 Open in App
Not now

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

• Difficulty Level : Easy
• Last Updated : 27 Nov, 2022

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.

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 Practice

## First String is a Subsequence of second by Recursion:

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

Below is the Implementation of the above idea:

## C++

 `// Recursive C++ program to check ` `// if a string is subsequence ` `// of another string ` `#include ` `#include ` `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 check whether str1 is subsequence of str2 or not. ` `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`

## Python3

 `# 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, n): ` `    ``# 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, ``len``(string1), ``len``(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

 ``

## Javascript

 ``

Output

`Yes `

Time Complexity: O(N), The recursion will call at most N times.
Auxiliary Space: O(N), Function call stack space

## First String is a Subsequence of second using Two Pointers (Iterative):

The idea is to use two pointers, one pointer will start from start of str1 and another will start from start of str2. If current character on both the indexes are same then increment both pointers otherwise increment the pointer which is pointing str2.

Follow the steps below to solve the problem:

• Initialize the pointers i and j with zero, where i is the pointer to str1 and j is the pointer to str2.
• If str1[i] = str2[j] then increment both i and j by 1.
• Otherwise, increment only j by 1.
• If i reaches the end of str1 then return TRUE else return FALSE.

Below is the implementation of the above approach

## C++

 `/*Iterative C++ program to check ` `If a string is subsequence of another string*/` ` `  `#include ` `using` `namespace` `std; ` ` `  `/*Returns true if s1 is subsequence of s2*/` `bool` `issubsequence(string& s1, string& s2) ` `{ ` `    ``int` `n = s1.length(), m = s2.length(); ` `    ``int` `i = 0, j = 0; ` `    ``while` `(i < n && j < m) { ` `        ``if` `(s1[i] == s2[j]) ` `            ``i++; ` `        ``j++; ` `    ``} ` `    ``/*If i reaches end of s1,that mean we found all ` `    ``characters of s1 in s2, ` `    ``so s1 is subsequence of s2, else not*/` `    ``return` `i == n; ` `} ` `int` `main() ` ` `  `{ ` `    ``string s1 = ``"gksrek"``; ` `    ``string s2 = ``"geeksforgeeks"``; ` `    ``if` `(issubsequence(s1, s2)) ` `        ``cout << ``"gksrek is subsequence of geekforgeeks"` `<< endl; ` `    ``else` `        ``cout << ``"gksrek is not a subsequence of geekforgeeks"` `<< endl; ` `    ``return` `0; ` `}`

## Java

 `/*package whatever //do not write package name here */` `import` `java.io.*; ` `import` `java.util.*; ` ` `  `class` `GFG  ` `{ ` `   `  `    ``/*Iterative Java program to check ` `If a String is subsequence of another String*/` ` `  `/*Returns true if s1 is subsequence of s2*/` `static` `boolean` `issubsequence(String s1, String s2) ` `{ ` `    ``int` `n = s1.length(), m = s2.length(); ` `    ``int` `i = ``0``, j = ``0``; ` `    ``while` `(i < n && j < m) { ` `        ``if` `(s1.charAt(i) == s2.charAt(j)) ` `            ``i++; ` `        ``j++; ` `    ``} ` `    ``/*If i reaches end of s1,that mean we found all ` `    ``characters of s1 in s2, ` `    ``so s1 is subsequence of s2, else not*/` `    ``return` `i == n; ` `} ` `     `  `public` `static` `void` `main(String args[]) ` `{ ` `    ``String s1 = ``"gksrek"``; ` `    ``String s2 = ``"geeksforgeeks"``; ` `    ``if` `(issubsequence(s1, s2)) ` `        ``System.out.println(``"gksrek is subsequence of geekforgeeks"``); ` `    ``else` `        ``System.out.println(``"gksrek is not a subsequence of geekforgeeks"``); ` `} ` `} ` ` `  `// This code is contributed by shinjanpatra.`

## Python3

 `# Iterative JavaScript program to check ` `# If a string is subsequence of another string ` ` `  `# Returns true if s1 is subsequence of s2 ` `def` `issubsequence(s1, s2): ` ` `  `    ``n,m ``=` `len``(s1),``len``(s2) ` `    ``i,j ``=` `0``,``0` `    ``while` `(i < n ``and` `j < m): ` `        ``if` `(s1[i] ``=``=` `s2[j]): ` `            ``i ``+``=` `1` `        ``j ``+``=` `1` `     `  `    ``# If i reaches end of s1,that mean we found all ` `    ``# characters of s1 in s2, ` `    ``# so s1 is subsequence of s2, else not ` `    ``return` `i ``=``=` `n ` ` `  ` `  `# driver code ` `s1 ``=` `"gksrek"` `s2 ``=` `"geeksforgeeks"` `if` `(issubsequence(s1, s2)): ` `    ``print``(``"gksrek is subsequence of geekforgeeks"``) ` `else``: ` `    ``print``(``"gksrek is not a subsequence of geekforgeeks"``) ` ` `  `# This code is contributed by shinjanpatra`

## C#

 `// C# code to implement the approach ` `using` `System; ` ` `  `class` `GFG { ` ` `  `  ``/*Returns true if s1 is subsequence of s2*/` `  ``static` `bool` `issubsequence(``string` `s1, ``string` `s2) ` `  ``{ ` `    ``int` `n = s1.Length, m = s2.Length; ` `    ``int` `i = 0, j = 0; ` `    ``while` `(i < n && j < m) { ` `      ``if` `(s1[i] == s2[j]) ` `        ``i++; ` `      ``j++; ` `    ``} ` `    ``/*If i reaches end of s1,that mean we found all ` `        ``characters of s1 in s2, ` `        ``so s1 is subsequence of s2, else not*/` `    ``return` `i == n; ` `  ``} ` ` `  `  ``public` `static` `void` `Main(``string``[] args) ` `  ``{ ` `    ``string` `s1 = ``"gksrek"``; ` `    ``string` `s2 = ``"geeksforgeeks"``; ` `    ``if` `(issubsequence(s1, s2)) ` `      ``Console.WriteLine(s1 + ``" is subsequence of "` `                        ``+ s2); ` `    ``else` `      ``Console.WriteLine( ` `      ``s1 + ``" is not a subsequence of "` `+ s2); ` `  ``} ` `} ` ` `  `// This code is contributed by phasing17.`

## Javascript

 ``

Output

`gksrek is subsequence of geekforgeeks`

Time Complexity: O(max(n,m)), where n,m are the length of given string s1 and s2 respectively.
Auxiliary Space: O(1)