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Check if string S2 can be obtained by appending subsequences of string S1
  • Difficulty Level : Hard
  • Last Updated : 04 Dec, 2020

Given two strings S1 and S2, the task is to check if it’s possible to generate string S2 by repeatedly appending subsequences of S1 to the end of an initially empty string. If possible, print “YES” and the minimum number of operations required. Otherwise, print “NO“.

Examples:

Input: S1 = “acebcd”, S2 = “acbcd” 
Output: 
YES 

Explanation: Append subsequence “acbc” followed by “d” to obtain S2.

Input: S1 = “aceasd”, S2 = “asdxds” 
Output: NO 
Explanation: Since character ‘x’ is not present in S1, S2 cannot be obtained.

Approach: Follow the steps below to solve the problem:



  • Iterate over characters of string S1 and store frequencies of each character in S1 in an array freq[].
  • Traverse the string S2 and check if there is any character in S2 which is not present in S1. If any such character is found, print “NO”.
  • Otherwise, iterate over characters in S1 and update indices of each character in a Set
  • Traverse string S2 and for each character, check if it can be included in the current subsequence of S1 that can be appended.
  • If found to be true, set the index of current character as that of the last character appended. Otherwise, increase count of subsequences and set the index of current character as that of the last character appended. Proceed to the next character.
  • finally, print “YES” and the count of such subsequences as the required answer.

Below is the implementation of the above approach:

C++

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// C++ Program to implement
// the above approach
#include "bits/stdc++.h"
using namespace std;
 
// Function for finding minimum
// number of operations
int findMinimumOperations(string s,
                          string s1)
{
 
    // Stores the length of strings
    int n = s.length(), m = s1.length();
 
    // Stores frequency of
    // characters in string s
    int frequency[26] = { 0 };
 
    // Update  frequencies of
    // character in s
    for (int i = 0; i < n; i++)
        frequency[s[i] - 'a']++;
 
    // Traverse string s1
    for (int i = 0; i < m; i++) {
 
        // If any character in s1
        // is not present in s
        if (frequency[s1[i] - 'a']
            == 0) {
            return -1;
        }
    }
 
    // Stores the indices of
    // each character in s
    set<int> indices[26];
 
    // Traverse string s
    for (int i = 0; i < n; i++) {
 
        // Store indices of characters
        indices[s[i] - 'a'].insert(i);
    }
 
    int ans = 1;
 
    // Stores index of last
    // appended characterr
    int last = (*indices[s1[0]
                      - 'a']
                     .begin());
 
    // Traaverse string s1
    for (int i = 1; i < m; i++) {
 
        int ch = s1[i] - 'a';
 
        // Find the index of next
        // character that can be appended
        auto it = indices[ch].upper_bound(
            last);
 
        // Check if the current
        // character be included
        // in the current subsequence
        if (it != indices[ch].end()) {
            last = (*it);
        }
 
        // Otherwise
        else {
 
            // Start a new subsequence
            ans++;
 
            // Update index of last
            // character appended
            last = (*indices[ch].begin());
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
 
    string S1 = "acebcd", S2 = "acbcde";
    int ans = findMinimumOperations(
        S1, S2);
 
    // If S2 cannot be obtained
    // from subsequences of S1
    if (ans == -1) {
        cout << "NO\n";
    }
    // Otherwise
    else {
        cout << "YES\n"
             << ans;
    }
 
    return 0;
}

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Python3

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# Python3 Program to implement
# the above approach
from bisect import bisect ,bisect_left,bisect_right
 
# Function for finding minimum
# number of operations
def findMinimumOperations(s,s1):
 
    #Stores the length of strings
    n = len(s)
    m = len(s1)
 
    # Stores frequency of
    # characters in string s
    frequency = [0]*26
 
    # Update  frequencies of
    # character in s
    for i in range(n):
        frequency[ord(s[i]) - ord('a')] += 1
 
    # Traverse string s1
    for i in range(m):
 
        # If any character in s1
        # is not present in s
        if (frequency[ord(s1[i]) - ord('a')] == 0):
            return -1
 
    # Stores the indices of
    # each character in s
    indices = [[] for i in range(26)]
 
    # Traverse string s
    for i in range(n):
 
        # Store indices of characters
        indices[ord(s[i]) - ord('a')].append(i)
 
    ans = 2
 
    # Stores index of last
    # appended characterr
    last =len(indices[ord(s1[0])- ord('a')]) - 1
 
    # Traaverse string s1
    for i in range(1,m):
 
        ch = ord(s1[i]) - ord('a')
 
        # Find the index of next
        # character that can be appended
        it = bisect_right(indices[ch],last)
        # print(it)
 
        # Check if the current
        # character be included
        # in the current subsequence
        if (it != len(indices[ch])):
            last = it
        # Otherwise
        else:
 
            # Start a new subsequence
            ans += 1
 
            # Update index of last
            # character appended
            last = len(indices[ch])
    return ans
 
# Driver Code
if __name__ == '__main__':
 
    S1 = "acebcd"
    S2 = "acbcde"
    ans = findMinimumOperations(S1, S2)
 
    # If S2 cannot be obtained
    # from subsequences of S1
    if (ans == -1):
        print("NO")
    # Otherwise
    else:
        print("YES")
        print(ans)
 
# This code is contributed by mohit kumar 29

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Output: 

YES
2

 

Time Complexity: O(Mlog(N))
Auxiliary Space: O(N)

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