Given two strings, find if first string is a subsequence of second
- Difficulty Level : Easy
- Last Updated : 22 Jul, 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 (source: wiki). The 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)
The idea is simple, we traverse both strings from one side to another 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 a Recursive Implementation of the above idea.
C++
// Recursive C++ program to check // if a string is subsequence // of another string #include <cstring> #include <iostream> 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
<?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 ?> |
Javascript
<script> // Recursive Javascript 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 let str1 = "gksrek" ; let str2 = "geeksforgeeks" ; let m = str1.length; let n = str2.length; let res = isSubSequence(str1, str2, m, n); if (res) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by divyesh072019 </script> |
Yes
Following is the Iterative Implementation of the Above Idea.
C++
/*Iterative C++ program to check If a string is subsequence of another string*/ #include <bits/stdc++.h> 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 a subsequence of " + s2); else Console.WriteLine( s1 + " not is a subsequence of " + s2); } } // This code is contributed by phasing17. |
Javascript
<script> /*Iterative JavaScript program to check If a string is subsequence of another string*/ /*Returns true if s1 is subsequence of s2*/ function issubsequence(s1, s2) { let n = s1.length, m = s2.length; let 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; } // driver code let s1 = "gksrek" ; let s2 = "geeksforgeeks" ; if (issubsequence(s1, s2)) document.write( "gksrek is subsequence of geekforgeeks" , "</br>" ); else document.write( "gksrek is not a subsequence of geekforgeeks" , "</br>" ); // This code is contributed by shinjanpatra </script> |
gksrek is subsequence of geekforgeeks
Time Complexity: O(max(m,n)).
Auxiliary Space: O(max(m,n)). This is because when string is passed to any function it is passed by value and creates a copy of itself in stack.
Memoization Technique
Here the idea is to check whether the size of the longest common subsequence is equal to the size of str1. If it’s equal it means there is a subsequence that exists in str2. Below is the implementation using the memoization technique.
C++
// memoization C++ program to check // if a string is subsequence // of another string #include <bits/stdc++.h> using namespace std; int dp[1001][1001]; // returns the length of longest common subsequence int isSubSequence(string& s1, string& s2, int i, int j) { if (i == 0 || j == 0) { return 0; } if (dp[i][j] != -1) { return dp[i][j]; } if (s1[i - 1] == s2[j - 1]) { return dp[i][j] = 1 + isSubSequence(s1, s2, i - 1, j - 1); } else { return dp[i][j] = isSubSequence(s1, s2, i, j - 1); } } /* Driver program to test above function */ int main() { string str1 = "gksrek" ; string str2 = "geeksforgeeks" ; int m = str1.size(); int n = str2.size(); if (m > n) { cout << "NO" << endl; return 0; } dp[m][n]; memset (dp, -1, sizeof (dp)); if (isSubSequence(str1, str2, m, n) == m) { cout << "YES" << endl; } else { cout << "NO" << endl; } return 0; } // this code is contributed by Arun Bang |
Java
// memoization Java program to check // if a string is subsequence // of another string class GFG { public static int [][] dp = new int [ 1001 ][ 1001 ]; // returns the length of longest common subsequence public static int isSubSequence(String s1, String s2, int i, int j) { if (i == 0 || j == 0 ) { return 0 ; } if (dp[i][j] != - 1 ) { return dp[i][j]; } if (s1.charAt(i - 1 ) == s2.charAt(j - 1 )) { return dp[i][j] = 1 + isSubSequence(s1, s2, i - 1 , j - 1 ); } else { return dp[i][j] = isSubSequence(s1, s2, i, j - 1 ); } } /* Driver program to test above function */ public static void main(String[] args) { String str1 = "gksrek" ; String str2 = "geeksforgeeks" ; int m = str1.length(); int n = str2.length(); if (m > n) { System.out.println( "NO" ); } for ( int i = 0 ; i <= 1000 ; i++) { for ( int j = 0 ; j <= 1000 ; j++) dp[i][j] = - 1 ; } if (isSubSequence(str1, str2, m, n) == m) { System.out.println( "YES" ); } else { System.out.println( "NO" ); } } } // This code is contributed by phasing17 |
Python3
# memoization Python program to check # if a string is subsequence # of another string dp = [[ - 1 ] * 1001 ] * 1001 # returns the length of longest common subsequence def isSubSequence(s1,s2,i,j): if (i = = 0 or j = = 0 ): return 0 if (dp[i][j] ! = - 1 ): return dp[i][j] if (s1[i - 1 ] = = s2[j - 1 ]): dp[i][j] = 1 + isSubSequence(s1, s2, i - 1 , j - 1 ) return dp[i][j] else : dp[i][j] = isSubSequence(s1, s2, i, j - 1 ) return dp[i][j] # Driver program to test above function str1 = "gksrek" str2 = "geeksforgeeks" m = len (str1) n = len (str2) if (m > n): print ( "NO" ) if (isSubSequence(str1, str2, m, n) = = m): print ( "YES" ) else : print ( "NO" ) # this code is contributed by shinjanpatra |
C#
// memoization C# program to check // if a string is subsequence // of another string using System; public class GFG { public static int [, ] dp = new int [1001, 1001]; // returns the length of longest common subsequence public static int isSubSequence( string s1, string s2, int i, int j) { if (i == 0 || j == 0) { return 0; } if (dp[i, j] != -1) { return dp[i, j]; } if (s1[i - 1] == s2[j - 1]) { return dp[i, j] = 1 + isSubSequence(s1, s2, i - 1, j - 1); } else { return dp[i, j] = isSubSequence(s1, s2, i, j - 1); } } // Driver Code public static void Main( string [] args) { string str1 = "gksrek" ; string str2 = "geeksforgeeks" ; int m = str1.Length; int n = str2.Length; if (m > n) { Console.WriteLine( "NO" ); } for ( int i = 0; i <= 1000; i++) { for ( int j = 0; j <= 1000; j++) dp[i, j] = -1; } if (isSubSequence(str1, str2, m, n) == m) { Console.WriteLine( "YES" ); } else { Console.WriteLine( "NO" ); } } } // This code is contributed by phasing17 |
Javascript
<script> // memoization JavaScript program to check // if a let is subsequence // of another let let dp = new Array(1001).fill(-1).map(() => new Array(1001).fill(-1)); // returns the length of longest common subsequence const isSubSequence = (s1, s2, i, j) => { if (i == 0 || j == 0) { return 0; } if (dp[i][j] != -1) { return dp[i][j]; } if (s1[i - 1] == s2[j - 1]) { return dp[i][j] = 1 + isSubSequence(s1, s2, i - 1, j - 1); } else { return dp[i][j] = isSubSequence(s1, s2, i, j - 1); } } /* Driver program to test above function */ let str1 = "gksrek" ; let str2 = "geeksforgeeks" ; let m = str1.length; let n = str2.length; if (m > n) document.write( "NO<br/>" ); if (isSubSequence(str1, str2, m, n) == m) { document.write( "YES<br/>" ); } else { document.write( "NO<br/>" ); } // This code is contributed by rakeshsahni </script> |
YES
Time complexity: O(m*n)
Auxiliary Space: O(m*n)
Following is the Iterative Implementation:
C++
// Iterative C++ program to check // if a string is subsequence // of another string #include <cstring> #include <iostream> 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 |
Python3
# 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. ?> |
Javascript
<script> // Iterative Javascript 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) { let 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 (let 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); } let str1 = "gksrek" ; let str2 = "geeksforgeeks" ; let m = str1.length; let n = str2.length; let res = isSubSequence(str1, str2, m, n); if (res) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by decode2207. </script> |
Yes
The Time Complexity of the implementation above is O(n) where n is the length of str2.
Auxiliary Space: O(1).
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