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Check if it is possible to form string B from A under the given constraints
  • Difficulty Level : Expert
  • Last Updated : 16 Dec, 2020

Given two strings A and B and two integers b and m. The task is to find that if it is possible to form string B from A such that A is divided into groups of b characters except the last group which will have characters ≤ b and you are allowed to pick atmost m characters from each group, and also order of characters in B must be same as that of A. If it is possible then print Yes else print No.

Examples: 

Input: A = abcbbcdefxyz, B = acdxyz, b = 5, m = 2 
Output: Yes 
Groups can be “abcbb”, “cdefx” and “yz” 
Now “acdxyz” can be used to pick “ac” and “dx” can be picked from “cdefx”. 
Finally, “yz” if the last group.

Input: A = abcbbcdefxyz, B = baz, b = 3, m = 2 
Output: No 

Approach: The idea is to use binary search. Iterate through string A and store the frequency of each of the characters of A in vector S. Now iterate through B and if the current character is not in the vector then print No since its not possible to form string B using A. Else, check the first occurrence of current character starting from the index of the last chosen character low, which denotes starting position in string A from where we want to match characters of string B. Keep track of number of characters stored in each group. If it exceeds, the given limit of characters in current block, we update the pointer low to the next block.



Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if it is possible
// to form B from A satisfying the given conditions
bool isPossible(string A, string B, int b, int m)
{
 
    // Vector to store the frequency
    // of characters in A
    vector<int> S[26];
 
    // Vector to store the count of characters
    // used from a particular group of characters
    vector<int> box(A.length(), 0);
 
    // Store the frequency of the characters
    for (int i = 0; i < A.length(); i++) {
        S[A[i] - 'a'].push_back(i);
    }
 
    int low = 0;
 
    for (int i = 0; i < B.length(); i++) {
        auto it = lower_bound(S[B[i] - 'a'].begin(),
                              S[B[i] - 'a'].end(), low);
 
        // If a character in B is not
        // present in A
        if (it == S[B[i] - 'a'].end())
            return false;
 
        int count = (*it) / b;
        box[count] = box[count] + 1;
 
        // If count of characters used from
        // a particular group of characters
        // exceeds m
        if (box[count] >= m) {
            count++;
 
            // Update low to the starting index
            // of the next group
            low = (count)*b;
        }
 
        // If count of characters used from
        // a particular group of characters
        // has not exceeded m
        else
            low = (*it) + 1;
    }
 
    return true;
}
 
// Driver code
int main()
{
    string A = "abcbbcdefxyz";
    string B = "acdxyz";
    int b = 5;
    int m = 2;
 
    if (isPossible(A, B, b, m))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java




// Java implementation of the approach
import java.io.*;
import java.util.*;
 
class GFG{
     
// Function that returns true if it is
// possible to form B from A satisfying
// the given conditions
static boolean isPossible(String A, String B,
                          int b, int m)
{
     
    // List to store the frequency
    // of characters in A
    List<List<Integer>> S = new ArrayList<List<Integer>>();
 
    for(int i = 0; i < 26; i++)
        S.add(new ArrayList<Integer>());
         
    // Vector to store the count of characters
    // used from a particular group of characters
    int[] box = new int[A.length()];
 
    // Store the frequency of the characters
    for(int i = 0; i < A.length(); i++)
    {
        S.get(A.charAt(i) - 'a').add(i);
    }
 
    int low = 0;
 
    for(int i = 0; i < B.length(); i++)
    {
        List<Integer> indexes = S.get(
            B.charAt(i) - 'a');
 
        int it = lower_bound(indexes, low);
 
        // If a character in B is not
        // present in A
        if (it == indexes.size())
            return false;
 
        int count = indexes.get(it) / b;
        box[count] = box[count] + 1;
 
        // If count of characters used from
        // a particular group of characters
        // exceeds m
        if (box[count] >= m)
        {
            count++;
 
            // Update low to the starting index
            // of the next group
            low = (count) * b;
        }
 
        // If count of characters used from
        // a particular group of characters
        // has not exceeded m
        else
            low = indexes.get(it) + 1;
    }
 
    return true;
}
 
static int lower_bound(List<Integer> indexes, int k)
{
    int low = 0, high = indexes.size() - 1;
 
    while (low < high)
    {
        int mid = (low + high) / 2;
        if (indexes.get(mid) < k)
            low = mid + 1;
        else
            high = mid;
    }
    return (indexes.get(low) < k) ? low + 1 : low;
}
 
// Driver code
public static void main(String[] args)
{
    String A = "abcbbcdefxyz";
    String B = "acdxyz";
    int b = 5;
    int m = 2;
 
    if (isPossible(A, B, b, m))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by jithin

Python3




# Python3 implementation of the approach
 
# Function that returns true if it is
# possible to form B from A satisfying
# the given conditions
def isPossible(A, B, b, m) :
       
    # List to store the frequency
    # of characters in A
    S = []
   
    for i in range(26) :
     
        S.append([])  
           
    # Vector to store the count of characters
    # used from a particular group of characters
    box = [0] * len(A)
   
    # Store the frequency of the characters
    for i in range(len(A)) :
     
        S[ord(A[i]) - ord('a')].append(i)
   
    low = 0
   
    for i in range(len(B)) :
     
        indexes = S[ord(B[i]) - ord('a')]
   
        it = lower_bound(indexes, low)
   
        # If a character in B is not
        # present in A
        if (it == len(indexes)) :
            return False
   
        count = indexes[it] // b
        box[count] = box[count] + 1
   
        # If count of characters used from
        # a particular group of characters
        # exceeds m
        if (box[count] >= m) :
         
            count += 1
   
            # Update low to the starting index
            # of the next group
            low = (count) * b
   
        # If count of characters used from
        # a particular group of characters
        # has not exceeded m
        else :
            low = indexes[it] + 1
   
    return True
   
def lower_bound(indexes, k) :
 
    low, high = 0, len(indexes) - 1
   
    while (low < high) :
     
        mid = (low + high) // 2
        if (indexes[mid] < k) :
            low = mid + 1
        else :
            high = mid
     
    if indexes[low] < k :
        return (low + 1)
    else :
        return low
         
A = "abcbbcdefxyz"
B = "acdxyz"
b = 5
m = 2
 
if (isPossible(A, B, b, m)) :
    print("Yes")
else :
    print("No")
 
    # This code is contributed by divyeshrabadiya07

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
     
    // Function that returns true if it is
    // possible to form B from A satisfying
    // the given conditions
    static bool isPossible(string A, string B, int b, int m)
    {
          
        // List to store the frequency
        // of characters in A
        List<List<int>> S = new List<List<int>>();
      
        for(int i = 0; i < 26; i++)
        {
            S.Add(new List<int>());
        }
              
        // Vector to store the count of characters
        // used from a particular group of characters
        int[] box = new int[A.Length];
      
        // Store the frequency of the characters
        for(int i = 0; i < A.Length; i++)
        {
            S[A[i] - 'a'].Add(i);
        }
      
        int low = 0;
      
        for(int i = 0; i < B.Length; i++)
        {
            List<int> indexes = S[B[i] - 'a'];
      
            int it = lower_bound(indexes, low);
      
            // If a character in B is not
            // present in A
            if (it == indexes.Count)
                return false;
      
            int count = indexes[it] / b;
            box[count] = box[count] + 1;
      
            // If count of characters used from
            // a particular group of characters
            // exceeds m
            if (box[count] >= m)
            {
                count++;
      
                // Update low to the starting index
                // of the next group
                low = (count) * b;
            }
      
            // If count of characters used from
            // a particular group of characters
            // has not exceeded m
            else
                low = indexes[it] + 1;
        }
      
        return true;
    }
      
    static int lower_bound(List<int> indexes, int k)
    {
        int low = 0, high = indexes.Count - 1;
      
        while (low < high)
        {
            int mid = (low + high) / 2;
            if (indexes[mid] < k)
                low = mid + 1;
            else
                high = mid;
        }
        return (indexes[low] < k) ? low + 1 : low;
    }
 
  static void Main() {
       
    string A = "abcbbcdefxyz";
    string B = "acdxyz";
    int b = 5;
    int m = 2;
  
    if (isPossible(A, B, b, m))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
  }
}
 
// This code is contributed by divyesh072019

 
 

Output: 
Yes

 

 

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