Longest Common Prefix | Set 4 (Binary Search)

Given a set of strings, find the longest common prefix.

```Input  : {“geeksforgeeks”, “geeks”, “geek”, “geezer”}
Output : "gee"

Input  : {"apple", "ape", "april"}
Output : "ap"
```

Previous Approaches – Word by Word Matching , Character by Character Matching, Divide and Conquer

Steps:

1. Find the string having the minimum length. Let this length be L.
2. Perform a binary search on any one string (from the input array of strings). Let us take the first string and do a binary search on the characters from the index – 0 to L-1.
3. Initially, take low = 0 and high = L-1 and divide the string into two halves – left (low to mid) and right (mid+1 to high).
4. Check whether all the characters in the left half is present at the corresponding indices (low to mid) of all the strings or not. If it is present then we append this half to our prefix string and we look in the right half in a hope to find a longer prefix.(It is guaranteed that a common prefix string is there.)
5. Otherwise, if all the characters in the left half is not present at the corresponding indices (low to mid) in all the strings, then we need not look at the right half as there is some character(s) in the left half itself which is not a part of the longest prefix string. So we indeed look at the left half in a hope to find a common prefix string. (It may be possible that we don’t find any common prefix string)

Algorithm Illustration considering strings as – “geeksforgeeks”, “geeks”, “geek”, “geezer”

C++

```//  A C++ Program to find the longest common prefix
#include<bits/stdc++.h>
using namespace std;

// A Function to find the string having the minimum
// length and returns that length
int findMinLength(string arr[], int n)
{
int min = INT_MAX;

for (int i=0; i<=n-1; i++)
if (arr[i].length() < min)
min = arr[i].length();
return(min);
}

bool allContainsPrefix(string arr[], int n, string str,
int start, int end)
{
for (int i=0; i<=n-1; i++)
for (int j=start; j<=end; j++)
if (arr[i][j] != str[j])
return (false);
return (true);
}

// A Function that returns the longest common prefix
// from the array of strings
string commonPrefix(string arr[], int n)
{
int index = findMinLength(arr, n);
string prefix; // Our resultant string

// We will do an in-place binary search on the
// first string of the array in the range 0 to
// index
int low = 0, high = index;

while (low <= high)
{
// Same as (low + high)/2, but avoids overflow
// for large low and high
int mid = low + (high - low) / 2;

if (allContainsPrefix (arr, n, arr[0], low, mid))
{
// If all the strings in the input array contains
// this prefix then append this substring to
prefix = prefix + arr[0].substr(low, mid-low+1);

// And then go for the right part
low = mid + 1;
}

else // Go for the left part
high = mid - 1;
}

return (prefix);
}

// Driver program to test above function
int main()
{
string arr[] = {"geeksforgeeks", "geeks",
"geek", "geezer"};
int n = sizeof (arr) / sizeof (arr[0]);

string ans = commonPrefix(arr, n);

if (ans.length())
cout << "The longest common prefix is "
<< ans;
else
cout << "There is no common prefix";
return (0);
}
```

Java

```// A Java Program to find the longest common prefix

class GFG {

// A Function to find the string having the
// minimum length and returns that length
static int findMinLength(String arr[], int n)
{
int min = Integer.MAX_VALUE;
for (int i = 0; i <= (n - 1); i++)
{
if (arr[i].length() < min) {
min = arr[i].length();
}
}
return min;
}

static boolean allContainsPrefix(String arr[], int n,
String str, int start, int end)
{
for (int i = 0; i <= (n - 1); i++)
{
String arr_i = arr[i];

for (int j = start; j <= end; j++)
if (arr_i.charAt(j) != str.charAt(j))
return false;
}
return true;
}

// A Function that returns the longest common prefix
// from the array of strings
static String commonPrefix(String arr[], int n)
{
int index = findMinLength(arr, n);
String prefix = ""; // Our resultant string

// We will do an in-place binary search on the
// first string of the array in the range 0 to
// index
int low = 0, high = index;
while (low <= high) {

// Same as (low + high)/2, but avoids
// overflow for large low and high
int mid = low + (high - low) / 2;

if (allContainsPrefix(arr, n, arr[0], low,
mid))
{
// If all the strings in the input array
// contains this prefix then append this
prefix = prefix + arr[0].substring(low,
mid - low + 1);

// And then go for the right part
low = mid + 1;
}
else // Go for the left part
{
high = mid - 1;
}
}

return prefix;
}

// Driver program to test above function
public static void main(String args[])
{
String arr[] = {"geeksforgeeks", "geeks",
"geek", "geezer"};
int n = arr.length;

String ans = commonPrefix(arr, n);

if (ans.length() > 0)
System.out.println("The longest common"
+ " prefix is " + ans);
else
System.out.println("There is no common"
+ " prefix");
}
}

// This code is contributed by Indrajit Sinha.
```

Output :

`The longest common prefix is - gee`

Time Complexity :
The recurrence relation is

`T(M) = T(M/2) + O(MN) `

where

```N = Number of strings
M = Length of the largest string string
```

So we can say that the time complexity is O(NM log M)

Auxiliary Space: To store the longest prefix string we are allocating space which is O(N) where, N = length of the largest string among all the strings

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.