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
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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.
C++
// 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 contaning 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
// 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 contaning 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
// 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 contaning 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
# 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): # lenlgth 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 contaning 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#
// 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 // contaning cols char [] col = new char [N]; for ( int i = 0; i < N; i++) { // Allocate memory for string // contaning 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 |
Output:
Horizontal
This article is contributed by Aditya Goel. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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