Minimum subsequences of a string A required to be appended to obtain the string B

Last Updated : 05 Feb, 2021

Given two strings A and B, the task is to count the minimum number of operations required to construct the string B by following operations:

Select a subsequence of the string A.
Append the subsequence at the newly formed string (initially empty).

Print the minimum count of operations required. If it is impossible to make the new string equal to B by applying the given operations, then print -1.

Examples:

Input: A = “abc”, B = “abac”
Output: 2
Explanation:
Initially, C = “”.
Step 1: Select subsequence “ab” from string A and append it to the empty string C, i.e. C = “ab”.
Step 2: Select subsequence “ac” from string A and append it to the end of string C, i.e. C = “abac”.
Now, the string C is same as string B.
Therefore, count of operations required is 2.
Input: A = “geeksforgeeks”, B = “programming”
Output: -1

Approach: Follow the below steps to solve this problem:

Initialize a Map to map characters present in the string A with their respective indices.
For each character in string A, keep track of all of its occurrences.
Initialize a variable, say ans, to store the count of operations required. As the number of operations must be greater than 1, set ans = 1.
Iterate over the characters of string B and check if the character is present in the string A or not by using the Map.
Lastly, maximize the length of the subsequence chosen from the string A for each operation.
Finally, print the minimum operations required.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the minimum
// subsequences of a string A required
// to be appended to obtain the string B
void countminOpsToConstructAString(string A,
                                   string B)
{
    // Size of the string
    int N = A.length();
 
    int i = 0;
 
    // Maps characters to their
    // respective indices
    map<char, set<int> > mp;
 
    // Insert indices of characters
    // into the sets
    for (i = 0; i < N; i++) {
        mp[A[i]].insert(i);
    }
 
    // Stores the position of the last
    // visited index in the string A.
    // Initially set it to -1.
    int previous = -1;
 
    // Stores the required count
    int ans = 1;
 
    // Iterate over the characters of B
    for (i = 0; i < B.length(); i++) {
        char ch = B[i];
 
        // If the character in B is
        // not present in A, return -1
        if (mp[ch].size() == 0) {
            cout << -1;
            return;
        }
 
        // Fetch the next index from B[i]'s set
        auto it = mp[ch].upper_bound(previous);
 
        // If the iterator points to
        // the end of that set
        if (it == mp[ch].end()) {
 
            previous = -1;
            ans++;
            --i;
            continue;
        }
 
        // If it doesn't point to the
        // end, update  previous
        previous = *it;
    }
 
    // Print the answer
    cout << ans;
}
 
// Driver Code
int main()
{
    string A = "abc", B = "abac";
    countminOpsToConstructAString(A, B);
 
    return 0;
}

Python3

# Python3 program for the above approac
from bisect import bisect_right
 
# Function to count the minimum
# subsequences of a A required
# to be appended to obtain the B
def countminOpsToConstructAString(A, B):
   
    # Size of the string
    N = len(A)
    i = 0
 
    # Maps characters to their
    # respective indices
    mp = [[] for i in range(26)]
 
    # Insert indices of characters
    # into the sets
    for i in range(N):
        mp[ord(A[i]) - ord('a')].append(i)
 
    # Stores the position of the last
    # visited index in the A.
    # Initially set it to -1.
    previous = -1
 
    # Stores the required count
    ans, i = 1, 0
 
    # Iterate over the characters of B
    while i < len(B):
        ch = B[i]
 
        # If the character in B is
        # not present in A, return -1
        if (len(mp[ord(ch) - ord('a')]) == 0):
            print(-1)
            return
 
        # Fetch the next index from B[i]'s set
        it = bisect_right(mp[ord(ch) - ord('a')], previous)
 
        # If the iterator points to
        # the end of that set
        if (it == len(mp[ord(ch) - ord('a')])):
            previous = -1
            ans += 1
            # i -= 1
            continue
 
        # If it doesn't poto the
        # end, update  previous
        previous = mp[ord(ch) - ord('a')][it]
        i += 1
 
    # Prthe answer
    print (ans)
 
# Driver Code
if __name__ == '__main__':
    A, B = "abc", "abac"
    countminOpsToConstructAString(A, B)
 
    # This code is contributed by mohit kumar 29.

Output:

Time Complexity: O(N * logN)
Auxiliary Space: O(N)

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