Given a matrix of characters and a pattern, find the orientation of pattern in the matrix. In other words, find if pattern appears in matrix in horizontal or vertical direction. Achieve this in minimum time possible.
Input: mat[N][N] = { {'a', 'b', 'c', 'd', 'e'}, {'f', 'g', 'h', 'i', 'j'}, {'k', 'l', 'm', 'n', 'o'}, {'p', 'q', 'r', 's', 't'}, {'u', 'v', 'w', 'x', 'y'}}; pattern = "pqrs"; Output: Horizontal
We strongly recommend you to minimize your browser and try this yourself first.
A simple solution is for each row and column, use Naive pattern searching algorithm to find the orientation of pattern in the matrix. The time complexity of Naive pattern searching algorithm for every row is O(NM) where N is size of the matrix and M is length of the pattern. So, the time complexity of this solution will be O(N*(NM)) as each of N rows and N columns takes O(NM) time.
Can we do better?
The idea is to use KMP pattern matching algorithm for each row and column. The KMP matching algorithm improves the worst case to O(N + M). The total cost of a KMP search is linear in the number of characters of string and pattern. For a N x N matrix and pattern of length M, complexity of this solution will be O(N*(N+M)) as each of N rows and N columns will take O(N + M) time.
Implementation:
// C++ program for finding orientation of the pattern // using KMP pattern searching algorithm #include<bits/stdc++.h> using namespace std;
#define N 5 // Used in KMP Search for preprocessing the pattern void computeLPSArray( char *pat, int M, int *lps)
{ // length of the previous longest prefix suffix
int len = 0;
int i = 1;
lps[0] = 0; // lps[0] is always 0
// the loop calculates lps[i] for i = 1 to M-1
while (i < M)
{
if (pat[i] == pat[len])
{
len++;
lps[i++] = len;
}
else // (pat[i] != pat[len])
{
if (len != 0)
{
// This is tricky. Consider the example
// AAACAAAA and i = 7.
len = lps[len - 1];
// Also, note that we do not increment i here
}
else // if (len == 0)
{
lps[i++] = 0;
}
}
}
} int KMPSearch( char *pat, char *txt)
{ int M = strlen (pat);
// create lps[] that will hold the longest
// prefix suffix values for pattern
int *lps = ( int *) malloc ( sizeof ( int )*M);
int j = 0; // index for pat[]
// Preprocess the pattern (calculate lps[] array)
computeLPSArray(pat, M, lps);
int i = 0; // index for txt[]
while (i < N)
{
if (pat[j] == txt[i])
{
j++;
i++;
}
if (j == M)
{
// return 1 is pattern is found
return 1;
}
// mismatch after j matches
else if (i < N && pat[j] != txt[i])
{
// Do not match lps[0..lps[j-1]] characters,
// they will match anyway
if (j != 0)
j = lps[j - 1];
else
i = i + 1;
}
}
free (lps); // to avoid memory leak
// return 0 is pattern is not found
return 0;
} // Function to find orientation of pattern in the matrix // It uses KMP pattern searching algorithm void findOrientation( char mat[][N], char *pat)
{ // allocate memory for string containing cols
char *col = ( char *) malloc (N);
for ( int i = 0; i < N; i++)
{
// search in row i
if (KMPSearch(pat, mat[i]))
{
cout << "Horizontal" << endl;
return ;
}
// Construct an array to store i'th column
for ( int j = 0; j < N; j++)
col[j] = *(mat[j] + i);
// Search in column i
if (KMPSearch(pat, col))
cout << "Vertical" << endl;
}
// to avoid memory leak
free (col);
} // Driver Code int main()
{ char mat[N][N] = {{ 'a' , 'b' , 'c' , 'd' , 'e' },
{ 'f' , 'g' , 'h' , 'i' , 'j' },
{ 'k' , 'l' , 'm' , 'n' , 'o' },
{ 'p' , 'q' , 'r' , 's' , 't' },
{ 'u' , 'v' , 'w' , 'x' , 'y' }};
char pat[] = "pqrs" ;
findOrientation(mat, pat);
return 0;
} // This code is contributed by kumar65 |
// C program for finding orientation of the pattern // using KMP pattern searching algorithm #include<stdio.h> #include<string.h> #include<stdlib.h> #define N 5 // Used in KMP Search for preprocessing the pattern void computeLPSArray( char *pat, int M, int *lps)
{ // length of the previous longest prefix suffix
int len = 0;
int i = 1;
lps[0] = 0; // lps[0] is always 0
// the loop calculates lps[i] for i = 1 to M-1
while (i < M)
{
if (pat[i] == pat[len])
{
len++;
lps[i++] = len;
}
else // (pat[i] != pat[len])
{
if (len != 0)
{
// This is tricky. Consider the example
// AAACAAAA and i = 7.
len = lps[len-1];
// Also, note that we do not increment i here
}
else // if (len == 0)
{
lps[i++] = 0;
}
}
}
} int KMPSearch( char *pat, char *txt)
{ int M = strlen (pat);
// create lps[] that will hold the longest prefix suffix
// values for pattern
int *lps = ( int *) malloc ( sizeof ( int )*M);
int j = 0; // index for pat[]
// Preprocess the pattern (calculate lps[] array)
computeLPSArray(pat, M, lps);
int i = 0; // index for txt[]
while (i < N)
{
if (pat[j] == txt[i])
{
j++;
i++;
}
if (j == M)
{
// return 1 is pattern is found
return 1;
}
// mismatch after j matches
else if (i < N && pat[j] != txt[i])
{
// Do not match lps[0..lps[j-1]] characters,
// they will match anyway
if (j != 0)
j = lps[j-1];
else
i = i+1;
}
}
free (lps); // to avoid memory leak
// return 0 is pattern is not found
return 0;
} // Function to find orientation of pattern in the matrix // It uses KMP pattern searching algorithm void findOrientation( char mat[][N], char *pat)
{ // allocate memory for string containing cols
char *col = ( char *) malloc (N);
for ( int i = 0; i < N; i++)
{
// search in row i
if (KMPSearch(pat, mat[i]))
{
printf ( "Horizontal\n" );
return ;
}
// Construct an array to store i'th column
for ( int j = 0; j < N; j++)
col[j] = *(mat[j] + i);
// Search in column i
if (KMPSearch(pat, col))
printf ( "Vertical\n" );
}
// to avoid memory leak
free (col);
} // Driver program to test above function int main()
{ char mat[N][N] =
{
{ 'a' , 'b' , 'c' , 'd' , 'e' },
{ 'f' , 'g' , 'h' , 'i' , 'j' },
{ 'k' , 'l' , 'm' , 'n' , 'o' },
{ 'p' , 'q' , 'r' , 's' , 't' },
{ 'u' , 'v' , 'w' , 'x' , 'y' }
};
char pat[] = "pqrs" ;
findOrientation(mat, pat);
return 0;
} |
// Java program for finding orientation of the pattern // using KMP pattern searching algorithm import java.io.*;
import java.util.*;
class GFG
{ public static int N = 5 ;
// Used in KMP Search for preprocessing the pattern
public static void computeLPSArray( char pat[],
int M, int lps[])
{
// length of the previous longest prefix suffix
int len = 0 ;
int i = 1 ;
lps[ 0 ] = 0 ; // lps[0] is always 0
// the loop calculates lps[i] for i = 1 to M-1
while (i < M)
{
if (pat[i] == pat[len])
{
len++;
lps[i++] = len;
}
else // (pat[i] != pat[len])
{
if (len != 0 )
{
// This is tricky. Consider the example
// AAACAAAA and i = 7.
len = lps[len - 1 ];
// Also, note that we do not increment i here
}
else // if (len == 0)
{
lps[i++] = 0 ;
}
}
}
}
public static int KMPSearch( char pat[], char txt[])
{
int M = pat.length;
// create lps[] that will hold the longest
// prefix suffix values for pattern
int [] lps = new int [M];
int j = 0 ; // index for pat[]
// Preprocess the pattern (calculate lps[] array)
computeLPSArray(pat, M, lps);
int i = 0 ; // index for txt[]
while (i < N)
{
if (pat[j] == txt[i])
{
j++;
i++;
}
if (j == M)
{
// return 1 is pattern is found
return 1 ;
}
// mismatch after j matches
else if (i < N && pat[j] != txt[i])
{
// Do not match lps[0..lps[j-1]] characters,
// they will match anyway
if (j != 0 )
{
j = lps[j - 1 ];
}
else
{
i = i + 1 ;
}
}
}
// return 0 is pattern is not found
return 0 ;
}
// Function to find orientation of pattern in the matrix
// It uses KMP pattern searching algorithm
public static void findOrientation( char mat[][], char pat[])
{
// allocate memory for string containing cols
char [] col = new char [N];
for ( int i = 0 ; i < N; i++)
{
// search in row i
if (KMPSearch(pat, mat[i]) == 1 )
{
System.out.println( "Horizontal" );
return ;
}
// Construct an array to store i'th column
for ( int j = 0 ; j < N; j++)
{
col[j] = mat[j][i];
}
// Search in column i
if (KMPSearch(pat, col) == 1 )
{
System.out.println( "Vertical" );
}
}
}
// Driver Code
public static void main (String[] args)
{
char [][] mat = {
{ 'a' , 'b' , 'c' , 'd' , 'e' },{ 'f' , 'g' , 'h' , 'i' , 'j' },
{ 'k' , 'l' , 'm' , 'n' , 'o' },{ 'p' , 'q' , 'r' , 's' , 't' },
{ 'u' , 'v' , 'w' , 'x' , 'y' }};
char pat[] = { 'p' , 'q' , 'r' , 's' };
findOrientation(mat, pat);
}
} // This code is contributed by avanitrachhadiya2155 |
# Python3 program for finding orientation of the pattern # using KMP pattern searching algorithm N = 5
# Used in KMP Search for preprocessing the pattern def computeLPSArray(pat, M, lps):
# length of the previous longest prefix suffix
lenl = 0
i = 1
lps[ 0 ] = 0 # lps[0] is always 0
# the loop calculates lps[i] for i = 1 to M-1
while (i < M):
if (pat[i] = = pat[lenl]):
lenl + = 1
lps[i] = lenl
i + = 1
else : # (pat[i] != pat[lenl])
if (lenl ! = 0 ) :
# This is tricky. Consider the example
# AAACAAAA and i = 7.
lenl = lps[lenl - 1 ]
# Also, note that we do not increment i here
# if (lenl == 0)
else :
lps[i] = 0
i + = 1
def KMPSearch(pat, txt):
M = len (pat)
# create lps[] that will hold the longest
# prefix suffix values for pattern
lps = [ 0 ] * M
j = 0 # index for pat[]
# Preprocess the pattern (calculate lps[] array)
computeLPSArray(pat, M, lps)
# index for txt[]
i = 0 while (i < N):
if (pat[j] = = txt[i]):
j + = 1
i + = 1
if (j = = M):
# return 1 is pattern is found
return 1
# mismatch after j matches
elif (i < N and pat[j] ! = txt[i]):
# Do not match lps[0..lps[j-1]] characters,
# they will match anyway
if (j ! = 0 ):
j = lps[j - 1 ]
else :
i = i + 1
# to amemory leak
# return 0 is pattern is not found
return 0
# Function to find orientation of pattern in the matrix # It uses KMP pattern searching algorithm def findOrientation(mat, pat):
# allocate memory for string containing cols
col = [ 'a' ] * (N)
for i in range (N):
# search in row i
if (KMPSearch(pat, mat[i])):
print ( "Horizontal" )
return
# Construct an array to store i'th column
for j in range (N):
col[j] = mat[j][i]
# Search in column i
if (KMPSearch(pat, col)):
print ( "Vertical" )
# Driver Code mat = [[ 'a' , 'b' , 'c' , 'd' , 'e' ],
[ 'f' , 'g' , 'h' , 'i' , 'j' ],
[ 'k' , 'l' , 'm' , 'n' , 'o' ],
[ 'p' , 'q' , 'r' , 's' , 't' ],
[ 'u' , 'v' , 'w' , 'x' , 'y' ]]
pat = "pqrs"
findOrientation(mat, pat) # This code is contributed by Mohit kumar 29 |
// C# program for finding orientation of // the pattern using KMP pattern searching // algorithm using System;
class GFG{
public static int N = 5;
// Used in KMP Search for preprocessing the pattern public static void computeLPSArray( char [] pat,
int M, int [] lps)
{ // Length of the previous longest
// prefix suffix
int len = 0;
int i = 1;
// lps[0] is always 0
lps[0] = 0;
// The loop calculates lps[i]
// for i = 1 to M-1
while (i < M)
{
if (pat[i] == pat[len])
{
len++;
lps[i++] = len;
}
else // (pat[i] != pat[len])
{
if (len != 0)
{
// This is tricky. Consider the
// example AAACAAAA and i = 7.
len = lps[len - 1];
// Also, note that we do not
// increment i here
}
// If (len == 0)
else {
lps[i++] = 0;
}
}
}
} public static int KMPSearch( char [] pat, char [] txt)
{ int M = pat.Length;
// Create lps[] that will hold the longest
// prefix suffix values for pattern
int [] lps = new int [M];
int j = 0; // index for pat[]
// Preprocess the pattern
// (calculate lps[] array)
computeLPSArray(pat, M, lps);
int i = 0; // index for txt[]
while (i < N)
{
if (pat[j] == txt[i])
{
j++;
i++;
}
if (j == M)
{
// Return 1 is pattern is found
return 1;
}
// Mismatch after j matches
else if (i < N && pat[j] != txt[i])
{
// Do not match lps[0..lps[j-1]]
// characters, they will match anyway
if (j != 0)
{
j = lps[j - 1];
}
else
{
i = i + 1;
}
}
}
// Return 0 is pattern is not found
return 0;
} // Function to find orientation of pattern // in the matrix. It uses KMP pattern // searching algorithm public static void findOrientation( char [,] mat,
char [] pat)
{ // Allocate memory for string
// containing cols
char [] col = new char [N];
for ( int i = 0; i < N; i++)
{
// Allocate memory for string
// containing rows
char [] row = new char [(mat.GetLength(1))];
for ( int j = 0; j < mat.GetLength(1); j++)
{
row[j] = mat[i, j];
}
// Search in row i
if (KMPSearch(pat, row) == 1)
{
Console.WriteLine( "Horizontal" );
return ;
}
// Construct an array to store
// i'th column
for ( int j = 0; j < N; j++)
{
col[j] = mat[j,i];
}
// Search in column i
if (KMPSearch(pat, col) == 1)
{
Console.WriteLine( "Vertical" );
}
}
} // Driver Code static public void Main()
{ char [,] mat = { { 'a' , 'b' , 'c' , 'd' , 'e' },
{ 'f' , 'g' , 'h' , 'i' , 'j' },
{ 'k' , 'l' , 'm' , 'n' , 'o' },
{ 'p' , 'q' , 'r' , 's' , 't' },
{ 'u' , 'v' , 'w' , 'x' , 'y' } };
char [] pat = { 'p' , 'q' , 'r' , 's' };
findOrientation(mat, pat);
} } // This code is contributed by rag2127 |
<script> // Javascript program for finding orientation of the pattern // using KMP pattern searching algorithm let N = 5;
// Used in KMP Search for preprocessing the pattern
function computeLPSArray(pat,M,lps)
{
// length of the previous longest prefix suffix
let len = 0;
let i = 1;
lps[0] = 0; // lps[0] is always 0
// the loop calculates lps[i] for i = 1 to M-1
while (i < M)
{
if (pat[i] == pat[len])
{
len++;
lps[i++] = len;
}
else // (pat[i] != pat[len])
{
if (len != 0)
{
// This is tricky. Consider the example
// AAACAAAA and i = 7.
len = lps[len - 1];
// Also, note that we do not increment i here
}
else // if (len == 0)
{
lps[i++] = 0;
}
}
}
}
function KMPSearch(pat,txt)
{
let M = pat.length;
// create lps[] that will hold the longest
// prefix suffix values for pattern
let lps = new Array(M);
let j = 0; // index for pat[]
// Preprocess the pattern (calculate lps[] array)
computeLPSArray(pat, M, lps);
let i = 0; // index for txt[]
while (i < N)
{
if (pat[j] == txt[i])
{
j++;
i++;
}
if (j == M)
{
// return 1 is pattern is found
return 1;
}
// mismatch after j matches
else if (i < N && pat[j] != txt[i])
{
// Do not match lps[0..lps[j-1]] characters,
// they will match anyway
if (j != 0)
{
j = lps[j - 1];
}
else
{
i = i + 1;
}
}
}
// return 0 is pattern is not found
return 0;
}
// Function to find orientation of pattern in the matrix
// It uses KMP pattern searching algorithm
function findOrientation(mat,pat)
{
// allocate memory for string containing cols
let col = new Array(N);
for (let i = 0; i < N; i++)
{
// search in row i
if (KMPSearch(pat, mat[i]) == 1)
{
document.write( "Horizontal" );
return ;
}
// Construct an array to store i'th column
for (let j = 0; j < N; j++)
{
col[j] = mat[j][i];
}
// Search in column i
if (KMPSearch(pat, col) == 1)
{
document.write( "Vertical" );
}
}
}
// Driver Code
let mat=[['a ', ' b ', ' c ', ' d ', ' e '],
[' f ', ' g ', ' h ', ' i ', ' j '],
[' k ', ' l ', ' m ', ' n ', ' o '],
[' p ', ' q ', ' r ', ' s ', ' t '],
[' u ', ' v ', ' w ', ' x ', ' y']];
let pat= "pqrs" ;
findOrientation(mat, pat);
// This code is contributed by unknown2108
</script> |
Horizontal
Time complexity: O(n2).
Auxiliary Space: O(m)