# Longest Common Extension / LCE | Set 2 ( Reduction to RMQ)

Prerequisites :

The Longest Common Extension (LCE) problem considers a string s and computes, for each pair (L , R), the longest sub string of s that starts at both L and R. In LCE, in each of the query we have to answer the length of the longest common prefix starting at indexes L and R.

Example:
String : “abbababba”
Queries: LCE(1, 2), LCE(1, 6) and LCE(0, 5)

Find the length of the Longest Common Prefix starting at index given as, (1, 2), (1, 6) and (0, 5).

The string highlighted “green” are the longest common prefix starting at index- L and R of the respective queries. We have to find the length of the longest common prefix starting at index- (1, 2), (1, 6) and (0, 5).

In Set 1, we explained about the naive method to find the length of the LCE of a string on many queries. In this set we will show how a LCE problem can be reduced to a RMQ problem, hence decreasing the asymptotic time complexity of the naive method.

Reduction of LCE to RMQ

Let the input string be S and queries be of the formLCE(L, R). Let the suffix array for s be Suff[] and the lcp array be lcp[].

The longest common extension between two suffixes SL and SR of S can be obtained from the lcp array in the following way.

• Let low be the rank of SL among the suffixes of S (that is, Suff[low] = L).
• Let high be the rank of SR among the suffixes of S. Without loss of generality, we assume that low < high.
• Then the longest common extension of SL and SR is lcp(low, high) = min (low<=k< high)lcp [k].

Proof: Let SL = SL…SL+C…sn and SR = SR…SR+c…sn, and let c be the longest common extension of SL and SR(i.e. SL…SL+C-1 = sn…SR+c-1). We assume that the string S has a sentinel character so that no suffix of S is a prefix of any other suffix of S but itself.

• If low = high – 1 then i = low and lcp[low] = c is the longest common extension of SL and SR and we are done.
• If low < high -1 then select i such lcp[i] is the minimum value in the interval [low, high] of the lcp array. We then have two possible cases:
• If c < lcp[i] we have a contradiction because SL . . . SL+lcp[i]-1 = SR. . . SR+lcp[i]-1 by the definition of the LCP table, and the fact that the entries of lcp correspond to sorted suffixes of S.
• if c > lcp[i], let high = Suff[i], so that Shigh is the suffix associated with position i. Si is such that shigh . . . shigh+lcp[i]-1 = SL . . . SL+lcp[i]-1 and shigh . . . shigh+lcp[i]-1 = SR . . . SR+lcp[i]-1, but since SL . . . SL+c-1 = SR. . . SR+c-1 we have that the lcp array should be wrongly sorted which is a contradiction.

Therefore we have c = lcp[i]

Thus we have reduced our longest common extension query to a range minimum-query over a range in lcp.

Algorithm

• To find low and high, we must have to compute the suffix array first and then from the suffix array we compute the inverse suffix array.
• We also need lcp array, hence we use Kasai’s Algorithm to find lcp array from the suffix array.
• Once the above things are done, we simply find the minimum value in lcp array from index – low to high (as proved above) for each query.

The minimum value is the length of the LCE for that query.

Implementation

```// A C++ Program to find the length of longest common
// extension using Direct Minimum Algorithm
#include<bits/stdc++.h>
using namespace std;

// Structure to represent a query of form (L,R)
struct Query
{
int L, R;
};

// Structure to store information of a suffix
struct suffix
{
int index;  // To store original index
int rank[2]; // To store ranks and next rank pair
};

// A utility function to get minimum of two numbers
int minVal(int x, int y) { return (x < y)? x: y; }

// A utility function to get minimum of two numbers
int maxVal(int x, int y) { return (x > y)? x: y; }

// A comparison function used by sort() to compare
// two suffixes Compares two pairs, returns 1 if
// first pair is smaller
int cmp(struct suffix a, struct suffix b)
{
return (a.rank[0] == b.rank[0])?
(a.rank[1] < b.rank[1]):
(a.rank[0] < b.rank[0]);
}

// This is the main function that takes a string 'txt'
// of size n as an argument, builds and return the
// suffix array for the given string
vector<int> buildSuffixArray(string txt, int n)
{
// A structure to store suffixes and their indexes
struct suffix suffixes[n];

// Store suffixes and their indexes in an array
// of structures.
// The structure is needed to sort the suffixes
// alphabatically and maintain their old indexes
// while sorting
for (int i = 0; i < n; i++)
{
suffixes[i].index = i;
suffixes[i].rank[0] = txt[i] - 'a';
suffixes[i].rank[1] =
((i+1) < n)? (txt[i + 1] - 'a'): -1;
}

// Sort the suffixes using the comparison function
// defined above.
sort(suffixes, suffixes+n, cmp);

// At his point, all suffixes are sorted according
// to first 2 characters.  Let us sort suffixes
// according to first 4/ characters, then first 8
// and so on

// This array is needed to get the index in suffixes[]
// from original index.  This mapping is needed to get
// next suffix.
int ind[n];

for (int k = 4; k < 2*n; k = k*2)
{
// Assigning rank and index values to first suffix
int rank = 0;
int prev_rank = suffixes[0].rank[0];
suffixes[0].rank[0] = rank;
ind[suffixes[0].index] = 0;

// Assigning rank to suffixes
for (int i = 1; i < n; i++)
{
// If first rank and next ranks are same as
// that of previous/ suffix in array, assign
// the same new rank to this suffix
if (suffixes[i].rank[0] == prev_rank &&
suffixes[i].rank[1] == suffixes[i-1].rank[1])
{
prev_rank = suffixes[i].rank[0];
suffixes[i].rank[0] = rank;
}
else // Otherwise increment rank and assign
{
prev_rank = suffixes[i].rank[0];
suffixes[i].rank[0] = ++rank;
}
ind[suffixes[i].index] = i;
}

// Assign next rank to every suffix
for (int i = 0; i < n; i++)
{
int nextindex = suffixes[i].index + k/2;
suffixes[i].rank[1] = (nextindex < n)?
suffixes[ind[nextindex]].rank[0]: -1;
}

// Sort the suffixes according to first k characters
sort(suffixes, suffixes+n, cmp);
}

// Store indexes of all sorted suffixes in the suffix array
vector<int>suffixArr;
for (int i = 0; i < n; i++)
suffixArr.push_back(suffixes[i].index);

// Return the suffix array
return  suffixArr;
}

/* To construct and return LCP */
vector<int> kasai(string txt, vector<int> suffixArr,
vector<int> &invSuff)
{
int n = suffixArr.size();

// To store LCP array
vector<int> lcp(n, 0);

// Fill values in invSuff[]
for (int i=0; i < n; i++)
invSuff[suffixArr[i]] = i;

// Initialize length of previous LCP
int k = 0;

// Process all suffixes one by one starting from
// first suffix in txt[]
for (int i=0; i<n; i++)
{
/* If the current suffix is at n-1, then we don’t
have next substring to consider. So lcp is not
defined for this substring, we put zero. */
if (invSuff[i] == n-1)
{
k = 0;
continue;
}

/* j contains index of the next substring to
be considered  to compare with the present
substring, i.e., next string in suffix array */
int j = suffixArr[invSuff[i]+1];

// Directly start matching from k'th index as
// at-least k-1 characters will match
while (i+k<n && j+k<n && txt[i+k]==txt[j+k])
k++;

lcp[invSuff[i]] = k; // lcp for the present suffix.

// Deleting the starting character from the string.
if (k>0)
k--;
}

// return the constructed lcp array
return lcp;
}

// A utility function to find longest common extension
// from index - L and index - R
int LCE(vector<int> lcp, vector<int>invSuff, int n,
int L, int R)
{
// Handle the corner case
if (L == R)
return (n-L);

int low = minVal(invSuff[L], invSuff[R]);
int high = maxVal(invSuff[L], invSuff[R]);

int length = lcp[low];

for (int i=low+1; i<high; i++)
{
if (lcp[i] < length)
length = lcp[i];
}

return (length);
}

// A function to answer queries of longest common extension
void LCEQueries(string str, int n, Query q[],
int m)
{
// Build a suffix array
vector<int>suffixArr = buildSuffixArray(str, str.length());

// An auxiliary array to store inverse of suffix array
// elements. For example if suffixArr[0] is 5, the
// invSuff[5] would store 0.  This is used to get next
// suffix string from suffix array.
vector<int> invSuff(n, 0);

// Build a lcp vector
vector<int>lcp = kasai(str, suffixArr, invSuff);

for (int i=0; i<m; i++)
{
int L = q[i].L;
int R = q[i].R;

printf ("LCE (%d, %d) = %d\n", L, R,
LCE(lcp, invSuff, n, L, R));
}

return;
}

// Driver Program to test above functions
int main()
{
string str = "abbababba";
int n = str.length();

Query q[] = {{1, 2}, {1, 6}, {0, 5}};
int m = sizeof(q)/sizeof(q[0]);

LCEQueries(str, n, q, m);

return (0);
}
```

Output:

```LCE (1, 2) = 1
LCE (1, 6) = 3
LCE (0, 5) = 4
```

Analysis of Reduction to RMQ method

Time Complexity :

• To construct the lcp and the suffix array it takes O(N.logN) time.
• To answer each query it takes O(|invSuff[R] – invSuff[L]|).
• Hence the overall time complexity is O(N.logN + Q. (|invSuff[R] – invSuff[L]|))
where,
Q = Number of LCE Queries.
N = Length of the input string.
invSuff[] = Inverse suffix array of the input string.
Although this may seems like an inefficient algorithm but this algorithm generally outperforms all other algorithms to answer the LCE queries.
We will give a detail description of the performance of this method in the next set.

Auxiliary Space: We use O(N) auxiliary space to store lcp, suffix and inverse suffix arrays.

Reference:

• http://www.sciencedirect.com/science/article/pii/S1570866710000377
•

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