Given two strings, find if first string is a subsequence of second

Given two strings str1 and str2, find if str1 is a subsequence of str2. A subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements (source: wiki). Expected time complexity is linear.

Examples :

Input: str1 = "AXY", str2 = "ADXCPY"
Output: True (str1 is a subsequence of str2)

Input: str1 = "AXY", str2 = "YADXCP"
Output: False (str1 is not a subsequence of str2)

Input: str1 = "gksrek", str2 = "geeksforgeeks"
Output: True (str1 is a subsequence of str2)

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is simple, we traverse both strings from one side to other side (say from rightmost character to leftmost). If we find a matching character, we move ahead in both strings. Otherwise we move ahead only in str2.

Following is Recursive Implementationof the above idea.

C/C++

// Recursive C++ program to check if a string is subsequence of another string 
#include<iostream> 
#include<cstring> 
using namespace std; 
  
// Returns true if str1[] is a subsequence of str2[]. m is 
// length of str1 and n is length of str2 
bool isSubSequence(char str1[], char str2[], int m, int n) 
{ 
    // Base Cases 
    if (m == 0) return true; 
    if (n == 0) return false; 
  
    // If last characters of two strings are matching 
    if (str1[m-1] == str2[n-1]) 
        return isSubSequence(str1, str2, m-1, n-1); 
  
    // If last characters are not matching 
    return isSubSequence(str1, str2, m, n-1); 
} 
  
// Driver program to test methods of graph class 
int main() 
{ 
    char str1[] = "gksrek"; 
    char str2[] = "geeksforgeeks"; 
    int m = strlen(str1); 
    int n = strlen(str2); 
    isSubSequence(str1, str2, m, n)? cout << "Yes ": 
                                     cout << "No"; 
    return 0; 
} 

Java

// Recursive Java program to check if a string 
// is subsequence of another string 
import java.io.*; 
  
class SubSequence 
{ 
    // Returns true if str1[] is a subsequence of str2[] 
    // m is length of str1 and n is length of str2 
    static boolean isSubSequence(String str1, String str2, int m, int n) 
    { 
        // Base Cases 
        if (m == 0)  
            return true; 
        if (n == 0)  
            return false; 
              
        // If last characters of two strings are matching 
        if (str1.charAt(m-1) == str2.charAt(n-1)) 
            return isSubSequence(str1, str2, m-1, n-1); 
  
        // If last characters are not matching 
        return isSubSequence(str1, str2, m, n-1); 
    } 
      
    // Driver program 
    public static void main (String[] args)  
    { 
        String str1 = "gksrek"; 
        String str2 = "geeksforgeeks"; 
        int m = str1.length(); 
        int n = str2.length(); 
        boolean res = isSubSequence(str1, str2, m, n); 
        if(res) 
            System.out.println("Yes"); 
        else
            System.out.println("No"); 
    } 
} 
  
// Contributed by Pramod Kumar 

Python

# Recursive Python program to check if a string is subsequence 
# of another string 
  
# Returns true if str1[] is a subsequence of str2[]. 
def isSubSequence(string1, string2): 
    m = len(string1) 
    n = len(string2) 
    # Base Cases 
    if m == 0:    return True
    if n == 0:    return False
  
    # If last characters of two strings are matching 
    if string1[m-1] == string2[n-1]: 
        return isSubSequence(string1, string2, m-1, n-1) 
  
    # If last characters are not matching 
    return isSubSequence(string1, string2, m, n-1) 
  
# Driver program to test the above function 
string1 = "gksrek"
string2 = "geeksforgeeks"
  
if isSubSequence(string1, string2): 
    print "Yes"
else: 
    print "No"
  
# This code is contributed by BHAVYA JAIN 

C#

// Recursive C# program to check if a string  
// is subsequence of another string 
using System; 
  
class GFG { 
      
    // Returns true if str1[] is a  
    // subsequence of str2[] m is  
    // length of str1 and n is length  
    // of str2 
    static bool isSubSequence(string str1, 
                  string str2, int m, int n) 
    { 
          
        // Base Cases 
        if (m == 0)  
            return true; 
        if (n == 0)  
            return false; 
              
        // If last characters of two strings 
        // are matching 
        if (str1[m-1] == str2[n-1]) 
            return isSubSequence(str1, str2, 
                                    m-1, n-1); 
  
        // If last characters are not matching 
        return isSubSequence(str1, str2, m, n-1); 
    } 
      
    // Driver program 
    public static void Main ()  
    { 
        string str1 = "gksrek"; 
        string str2 = "geeksforgeeks"; 
        int m = str1.Length; 
        int n = str2.Length; 
        bool res = isSubSequence(str1, str2, m, n); 
          
        if(res) 
            Console.Write("Yes"); 
        else
            Console.Write("No"); 
    } 
} 
  
// This code is contributed by nitin mittal. 

PHP

<?php 
// Recursive PHP program to check 
// if a string is subsequence of  
// another string 
  
// Returns true if str1[] is a  
// subsequence of str2[]. m is 
// length of str1 and n is  
// length of str2 
  
function isSubSequence($str1, $str2,  
                             $m, $n) 
{ 
    // Base Cases 
    if ($m == 0) return true; 
    if ($n == 0) return false; 
  
    // If last characters of two 
    // strings are matching 
    if ($str1[$m - 1] == $str2[$n - 1]) 
        return isSubSequence($str1, $str2, 
                          $m - 1, $n - 1); 
  
    // If last characters  
    // are not matching 
    return isSubSequence($str1, $str2,  
                          $m, $n - 1); 
} 
  
// Driver Code 
$str1= "gksrek"; 
$str2 = "geeksforgeeks"; 
$m = strlen($str1); 
$n = strlen($str2); 
  
$t = isSubSequence($str1, $str2, $m, $n) ?  
                                   "Yes ": 
                                     "No"; 
  
if($t = true) 
    echo "Yes"; 
else 
    echo "No"; 
  
// This code is contributed by ajit 
?> 

Output :

Yes

Following is the Iterative Implementation:

C/C++

// Iterative C++ program to check if a string is subsequence of another string 
#include<iostream> 
#include<cstring> 
using namespace std; 
  
// Returns true if str1[] is a subsequence of str2[]. m is 
// length of str1 and n is length of str2 
bool isSubSequence(char str1[], char str2[], int m, int n) 
{ 
   int j = 0; // For index of str1 (or subsequence 
  
   // Traverse str2 and str1, and compare current character  
   // of str2 with first unmatched char of str1, if matched  
   // then move ahead in str1 
   for (int i=0; i<n&&j<m; i++) 
       if (str1[j] == str2[i]) 
         j++; 
  
   // If all characters of str1 were found in str2 
   return (j==m); 
} 
  
// Driver program to test methods of graph class 
int main() 
{ 
    char str1[] = "gksrek"; 
    char str2[] = "geeksforgeeks"; 
    int m = strlen(str1); 
    int n = strlen(str2); 
    isSubSequence(str1, str2, m, n)? cout << "Yes ": 
                                     cout << "No"; 
    return 0; 
} 

Java

// Iterative Java program to check if a string  
// is subsequence of another string 
import java.io.*; 
  
class GFG { 
      
    // Returns true if str1[] is a subsequence  
    // of str2[] m is length of str1 and n is 
    // length of str2 
    static boolean isSubSequence(String str1,  
                    String str2, int m, int n) 
    { 
        int j = 0; 
          
        // Traverse str2 and str1, and compare  
        // current character of str2 with first 
        // unmatched char of str1, if matched  
        // then move ahead in str1 
        for (int i = 0; i < n && j < m; i++) 
            if (str1.charAt(j) == str2.charAt(i)) 
                j++; 
  
        // If all characters of str1 were found 
        // in str2 
        return (j == m);  
    } 
      
    // Driver program to test methods of 
    // graph class 
    public static void main (String[] args)  
    { 
        String str1 = "gksrek"; 
        String str2 = "geeksforgeeks"; 
        int m = str1.length(); 
        int n = str2.length(); 
        boolean res = isSubSequence(str1, str2, m, n); 
          
        if(res) 
            System.out.println("Yes"); 
        else
            System.out.println("No"); 
    } 
} 
  
// This code is contributed by Pramod Kumar 

Python

# Iterative Python program to check if a string is subsequence of another string 
  
# Returns true if str1 is a subsequence of str2 
def isSubSequence(str1,str2): 
    m = len(str1) 
    n = len(str2) 
      
    j = 0    # Index of str1 
    i = 0    # Index of str2 
      
    # Traverse both str1 and str2 
    # Compare current character of str2 with  
    # first unmatched character of str1 
    # If matched, then move ahead in str1 
      
    while j<m and i<n: 
        if str1[j] == str2[i]:     
            j = j+1    
        i = i + 1
          
    # If all characters of str1 matched, then j is equal to m 
    return j==m 
      
# Driver Program 
  
str1 = "gksrek"
str2 = "geeksforgeeks"
  
print "Yes" if isSubSequence(str1,str2) else "No"
  
# Contributed by Harshit Agrawal 

C#

// Iterative C# program to check if a string  
// is subsequence of another string 
using System; 
  
class GFG { 
      
    // Returns true if str1[] is a subsequence 
    // of str2[] m is length of str1 and n is  
    // length of str2 
    static bool isSubSequence(string str1,  
                     string str2, int m, int n) 
    { 
        int j = 0; 
          
        // Traverse str2 and str1, and compare 
        // current character of str2 with first 
        // unmatched char of str1, if matched  
        // then move ahead in str1 
        for (int i = 0; i < n && j < m; i++) 
            if (str1[j] == str2[i]) 
                j++; 
  
        // If all characters of str1 were found 
        // in str2 
        return (j == m);  
    } 
      
    // Driver program to test methods of  
    // graph class 
    public static void Main ()  
    { 
        String str1 = "gksrek"; 
        String str2 = "geeksforgeeks"; 
        int m = str1.Length; 
        int n = str2.Length; 
        bool res = isSubSequence(str1, str2, m, n); 
          
        if(res) 
            Console.WriteLine("Yes"); 
        else
            Console.WriteLine("No"); 
    } 
} 
  
// This code is contributed by anuj_67. 

PHP

<?php 
// Iterative PHP program to check if  
// a string is subsequence of another 
// string 
  
// Returns true if str1[] is  
// a subsequence of str2[]. 
// m is length of str1 and n  
// is length of str2 
function isSubSequence($str1, $str2, 
                             $m, $n) 
{ 
    // For index of str1 
    $j = 0;  
      
    // Traverse str2 and str1,  
    // and compare current  
    // character of str2 with  
    // first unmatched char of 
    // str1, if matched then  
    // move ahead in str1 
    for($i = 0; $i < $n and
        $j < $m; $i++) 
        if ($str1[$j] == $str2[$i]) 
            $j++; 
      
    // If all characters of  
    // str1 were found in str2 
    return ($j == $m); 
} 
  
    // Driver Code 
    $str1 = "gksrek"; 
    $str2 = "geeksforgeeks"; 
    $m = strlen($str1); 
    $n = strlen($str2); 
      
    if(isSubSequence($str1, $str2, $m, $n)) 
        echo "Yes "; 
    else
        echo "No"; 
  
// This code is contributed by anuj_67. 
?> 

Output:

Yes

Asked in: Accolite,Tesco

Time Complexity of both implementations above is O(n) where n is the length of str2.

This article is contributed by Sachin Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C/C++

Java

Python

C#

PHP

C/C++

Java

Python

C#

PHP

Asked in: Accolite,Tesco

Recommended Posts: