# Longest Common Extension / LCE | Set 1 (Introduction and Naive Method)

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)**.

**Algorithm (Naive Method)**

- For each of the LCE queries of the form – LCE(L, R) do the following:
- Initialise the LCE ‘length’ as 0
- Start comparing the prefix starting from index- L and R character by character.
- If the characters matches, then this character is in our Longest Common Extension. So increment ‘length’ (length++).
- Else if the characters mismatch, then return this ‘length’.

- The returned ‘length’ will be the required LCE(L, R).

**Implementation :**

Below is C++ implementation of above Naive algorithm.

`// A C++ Program to find the length of longest ` `// common extension using Naive Method ` `#include<bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Structure to represent a query of form (L,R) ` `struct` `Query ` `{ ` ` ` `int` `L, R; ` `}; ` ` ` `// A utility function to find longest common ` `// extension from index - L and index - R ` `int` `LCE(string str, ` `int` `n, ` `int` `L, ` `int` `R) ` `{ ` ` ` `int` `length = 0; ` ` ` ` ` `while` `(str[L+length] == str[R+length] && ` ` ` `R+length < n) ` ` ` `length++; ` ` ` ` ` `return` `(length); ` `} ` ` ` `// A function to answer queries of longest ` `// common extension ` `void` `LCEQueries(string str, ` `int` `n, Query q[], ` ` ` `int` `m) ` `{ ` ` ` `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(str, n, L, R)); ` ` ` `} ` ` ` `return` `; ` `} ` ` ` `// Driver Program to test above functions ` `int` `main() ` `{ ` ` ` `string str = ` `"abbababba"` `; ` ` ` `int` `n = str.length(); ` ` ` ` ` `// LCA Queries to answer ` ` ` `Query q[] = {{1, 2}, {1, 6}, {0, 5}}; ` ` ` `int` `m = ` `sizeof` `(q)/` `sizeof` `(q[0]); ` ` ` ` ` `LCEQueries(str, n, q, m); ` ` ` ` ` `return` `(0); ` `} ` |

*chevron_right*

*filter_none*

Output:

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

**Analysis of Naive Method**

Time Complexity: The time complexity is O(Q.N), where

**Q**= Number of LCE Queries

**N**= Length of the input string

One may be surprised that the although having a greater asymptotic time complexity, the naive method outperforms other efficient method(asymptotically) in practical uses. We will be discussing this in coming sets on this topic.

Auxiliary Space: O(1), in-place algorithm.

**Applications:**

- K-Mismatch Problem->Landau-Vishkin uses LCE as a subroutine to solve k-mismatch problem
- Approximate String Searching.
- Palindrome Matching with Wildcards.
- K-Difference Global Alignment.

In the next sets we will discuss how LCE (Longest Common Extension) problem can be reduced to a RMQ (Range Minimum Query). We will also discuss more efficient methods to find the longest common extension.

**Reference :**

This article is contributed by **Rachit Belwariar**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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