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)

We strongly recommend to minimize the browser and try this yourself first.

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 Implementation in C++ and Python of 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;
}

Python

# Recursive Python 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
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"
m = len(string1)
n = len(string2)
if isSubSequence(string1, string2, m, n):
    print "Yes"
else:
    print "No"

# This code is contributed by BHAVYA JAIN


Output:
Yes

Following is Iterative Implementation in C++ for the same.

// 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;
}

Output:

Yes

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

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