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day 12 - Substring Search Problem

Question -- Given a text, and a pattern, find whether the given pattern exists in the text or not.

Implement various algorithms

    a) Brute Force Search
    b) Knuth-Morris-Pratt Algorithm
    c) Z Algorithm
    d) Rabin Karp Algorithm
    e) Boyer Moore Algorithm

Example

input: text: helloworld, substring: hello
output: 0 (index of 'h', where the substring starts)

input: text: helloworld, substring: hop
output: -1

input: text: abcrxyzgf, substring: xyz
output: 4

A) Brute Force Search

JavaScript Implementation

Solution

Substring search can be done using brute force. In this case, the worst case Time Complexity will be O(m.n), m and n are the lengths of string and pattern respectively.

/**
 * @author MadhavBahlMD
 * @date 04/01/2018
 * In this case, the worst case Time Complexity will be O(m.n), 
 * m and n are the lengths of string and pattern respectively.
 */

function substringSearch (str, pattern) {
    let strLen = str.length,
        patLen = pattern.length,
        flag = 0;

    for (let i=0; i<(strLen-patLen+1); i++) {
        if (str[i] === pattern[0]) {
            flag = 1;
            for (let j=1; j<patLen; j++) {
                if (str[i+j] !== pattern[j]) {
                    flag = 0;
                    break;
                }
            }
            if (flag === 1) {
                console.log (i);
                return i;
            } 
        }
    }

    console.log (-1);
    return -1;
}

substringSearch ("helloworld", "world");
substringSearch ("abcrxyzgf", "xyz");

Using Inbuilt JavaScript Methods

Using String.indexOf()

/**
 * @author MadhavBahlMD
 * @date 05/01/2018
 * METHOD - Using String.indexOf() method
 */

function substringSearch (str, subStr) {
    return str.indexOf(subStr);
}

console.log (substringSearch ("helloworld", "world"));
console.log (substringSearch ("abcrxyzgf", "xyz"));

Using String.search()

/**
 * @author MadhavBahlMD
 * @date 05/01/2018
 * METHOD - Using String.search() method
 */

function substringSearch (str, subStr) {
    return str.search (subStr);
}

console.log (substringSearch ("helloworld", "world"));
console.log (substringSearch ("abcrxyzgf", "xyz"));

Python Implementation

Solution

'''
  @author prateek3255
  @date 05/01/2018
 '''

def subStringSearch(string,pattern):
    for i in range(len(string)-len(pattern)):
        if string[i:i+len(pattern)]==pattern:
            return i
    return -1

print(subStringSearch("helloworld","hello"))
print(subStringSearch("helloworld","hop"))
print(subStringSearch("abcrxyzgf","xyz"))

C++ Implementation

Solution

/**
 * @date 04/01/18
 * @author SPREEHA DUTTA
 */
#include <bits/stdc++.h>
using namespace std;
int search(string s,string w)
{
    int i,j;int k=-1;int c=0;
    int m=w.length();
    int n=s.length();
    for(i=0;i<=n-m;i++)
    {
         for(j=0,c=i;j<m;j++,c++)
            if(s[c]!=w[j])
               break;
        if(j==m)
        {
            k=i;
            break;
        }
    }
    return k;
}
int main()
{
    string s,w;
    getline(cin,s);
    getline(cin,w);
    int t=search (w,s);
    cout<<"\n"<<t<<endl;
}

Java Implementation

Solution

import java.util.Scanner;

/**
 * Daily Codes Day 12 -- Substring Search (Brute Force Search)
 * @author MadhavBahlMD
 * @date 05/01/2018
 */

public class Bruteforce {
    public static void main(String[] args) {
        // Take input
        Scanner input = new Scanner (System.in);
        System.out.print("Enter the string: ");
        String str = input.next();
        System.out.print("Enter the substrinig to search: ");
        String subStr = input.next();

        // Search the string
        int i, j, flag,
            strLen = str.length(),
            subStrLen = subStr.length();

        for (i=0; i<(strLen-subStrLen+1); i++) {
            flag = 1;
            if (str.charAt(i) == subStr.charAt(0)) {
                for (j=1; j<subStrLen; j++) {
                    if (str.charAt(i+j) != subStr.charAt(j)) {
                        flag = 0;
                        break;
                    }
                }

                // If substring is found, print the index
                if (flag == 1) {
                    System.out.println("Substring \"" + subStr + "\" was found in string \"" + str + "\" at index " + i);
                    System.exit(0);
                }
            }
        }

        System.out.println("Substring \"" + subStr + "\" was not found in string \"" + str);
        System.exit(0);
    }
}

B) Knuth-Morris-Pratt Algorithm

JavaScript Implementation

Solution

To Be Added

Python Implementation

Solution

def kmp(string,pattern):
    n=len(string)
    m=len(pattern)
    lps=calculateLPS(pattern)
    i=0
    j=0
    while i<n:
        if pattern[j]==string[i]:
            i+=1
            j+=1
        if j==m:
            return i-j
        elif i<n and pattern[j]!=string[i]:
            if j!=0:
                j=lps[j-1]
            else:
                i+=1
    return -1

def calculateLPS(pattern):
    lps=[0]*len(pattern)
    i=1
    lenPat=0
    while i<len(pattern):
        if pattern[i]==pattern[lenPat]:
            lenPat+=1
            lps[i]=lenPat
            i+=1
        else:
            if lenPat!=0:
                lenPat=lps[lenPat-1]
            else:
                lps[i]=0
                i+=1
    return lps

print(kmp("helloworld","hello"))
print(kmp("helloworld","hop"))
print(kmp("ABABDABACDABABCABAB","ABABCABAB"))

Java Implementation

Solution

/**
 * @date 06/01/1998
 * @author spattk (Sitesh Pattanaik)
 */

import java.util.*;
class StringMatching
{
	static int KMPSearch(String text, String subs){

		int n = subs.length();
		int[] lps = new int[n];

		generateLPSArray(lps,subs);
		n = text.length();
		int m = subs.length();

		int i=0,j=0;
		while(i<n){
			if(text.charAt(i)==subs.charAt(j)){
				i++;
				j++;
			}

			if(j==m){
				//Pattern Found
				return (i-j);
			}

			else if(i<n && text.charAt(i)!=subs.charAt(j)){
				if(j!=0){
					j = lps[j-1];
				}
				else{
					i++;
				}
			}
		}

		return -1;
	}

	static void generateLPSArray(int[] lps, String subs)
	{
		int j = 0;
		int i = 1;
		int n = subs.length();

		while(i<n){
			if(subs.charAt(i)==subs.charAt(j)){
				j++;
				lps[i] = j;
				i++;
			}
			else{
				if(j!=0){
					j = lps[j-1];
				}
				else{
					lps[i] = 0;
					i++;
				}
			}
		}
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String text = sc.next();
		String subs = sc.next();

		System.out.println(KMPSearch(text,subs));
		
	}
}

C++ Implementation

Solution

/**
 * @date 04/01/18
 * @author SPREEHA DUTTA
 */

#include <bits/stdc++.h>
using namespace std;
void calc(string w,int m,int p[])
{
    int l=0,i=1;p[0]=0;
    while(i<m)
    {
        if(p[i]==p[l])
        {
            l++;
            p[i]=l;
            i++;
        }
        else
        {
            if(l!=0)
                l=p[l-1];
            else
            {
                p[i]=l;
                i++;
            }
        }
    }
}
int search(string s,string w)
{
    int i=0,j=0;int k=-1;
    int m=w.length();
    int n=s.length();
    int arr[m];
    calc(w,m,arr);
    while(i<n)
    {
        if(w[j]==s[i])
        {
            j++;
            i++;
        }
        if(j==m)
        {
            k=i-j;
            break;
        }
        else if(i<n && w[j]!=s[i])
        {
            if(j!=0)
                j=arr[j-1];
            else
                i++;
        }
    }
    return k;
}
int main()
{
    string s,w;
    getline(cin,s);
    getline(cin,w);
    int t=search (w,s);
    cout<<"\n"<<t<<endl;
}

C) Z Algorithm

JavaScript Implementation

Solution

To Be Added

D) Rabin Karp Algorithm

JavaScript Implementation

Solution

To Be Added

Python Implementation

Solution

'''
  @author prateek3255
  @date 05/01/2018
 '''

def rabinKarp(string,pattern,q=153):
    n=len(string)
    m=len(pattern)
    d=256
    h=d**(m-1)%q
    p=0
    t=0
    
    for i in range(0,m):
        p=(d*p+ord(pattern[i]))%q
        t=(d*t+ord(string[i]))%q
    for i in range(n-m+1):
        if p==t and string[i:i+m]==pattern:
            return i
        
        if i<(n-m):
            t= (d*(t-ord(string[i])*h)+ord(string[i+m]))%q
            if t<0:
                t+=q
    return -1

print(rabinKarp("helloworld","hello"))
print(rabinKarp("helloworld","hop"))
print(rabinKarp("ABABDABACDABABCABAB","ABABCABAB"))

C++ Implementation

Solution

/**
 * @date 04/01/18
 * @author SPREEHA DUTTA
 */
#include <bits/stdc++.h>
using namespace std;

int search(string s,string w,int q)
{
    int i=0,j=0;int k=-1;
    int m=w.length();
    int n=s.length();
    int sc=0,wc=0,h=1;
    int d=256;
    for(i=0;i<m-1;i++)
        h=(h*d)%q;
    for(i=0;i<m;i++)
    {
        wc=(d*wc+w[i])%q;
        sc=(d*sc+s[i])%q;
    }
    for(i=0;i<n-m;i++)
    {
        if(sc==wc)
        {
            for(j=0;j<m;j++)
                if(s[i+j]!=w[j])
                    break;
            if(j==m)
            {
                k=i;
                break;
            }
        }
        if(i<n-m)
        {
            sc=(d*(sc-s[i]*h)+s[i+m])%q;
            if(sc<0)
                sc+=q;
        }
    }
    return k;
}
int main()
{
    string s,w;
    getline(cin,s);
    getline(cin,w);
    int t=search (w,s,101);
    cout<<"\n"<<t<<endl;
}

E) Boyer Moore Algorithm

JavaScript Implementation

Solution

To Be Added

Python Implementation

Solution

'''
  @author prateek3255
  @date 05/01/2018
 '''
def boyerMoore(string,pattern):
    n=len(string)
    m=len(pattern)
    i=0
    while i<=n-m:
        k=m-1
        j=m+i-1
        while string[j]==pattern[k]:
            k=k-1
            j=j-1
            if k==0:
                return i
        if pattern.rfind(string[j])==-1:
            i=j+1
        else:
            i=max(1,j-pattern.rfind(string[j]))
    return -1

print(boyerMoore("helloworld","hello"))
print(boyerMoore("helloworld","hop"))
print(boyerMoore("abcrxyzgf","xyz"))
print(boyerMoore("ABABDABACDABABCABAB","ABABCABAB"))

Have Another solution?

The beauty of programming lies in the fact that there is never a single solution to any problem.

In case you have an alternative way to solve this problem, do contribute to this repository (https://github.com/CodeToExpress/dailycodebase)

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day 12 - Substring Search Problem

Example

A) Brute Force Search

JavaScript Implementation

Using Inbuilt JavaScript Methods

Python Implementation

C++ Implementation

Java Implementation

B) Knuth-Morris-Pratt Algorithm

JavaScript Implementation

Python Implementation

Java Implementation

C++ Implementation

C) Z Algorithm

JavaScript Implementation

D) Rabin Karp Algorithm

JavaScript Implementation

Python Implementation

C++ Implementation

E) Boyer Moore Algorithm

JavaScript Implementation

Python Implementation

Have Another solution?