Given text string with length n and a pattern with length m, the task is to prints all occurrences of pattern in text.
Note: You may assume that n > m.
Examples:
Input: text = "THIS IS A TEST TEXT", pattern = "TEST"
Output: Pattern found at index 10Input: text = "AABAACAADAABAABA", pattern = "AABA"
Output: Pattern found at index 0, Pattern found at index 9, Pattern found at index 12
Pattern searching
Naive Pattern Searching algorithm:
Slide the pattern over text one by one and check for a match. If a match is found, then slide by 1 again to check for subsequent matches.
C++
#include <iostream>
#include <string>
using namespace std;
void search(string& pat, string& txt) {
int M = pat.size();
int N = txt.size();
// A loop to slide pat[] one by one
for (int i = 0; i <= N - M; i++) {
int j;
// For current index i, check for pattern match
for (j = 0; j < M; j++) {
if (txt[i + j] != pat[j]) {
break;
}
}
// If pattern matches at index i
if (j == M) {
cout << "Pattern found at index " << i << endl;
}
}
}
// Driver's Code
int main() {
// Example 1
string txt1 = "AABAACAADAABAABA";
string pat1 = "AABA";
cout << "Example 1: " << endl;
search(pat1, txt1);
// Example 2
string txt2 = "agd";
string pat2 = "g";
cout << "\nExample 2: " << endl;
search(pat2, txt2);
return 0;
}
C
#include <stdio.h>
#include <string.h>
void search(char* pat, char* txt) {
int M = strlen(pat);
int N = strlen(txt);
// A loop to slide pat[] one by one
for (int i = 0; i <= N - M; i++) {
int j;
// For current index i, check for pattern match
for (j = 0; j < M; j++) {
if (txt[i + j] != pat[j]) {
break;
}
}
// If pattern matches at index i
if (j == M) {
printf("Pattern found at index %d\n", i);
}
}
}
int main() {
// Example 1
char txt1[] = "AABAACAADAABAABA";
char pat1[] = "AABA";
printf("Example 1:\n");
search(pat1, txt1);
// Example 2
char txt2[] = "agd";
char pat2[] = "g";
printf("\nExample 2:\n");
search(pat2, txt2);
return 0;
}
Java
public class PatternSearch {
public static void search(String pat, String txt) {
int M = pat.length();
int N = txt.length();
// A loop to slide pat[] one by one
for (int i = 0; i <= N - M; i++) {
int j;
// For current index i, check for pattern match
for (j = 0; j < M; j++) {
if (txt.charAt(i + j) != pat.charAt(j)) {
break;
}
}
// If pattern matches at index i
if (j == M) {
System.out.println("Pattern found at index " + i);
}
}
}
public static void main(String[] args) {
// Example 1
String txt1 = "AABAACAADAABAABA";
String pat1 = "AABA";
System.out.println("Example 1:");
search(pat1, txt1);
// Example 2
String txt2 = "agd";
String pat2 = "g";
System.out.println("\nExample 2:");
search(pat2, txt2);
}
}
Python3
def search_pattern(pattern, text):
# Get the lengths of the pattern and the text
m = len(pattern)
n = len(text)
# A loop to slide pattern over text one by one
for i in range(n - m + 1):
# For current index i, check for pattern match
j = 0
while j < m and text[i + j] == pattern[j]:
j += 1
# If the entire pattern matches the text starting at index i
if j == m:
print(f"Pattern found at index {i}")
# Example usage
if __name__ == "__main__":
# Example 1
text1 = "AABAACAADAABAABA"
pattern1 = "AABA"
print("Example 1:")
search_pattern(pattern1, text1)
# Example 2
text2 = "agd"
pattern2 = "g"
print("\nExample 2:")
search_pattern(pattern2, text2)
C#
using System;
class PatternSearch
{
static void Search(string pat, string txt)
{
int M = pat.Length;
int N = txt.Length;
// A loop to slide pat[] one by one
for (int i = 0; i <= N - M; i++)
{
int j;
// For current index i, check for pattern match
for (j = 0; j < M; j++)
{
if (txt[i + j] != pat[j])
{
break;
}
}
// If pattern matches at index i
if (j == M)
{
Console.WriteLine($"Pattern found at index {i}");
}
}
}
static void Main()
{
// Example 1
string txt1 = "AABAACAADAABAABA";
string pat1 = "AABA";
Console.WriteLine("Example 1:");
Search(pat1, txt1);
// Example 2
string txt2 = "agd";
string pat2 = "g";
Console.WriteLine("\nExample 2:");
Search(pat2, txt2);
}
}
Javascript
function search(pat, txt) {
const M = pat.length;
const N = txt.length;
// A loop to slide pat[] one by one
for (let i = 0; i <= N - M; i++) {
let j = 0;
// For current index i, check for pattern match
while (j < M && txt[i + j] === pat[j]) {
j++;
}
// If pattern matches at index i
if (j === M) {
console.log(`Pattern found at index ${i}`);
}
}
}
// Example 1
const txt1 = "AABAACAADAABAABA";
const pat1 = "AABA";
console.log("Example 1:");
search(pat1, txt1);
// Example 2
const txt2 = "agd";
const pat2 = "g";
console.log("\nExample 2:");
search(pat2, txt2);
PHP
function search($pat, $txt) {
$M = strlen($pat);
$N = strlen($txt);
// A loop to slide pat[] one by one
for ($i = 0; $i <= $N - $M; $i++) {
$j = 0;
// For current index i, check for pattern match
while ($j < $M && $txt[$i + $j] === $pat[$j]) {
$j++;
}
// If pattern matches at index i
if ($j == $M) {
echo "Pattern found at index $i\n";
}
}
}
// Example 1
$txt1 = "AABAACAADAABAABA";
$pat1 = "AABA";
echo "Example 1:\n";
search($pat1, $txt1);
// Example 2
$txt2 = "agd";
$pat2 = "g";
echo "\nExample 2:\n";
search($pat2, $txt2);
Output
Pattern found at index 0 Pattern found at index 9 Pattern found at index 13
Time Complexity: O(N2)
Auxiliary Space: O(1)
Complexity Analysis of Naive algorithm for Pattern Searching:
Best Case: O(n)
- When the pattern is found at the very beginning of the text (or very early on).
- The algorithm will perform a constant number of comparisons, typically on the order of O(n) comparisons, where n is the length of the pattern.
Worst Case: O(n2)
- When the pattern doesn't appear in the text at all or appears only at the very end.
- The algorithm will perform O((n-m+1)*m) comparisons, where n is the length of the text and m is the length of the pattern.
- In the worst case, for each position in the text, the algorithm may need to compare the entire pattern against the text.