check if 2 strings in string

text-preprocessing

@QuinnFox12

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regex

@QuinnFox12

Python regex related snippets preprocessing Text Full and path preprocessing Text Full and path check if 2 strings in string

#javascript #javascript #string #html

Inject HTML From a String of HTML

  document.body.innerHTML = string_of_html;

// Append it instead
document.body.innerHTML += string_of_html;

#inputstream #string #java

inputstream-into-a-string

https://stackoverflow.com/questions/309424/how-do-i-read-convert-an-inputstream-into-a-string-in-java

#python #string #format

string — Common string operations — Python 3.9.5 documentation

>>> 'Coordinates: {latitude}, {longitude}'.format(latitude='37.24N', longitude='-115.81W')
'Coordinates: 37.24N, -115.81W'
>>> coord = {'latitude': '37.24N', 'longitude': '-115.81W'}
>>> 'Coordinates: {latitude}, {longitude}'.format(**coord)
'Coordinates: 37.24N, -115.81W'

#javascript #url #validation #string

Verify if some string is valid URL, JavaScript

/**
 * Provides quick & simple way to verify if some
 * string is valid URL
 *
 * @param   <string>  URL to verify
 * @returns <boolean> Returns true if passed string
 *                    is valid URL, false otherwise
 */
function validateUrl (url) {
    try {
        new URL(url)
    } catch (error) {
        return false
    }
  
    return true
}

#java #gfg #geeksforgeeks #lecture #string #palindrome

Check whether a String is a Palindrome

// Efficient : Time Complexity : O(n), Space Complexity : Θ(1)

    static boolean isPalindrome(String str)
    {
 
        // Pointers pointing to the beginning
        // and the end of the string
        int begin = 0, end = str.length() - 1;
 
        // While there are characters to compare
        while (begin < end) {
 
            // If there is a mismatch
            if (str.charAt(begin) != str.charAt(end))
                return false;
 
            // Increment first pointer and
            // decrement the other
            begin++;
            end--;
        }
 
        // Given string is a palindrome
        return true;
    }


// Naive : Time Complexity : Θ(n), Space Complexity : Θ(n)

    static boolean isPalindrome(String str)
    {
      StringBuilder rev = new StringBuilder(str);
      rev.reverse();  // StringBuilder is mutable & has a function called reverse
      
      return str.equals(rev.toString());
    }

#java #gfg #geeksforgeeks #lecture #string #subsequence

Check if a String is Subsequence of Other

// Iterative Solution : Time Complexity : O(n+m), Space Complexity : Θ(1)

    static boolean isSubSeq(String s1, String s2, int n, int m){
        int j = 0;
        for(int i = 0; i < n && j < m; i++){
            if(s1.charAt(i) == s2.charAt(j))
            j++;
        }
        
        return j == m;
    }



// Recursive Solution : Time Complexity : O(n+m), Space Complexity : Θ(n+m)

    static boolean isSubSeq(String s1, String s2, int n, int m){
        if( m == 0 )
            return true;
        
        if( n == 0 )
            return false;
            
        if ( s1.charAt(n-1) == s2.charAt(m-1) )
            return isSubSeq(s1, s2, n-1, m-1);
        
        else
            return isSubSeq(s1, s2, n-1, m);
    }

#java #gfg #geeksforgeeks #lecture #string #anagram

Check for Anagram

// Efficient : Time Complexity : O(n)
 
import java.util.*;
import java.io.*;
  
class GFG { 
    
    static final int CHAR=256;
        
    static boolean areAnagram(String s1, String s2) 
    { 
       if (s1.length() != s2.length()) 
            return false; 
  
        int[] count=new int[CHAR];
        for(int i=0;i<s1.length();i++){
            count[s1.charAt(i)]++;
            count[s2.charAt(i)]--;
        }
    
        for(int i=0;i<CHAR;i++){
            if(count[i]!=0)return false;
        }
        return true;
    }
  
    public static void main(String args[]) 
    { 
        String str1 = "abaac"; 
        String str2 = "aacba";  
        if (areAnagram(str1, str2)) 
            System.out.println("The two strings are" + " anagram of each other"); 
        else
            System.out.println("The two strings are not" + " anagram of each other"); 
    } 
} 
 
 

// Naive : Time Complexity : Θ(nlogn)
 
    static boolean areAnagram(String s1, String s2) 
    { 
       
        if (s1.length() != s2.length()) 
            return false; 
  
       char a1[]=s1.toCharArray();
        Arrays.sort(a1);
        s1=new String(a1);
        char a2[]=s2.toCharArray();
        Arrays.sort(a2);
        s2=new String(a2);
        
        return s1.equals(s2);
    }

#java #gfg #geeksforgeeks #lecture #string #leftmostrepeatingcharacter #index

Index of Leftmost Repeating Character

// Efficient Approach-2 : Time & Space similar to previous method

    static final int CHAR=256;
    static int leftMost(String str) 
    {
        boolean[] visited=new boolean[CHAR];
        int res=-1;
        for(int i=str.length()-1;i>=0;i--){
            if(visited[str.charAt(i)])
            res=i;
            else
            visited[str.charAt(i)]=true;
        }
        
        return res;
    } 



// One Traversal
// Efficient Approach-1 : Time Complexity : O(n + CHAR), Auxiliary Space : O(CHAR)

    static final int CHAR=256;
    static int leftMost(String str) 
    {
        int[] fIndex=new int[CHAR];
        Arrays.fill(fIndex,-1);
        int res=Integer.MAX_VALUE;
        for(int i=0;i<str.length();i++){
            int fi=fIndex[str.charAt(i)];
            if(fi==-1)
            fIndex[str.charAt(i)]=i;
            else
            res=Math.min(res,fi);
        }
        
        return (res==Integer.MAX_VALUE)?-1:res;
    } 



// Better Approach : Time Complexity : O(n) but requires two loops for input string 

    static final int CHAR=256;
    static int leftMost(String str) 
    {
        int[] count=new int[CHAR];
        for(int i=0;i<str.length();i++){
            count[str.charAt(i)]++;
        }
        for(int i=0;i<str.length();i++){
            if(count[str.charAt(i)]>1)return i;
        }
        return -1;
    }


// Naive Code : Time Complexity : O(n^2)

    static int leftMost(String str) 
    {
        for(int i=0;i<str.length();i++){
            for(int j=i+1;j<str.length();j++){
                if(str.charAt(i)==str.charAt(j))return i;
            }
        }
        return -1;
    }

#java #gfg #geeksforgeeks #lecture #string #index #leftmostnonrepeatingcharacter

Index of Leftmost Non-repeating Element

// One Traversal
// Efficient Approach-1 : Time Complexity : O(n)
 
    static final int CHAR=256;
    static int nonRep(String str) 
    {
        int[] fI=new int[CHAR];
        Arrays.fill(fI,-1);
    
        for(int i=0;i<str.length();i++){
            if(fI[str.charAt(i)]==-1)
            fI[str.charAt(i)]=i;
            else
            fI[str.charAt(i)]=-2;
        }
        int res=Integer.MAX_VALUE;
        for(int i=0;i<CHAR;i++){
            if(fI[i]>=0)res=Math.min(res,fI[i]);
        }
        return (res==Integer.MAX_VALUE)?-1:res;
    }
 
 
// Two Traversal
// Better Approach : Time Complexity : O(n) 
 
    static final int CHAR=256;
    static int nonRep(String str) 
    {
        int[] count=new int[CHAR];
        for(int i=0;i<str.length();i++){
            count[str.charAt(i)]++;
        }
        for(int i=0;i<str.length();i++){
            if(count[str.charAt(i)]==1)return i;
        }
        return -1;
    } 
 
 
// Naive Code : Time Complexity : O(n^2)
 
    static int nonRep(String str) 
    {
        for(int i=0;i<str.length();i++){
            boolean flag=false;
            for(int j=0;j<str.length();j++){
                if(i!=j&&str.charAt(i)==str.charAt(j)){
                    flag=true;
                    break;
                }
            }
            if(flag==false)return i;
        }
        return -1;
    }

#java #gfg #geeksforgeeks #lecture #string #reversewords

Reverse words in a string

// Efficient Approach :
// NOTE : The code doesn’t handle the cases when the string starts with space. 

import java.util.*;
import java.io.*;
  
class GFG { 
       
    static void reverse(char str[],int low, int high)
    {
        while(low<=high)
        {
            //swap
            char temp=str[low];
            str[low]=str[high];
            str[high]=temp;

            low++;
            high--;
        }
    }

    static void reverseWords(char str[],int n){
    int start=0;
    for(int end=0;end<n;end++){
        if(str[end]==' '){
            reverse(str,start,end-1);
            start=end+1;
        }
    }
    reverse(str,start,n-1);
    reverse(str,0,n-1);
    }
  
    public static void main(String args[]) 
    {   String s = "Welcome to Gfg";int n=s.length();
        char[] str = s.toCharArray();
        System.out.println("After reversing words in the string:");
        reverseWords(str,n);
        System.out.println(str);  
    } 
}

#java #gfg #geeksforgeeks #lecture #string #patternsearching #overview

Overview of Pattern Searching in strings

m -> Pattern length
n -> Text length
1 <= m <=n
---------------------------------------------------------------------------------------------------

// NO PREPROCESSING

Naive : O((n-m+1)*m)

Naive (When all characters of Pattern are distinct) : O(n)
---------------------------------------------------------------------------------------------------

// PREPROCESS PATTERN
  
Rabin Karp : O((n-m+1)*m)  // But, better then naive on average

KMP Algorithm : O(n)
---------------------------------------------------------------------------------------------------

// PREPROCESS TEXT
  
Suffix Tree : O(m)

#java #gfg #geeksforgeeks #lecture #string #patternsearching #naive #index

Naive Pattern Searching in String

import java.util.*;
import java.io.*;
  
class GFG { 
       
    static void patSearchinng(String txt,String pat)
    {
        int m=pat.length();
        int n=txt.length();
        for(int i=0;i<=(n-m);i++){
      	    int j;
            for(j=0;j<m;j++)
                if(pat.charAt(j)!=txt.charAt(i+j))
              	    break;
            
        if(j==m)
            System.out.print(i+" ");
        }
    }
  
    public static void main(String args[]) 
    {   String txt = "ABCABCD";String pat="ABCD";
        System.out.print("All index numbers where pattern found: ");
        patSearchinng(txt,pat);  
    } 
}

#java #gfg #geeksforgeeks #lecture #string #patternsearching #index #improvednaive #distinct

Improved Naive Pattern Searching for Distinct

import java.util.*;
import java.io.*;
  
class GFG { 
       
    static void patSearchinng(String txt,String pat)
    {
        int m=pat.length();
        int n=txt.length();
        for(int i=0;i<=(n-m); )
        {
            int j;
            for(j=0;j<m;j++)
                if(pat.charAt(j)!=txt.charAt(i+j))
                    break;
            
            if(j==m)
                System.out.print(i+" ");
            if(j==0)
                i++;
            else
                i=(i+j);
        }
    }
  
    public static void main(String args[]) 
    {   String txt = "ABCABCD";String pat="ABCD";
        System.out.print("All index numbers where pattern found: ");
        patSearchinng(txt,pat);  
    } 
}

#java #gfg #geeksforgeeks #lecture #string #patternsearching #rabinkarp

Rabin Karp - Pattern Searching

import java.util.*;
import java.io.*;
  
class GFG { 
    static final int d=256;
    static final int q=101;   
    static void RBSearch(String pat,String txt,int M, int N)
    {
        //Compute (d^(M-1))%q
        int h=1;
        for(int i=1;i<=M-1;i++)
            h=(h*d)%q;
        
        //Compute p and to
        int p=0,t=0;
        for(int i=0;i<M;i++){
            p=(p*d+pat.charAt(i))%q;
            t=(t*d+txt.charAt(i))%q;
        }
        
        for(int i=0;i<=(N-M);i++){
           //Check for hit
           if(p==t){
               boolean flag=true;
               for(int j=0;j<M;j++)
                    if(txt.charAt(i+j)!=pat.charAt(j)){flag=false;break;}
                if(flag==true)System.out.print(i+" ");
           }
           //Compute ti+1 using ti
           if(i<N-M){
               t=((d*(t-txt.charAt(i)*h))+txt.charAt(i+M))%q;
            if(t<0)t=t+q;
           }
        }
        
    }
  
    public static void main(String args[]) 
    {   String txt = "GEEKS FOR GEEKS";String pat="GEEK";
        System.out.print("All index numbers where pattern found: ");
        RBSearch(pat,txt,4,15);  
    } 
}

#java #gfg #geeksforgeeks #lecture #string #patternsearching #kmp #lps

KMP Agorithm (Part 1 : Constructing LPS Array)

// Efficient Code O(n)

import java.util.*;
import java.io.*;
  
class GFG { 
  
    static void fillLPS(String str, int lps[])
    {
        int n=str.length(),len=0;
        lps[0]=0;
        int i=1;
        while(i<n){
            if(str.charAt(i)==str.charAt(len))
            {len++;lps[i]=len;i++;}
            else
            {if(len==0){lps[i]=0;i++;}
                else{len=lps[len-1];}
            }
        }
    }
  
    public static void main(String args[]) 
    {   String txt = "abacabad";int[] lps=new int[txt.length()];
        fillLPS(txt,lps);
        for(int i=0;i<txt.length();i++){
            System.out.print(lps[i]+" ");
        } 
    } 
} 




// Naive Code O(n^3)

import java.util.*;
import java.io.*;
  
class GFG { 

    static int longPropPreSuff(String str, int n)
    {
        for(int len=n-1;len>0;len--){
            boolean flag=true;
            for(int i=0;i<len;i++)
                if(str.charAt(i)!=str.charAt(n-len+i))
                    flag=false;
                    
            if(flag==true)
                return len;
        }
        return 0;
    }

    static void fillLPS(String str, int lps[]){
        for(int i=0;i<str.length();i++){
        lps[i]=longPropPreSuff(str,i+1);
        }
    }
  
    public static void main(String args[]) 
    {   String txt = "abacabad";int[] lps=new int[txt.length()];
        fillLPS(txt,lps);
        for(int i=0;i<txt.length();i++){
            System.out.print(lps[i]+" ");
    }  
    } 
}

check if 2 strings in string

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