<p>You are <strong>building</strong> a string <code>s</code> of length <code>n</code> <strong>one</strong> character at a time, <strong>prepending</strong> each new character to the <strong>front</strong> of the string. The strings are labeled from <code>1</code> to <code>n</code>, where the string with length <code>i</code> is labeled <code>s<sub>i</sub></code>.</p>

<ul>
	<li>For example, for <code>s = &quot;abaca&quot;</code>, <code>s<sub>1</sub> == &quot;a&quot;</code>, <code>s<sub>2</sub> == &quot;ca&quot;</code>, <code>s<sub>3</sub> == &quot;aca&quot;</code>, etc.</li>
</ul>

<p>The <strong>score</strong> of <code>s<sub>i</sub></code> is the length of the <strong>longest common prefix</strong> between <code>s<sub>i</sub></code> and <code>s<sub>n</sub></code> (Note that <code>s == s<sub>n</sub></code>).</p>

<p>Given the final string <code>s</code>, return<em> the <strong>sum</strong> of the <strong>score</strong> of every </em><code>s<sub>i</sub></code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;babab&quot;
<strong>Output:</strong> 9
<strong>Explanation:</strong>
For s<sub>1</sub> == &quot;b&quot;, the longest common prefix is &quot;b&quot; which has a score of 1.
For s<sub>2</sub> == &quot;ab&quot;, there is no common prefix so the score is 0.
For s<sub>3</sub> == &quot;bab&quot;, the longest common prefix is &quot;bab&quot; which has a score of 3.
For s<sub>4</sub> == &quot;abab&quot;, there is no common prefix so the score is 0.
For s<sub>5</sub> == &quot;babab&quot;, the longest common prefix is &quot;babab&quot; which has a score of 5.
The sum of the scores is 1 + 0 + 3 + 0 + 5 = 9, so we return 9.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;azbazbzaz&quot;
<strong>Output:</strong> 14
<strong>Explanation:</strong> 
For s<sub>2</sub> == &quot;az&quot;, the longest common prefix is &quot;az&quot; which has a score of 2.
For s<sub>6</sub> == &quot;azbzaz&quot;, the longest common prefix is &quot;azb&quot; which has a score of 3.
For s<sub>9</sub> == &quot;azbazbzaz&quot;, the longest common prefix is &quot;azbazbzaz&quot; which has a score of 9.
For all other s<sub>i</sub>, the score is 0.
The sum of the scores is 2 + 3 + 9 = 14, so we return 14.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

#include<bits/stdc++.h>
using namespace std;
// #define int long long 
const int MOD = 1e9+7;
const int BASE1 = 5689;
const int BASE2 =     8861;

struct Hash
{ 
         //if you get overflow change the int,int to long long just
//o(n)
    vector<pair<long long ,long long>>hashes,pow;


    Hash(string s){
        hashes.assign(s.size()+1,{1,1});
        pow.assign(s.size()+1,{1,1});

        for(int i=0;i<s.size();i++){
            hashes[i+1]={(hashes[i].first*BASE1+(s[i]-'a'+1))%MOD,
                        (hashes[i].second*BASE2+(s[i]-'a'+1))%MOD};

            pow[i+1]={(pow[i].first*BASE1)%MOD,(pow[i].second*BASE2)%MOD};
        }
    }


    //converting the input 0 based indexing to 1 based
    pair<int,int>get(int l,int r){ // o(1)
        l++,r++;

        int hash1 =(hashes[r].first-(hashes[l-1].first*pow[r-l+1].first)%MOD+MOD)%MOD;
        int hash2 =(hashes[r].second-(hashes[l-1].second*pow[r-l+1].second)%MOD+MOD)%MOD;
    
        return {hash1,hash2};
    }
};
class Solution {
public:
    long long sumScores(string s) {
        int n= s.size();
        Hash a(s);
        long long ans=0;
        for(int i=0;i<n;i++){
            int start = 0,end= n-i;
            long long curr=0;
            while(start<=end){
                int mid= start+(end-start)/2;
                // int leng = mid-i+1;
                if(a.get(0,mid-1)==a.get(i,i+mid-1)){
                    curr= mid;
                    start=mid+1;
                }
                else{
                    end= mid-1;
                }
            }
            ans+=curr;
        }
        return ans;
    }
};
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