";
        for(i=len-1;i>=0;i--){
            hatbc[str[i]]++;
            pp=str.substr(i,len-i);
            vv.push_back(make_pair(pp,i));
            }
        if(len==1 || (int)hatbc.size()==len){
            printf("0\n");
            continue;
            }
        if(hatbc.size()==1){
            printf("%d\n",len/2);
            continue;
            }
        char prev=str[0];
        int foo=1,koo=0;
        for(i=1;i<len;){
            while(str[i]==prev && i<len){i++;foo++;}
            prev=str[i];
            i+=1;
            if(koo<foo)koo=foo;
            foo=1;
            }

        sort(vv.begin(),vv.end());
        int ans=0;
        ans=koo/2;
        for(i=1;i<(int)vv.size();i++){
            int j=i-1;
            int a=vv[j].second,b=vv[i].second;
            string sa=vv[j].first,sb=vv[i].first;
            int cpl;
            cpl=cplen(sa,sb);
            if(abs(a-b)>=cpl)
                ans=max(ans,cpl);
            }
        printf("%d\n",ans);
        }
    return 0;
    }

Complexity : O(n*log(n)) (due to sorting)