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Check if string S2 can be obtained by appending subsequences of string S1

Last Updated : 23 Jul, 2025
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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 
2 
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++ 
// 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 character
    int last = (*indices[s1[0]
                      - 'a']
                     .begin());

    // Traverse 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;
}
Java 
// Java Program to implement
// the above approach
import java.util.*;

// Class for finding minimum
// number of operations
public class Main {


    public static int findMinimumOperations(String s, String s1) {
    
        // Stores the length of strings
        int n = s.length();
        int m = s1.length();
    
        // Stores frequency of
        // characters in string s
        int[] frequency = new int[26];
    
        // Update frequencies of
        // character in s
        for (int i = 0; i < n; i++) {
            frequency[s.charAt(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.charAt(i) - 'a'] == 0) {
                return -1;
            }
        }
    
        // Stores the indices of
        // each character in s
        List<List<Integer>> indices = new ArrayList<>();
        for (int i = 0; i < 26; i++) {
            indices.add(new ArrayList<Integer>());
        }
    
        // Traverse string s
        for (int i = 0; i < n; i++) {
    
            // Store indices of characters
            indices.get(s.charAt(i) - 'a').add(i);
        }
    
        int ans = 2;
    
        // Stores index of last
        // appended character
        int last = indices.get(s1.charAt(0) - 'a').size() - 1;
    
        // Traverse string s1
        for (int i = 1; i < m; i++) {
    
            int ch = s1.charAt(i) - 'a';
    
            // Find the index of next
            // character that can be appended
            int it = binarySearchRight(indices.get(ch), last);
    
            // Check if the current
            // character be included
            // in the current subsequence
            if (it != indices.get(ch).size()) {
                last = it;
            }
            // Otherwise
            else {
    
                // Start a new subsequence
                ans++;
    
                // Update index of last
                // character appended
                last = indices.get(ch).size();
            }
        }
        return ans;
    }
    
    // Function to perform binary search
    // to get the rightmost position of
    // the element in the given array
    public static int binarySearchRight(List<Integer> arr, int x) {
        int left = 0, right = arr.size();
        while (left < right) {
            int mid = (left + right) / 2;
            if (arr.get(mid) <= x) {
                left = mid + 1;
            } else {
                right = mid;
            }
        }
        return left;
    }
    
    // Driver Code
    public static void main(String[] args) {
    
        String S1 = "acebcd";
        String S2 = "acbcde";
        int ans = findMinimumOperations(S1, S2);
    
        // If S2 cannot be obtained
        // from subsequences of S1
        if (ans == -1) {
            System.out.println("NO");
        }
        // Otherwise
        else {
            System.out.println("YES");
            System.out.println(ans);
        }
    }
}

// This code is contributed by phasing17
Python3 
# 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 character
    last =len(indices[ord(s1[0])- ord('a')]) - 1

    # Traverse 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
C# 
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;

class GFG {
  public static int FindMinimumOperations(string s,
                                          string s1)
  {
    // Stores the length of strings
    int n = s.Length;
    int m = s1.Length;

    // Stores frequency of
    // characters in string s
    int[] frequency = new int[26];

    // 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
    List<List<int> > indices = new List<List<int> >();
    for (int i = 0; i < 26; i++) {
      indices.Add(new List<int>());
    }

    // Traverse string s
    for (int i = 0; i < n; i++) {
      // Store indices of characters
      indices[s[i] - 'a'].Add(i);
    }

    int ans = 2;

    // Stores index of last
    // appended character
    int last = indices[s1[0] - 'a'].Count - 1;

    // Traverse string s1
    for (int i = 1; i < m; i++) {
      int ch = s1[i] - 'a';

      // Find the index of next
      // character that can be appended
      int it = BinarySearchRight(indices[ch], last);

      // Check if the current
      // character be included
      // in the current subsequence
      if (it != indices[ch].Count) {
        last = it;
      }
      // Otherwise
      else {
        // Start a new subsequence
        ans++;

        // Update index of last
        // character appended
        last = indices[ch].Count;
      }
    }

    return ans;
  }

  // Function to perform binary search
  // to get the rightmost position of
  // the element in the given array
  public static int BinarySearchRight(List<int> arr,
                                      int x)
  {
    int left = 0, right = arr.Count;
    while (left < right) {
      int mid = (left + right) / 2;
      if (arr[mid] <= x) {
        left = mid + 1;
      }
      else {
        right = mid;
      }
    }
    return left;
  }

  // Driver Code
  public static void Main(string[] args)
  {
    string S1 = "acebcd";
    string S2 = "acbcde";
    int ans = FindMinimumOperations(S1, S2);

    // If S2 cannot be obtained
    // from subsequences of S1
    if (ans == -1) {
      Console.WriteLine("NO");
    }
    // Otherwise
    else {
      Console.WriteLine("YES");
      Console.WriteLine(ans);
    }
  }
}

// This code is contributed by phasing17
JavaScript 
// Define function to find minimum operations
function findMinimumOperations(s, s1) {
    // Get length of strings
    const n = s.length,
        m = s1.length;

    // Create frequency array to store frequency of characters in s
    const frequency = new Array(26).fill(0);

    // Update frequency of each character in s
    for (let i = 0; i < n; i++) {
        frequency[s[i].charCodeAt() - 'a'.charCodeAt()]++;
    }

    // Check if all characters in s1 are present in s
    for (let i = 0; i < m; i++) {
        if (frequency[s1[i].charCodeAt() - 'a'.charCodeAt()] === 0) {
            return -1;
        }
    }

    // Create indices array to store indices of each character in s
    const indices = new Array(26);

    // Initialize each element of indices array as a Set
    for (let i = 0; i < 26; i++) {
        indices[i] = new Set();
    }

    // Update indices array with indices of each character in s
    for (let i = 0; i < n; i++) {
        indices[s[i].charCodeAt() - 'a'.charCodeAt()].add(i);
    }

    // Initialize variables to keep track of operations
    let ans = 1;
    let last = [...indices[s1[0].charCodeAt() - 'a'.charCodeAt()]][0];

    // Iterate through each character in s1
    for (let i = 1; i < m; i++) {
        // Get character code of current character in s1
        const ch = s1[i].charCodeAt() - 'a'.charCodeAt();
        // Get iterator of indices for current character in s
        const values = indices[ch].values();

        // Find first index greater than last
        let it = values.next();
        while (!it.done && it.value <= last) {
            it = values.next();
        }

        // If all indices are less than or equal to last, increment operations
        if (it.done) {
            ans++;
            last = [...indices[ch]][0];
        } else {
            // Else, update last with the next index
            last = it.value;
        }
    }

    // Return minimum number of operations required
    return ans;
}

// Define two strings S1 and S2
const S1 = "acebcd",
    S2 = "acbcde";

// Find minimum number of operations required to convert S1 to S2
const ans = findMinimumOperations(S1, S2);

// Print "NO" if S1 cannot be converted to S2, else print "YES" followed by the minimum number of operations
if (ans === -1) {
    console.log("NO");
} else {
    console.log("YES");
    console.log(ans);
}

Output: 
YES
2

 

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


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