Check if a string can be formed from another string using given constraints

Given two strings S1 and S2(all characters are in lower-case). The task is to check if S2 can be formed from S1 using given constraints: 
1. Characters of S2 is there in S1 if there are two ‘a’ in S2, then S1 should have two ‘a’ also. 
2. If any character of S2 is not present in S1, check if the previous two ASCII characters are there in S1. e.g., if ‘e’ is there in S2 and not in S1, then ‘c’ and ‘d’ can be used from S1 to make ‘e’. 
Note: All characters from S1 can be used only once. 

Examples:

Input: S= abbat, W= cat 
Output: YES 
‘c’ is formed from ‘a’ and ‘b’, ‘a’ and ‘t’ is present in S1. 

Input: S= abbt, W= cat 
Output: NO 
‘c’ is formed from ‘a’ and ‘b’, but to form the next character 
‘a’ in S2, there is no more unused ‘a’ left in S1. 

Approach: The above problem can be solved using hashing. The count of all the characters in S1 is stored in a hash-table. Traverse in the string, and check if the character in S2 is there in the hash-table, reduce the count of that particular character in the hash-table. If the character is not there in the hash-table, check if the previous two ASCII characters are there in the hash-table, then reduce the count of the previous two ASCII characters in the hash-table. If all the characters can be formed from S1 using the given constraints, the string S2 can be formed from S1, else it cannot be formed. 

Below is the implementation of the above approach:  

C++

// CPP program to Check if a given
// string can be formed from another
// string using given constraints
#include <bits/stdc++.h>
using namespace std;

// Function to check if S2 can be formed of S1
bool check(string S1, string S2)
{
    // length of strings
    int n1 = S1.size();
    int n2 = S2.size();

    // hash-table to store count
    unordered_map<int, int> mp;

    // store count of each character
    for (int i = 0; i < n1; i++) {
        mp[S1[i]]++;
    }

    // traverse and check for every character
    for (int i = 0; i < n2; i++) {

        // if the character of s2 is present in s1
        if (mp[S2[i]]) {
            mp[S2[i]]--;
        }

        // if the character of s2 is not present in
        // S1, then check if previous two ASCII characters
        // are present in S1
        else if (mp[S2[i] - 1] && mp[S2[i] - 2]) {

            mp[S2[i] - 1]--;
            mp[S2[i] - 2]--;
        }
        else {
            return false;
        }
    }

   return true;
}

// Driver Code
int main()
{
    string S1 = "abbat";
    string S2 = "cat";

    // Calling function to check
    if (check(S1, S2))
        cout << "YES";
    else
        cout << "NO";
}

Java

// JAVA program to Check if a given
// String can be formed from another
// String using given constraints
import java.util.*;

class GFG
{

// Function to check if S2 can be formed of S1
static boolean check(String S1, String S2)
{
    // length of Strings
    int n1 = S1.length();
    int n2 = S2.length();

    // hash-table to store count
    HashMap<Integer,Integer> mp =
        new HashMap<Integer,Integer>();

    // store count of each character
    for (int i = 0; i < n1; i++)
    {
        if(mp.containsKey((int)S1.charAt(i)))
        {
            mp.put((int)S1.charAt(i),
            mp.get((int)S1.charAt(i)) + 1);
        }
        else
        {
            mp.put((int)S1.charAt(i), 1);
        }
    }

    // traverse and check for every character
    for (int i = 0; i < n2; i++)
    {

        // if the character of s2 is present in s1
        if(mp.containsKey((int)S2.charAt(i)))
        {
            mp.put((int)S2.charAt(i),
            mp.get((int)S2.charAt(i)) - 1);
        }

        // if the character of s2 is not present in
        // S1, then check if previous two ASCII characters
        // are present in S1
        else if (mp.containsKey(S2.charAt(i)-1) &&
                    mp.containsKey(S2.charAt(i)-2))
        {
            mp.put((S2.charAt(i) - 1),
            mp.get(S2.charAt(i) - 1) - 1);
            mp.put((S2.charAt(i) - 2),
            mp.get(S2.charAt(i) - 2) - 1);
        }
        else
        {
            return false;
        }
    }

    return true;
}

// Driver Code
public static void main(String[] args)
{
    String S1 = "abbat";
    String S2 = "cat";

    // Calling function to check
    if (check(S1, S2))
        System.out.print("YES");
    else
        System.out.print("NO");
}
}

// This code is contributed by PrinciRaj1992

Python3

# Python3 program to Check if a given string
# can be formed from another string using
# given constraints
from collections import defaultdict

# Function to check if S2 can
# be formed of S1
def check(S1, S2):

    # length of strings
    n1 = len(S1)
    n2 = len(S2)

    # hash-table to store count
    mp = defaultdict(lambda:0)

    # store count of each character
    for i in range(0, n1):
        mp[S1[i]] += 1

    # traverse and check for every character
    for i in range(0, n2):

        # if the character of s2 is
        # present in s1
        if mp[S2[i]]:
            mp[S2[i]] -= 1

        # if the character of s2 is not present
        # in S1, then check if previous two ASCII
        # characters are present in S1
        elif (mp[chr(ord(S2[i]) - 1)] and
              mp[chr(ord(S2[i]) - 2)]):

            mp[chr(ord(S2[i]) - 1)] -= 1
            mp[chr(ord(S2[i]) - 2)] -= 1

        else:
            return False

    return True

# Driver Code
if __name__ == "__main__":

    S1 = "abbat"
    S2 = "cat"

    # Calling function to check
    if check(S1, S2):
        print("YES")
    else:
        print("NO")

# This code is contributed by Rituraj Jain

C#

// C# program to Check if a given
// String can be formed from another
// String using given constraints
using System;
using System.Collections.Generic;

class GFG
{

// Function to check if S2 can be formed of S1
static bool check(String S1, String S2)
{
    // length of Strings
    int n1 = S1.Length;
    int n2 = S2.Length;

    // hash-table to store count
    Dictionary<int,int> mp =
        new Dictionary<int,int>();

    // store count of each character
    for (int i = 0; i < n1; i++)
    {
        if(mp.ContainsKey((int)S1[i]))
        {
            mp[(int)S1[i]] = mp[(int)S1[i]] + 1;
        }
        else
        {
            mp.Add((int)S1[i], 1);
        }
    }

    // traverse and check for every character
    for (int i = 0; i < n2; i++)
    {

        // if the character of s2 is present in s1
        if(mp.ContainsKey((int)S2[i]))
        {
            mp[(int)S2[i]] = mp[(int)S2[i]] - 1;
        }

        // if the character of s2 is not present in
        // S1, then check if previous two ASCII characters
        // are present in S1
        else if (mp.ContainsKey(S2[i] - 1) &&
                    mp.ContainsKey(S2[i] - 2))
        {
            mp[S2[i] - 1] = mp[S2[i] - 1] - 1;
            mp[S2[i] - 2] = mp[S2[i] - 2] - 1;
        }
        else
        {
            return false;
        }
    }

    return true;
}

// Driver Code
public static void Main(String[] args)
{
    String S1 = "abbat";
    String S2 = "cat";

    // Calling function to check
    if (check(S1, S2))
        Console.Write("YES");
    else
        Console.Write("NO");
}
}

// This code is contributed by PrinciRaj1992

Javascript

<script>

// JavaScript program to Check if a given
// String can be formed from another
// String using given constraints

// Function to check if S2 can be formed of S1
function check(S1, S2)
{

    // Length of Strings
    var n1 = S1.length;
    var n2 = S2.length;

    // hash-table to store count
    var mp = {};

    // Store count of each character
    for(var i = 0; i < n1; i++)
    {
        if (mp.hasOwnProperty(S1[i]))
        {
            mp[S1[i]] = mp[S1[i]] + 1;
        }
        else
        {
            mp[S1[i]] = 1;
        }
    }

    // Traverse and check for every character
    for(var i = 0; i < n2; i++)
    {

        // If the character of s2 is present in s1
        if (mp.hasOwnProperty(S2[i]))
        {
            mp[S2[i]] = mp[S2[i]] - 1;
        }

        // If the character of s2 is not present
        // in S1, then check if previous two ASCII
        // characters are present in S1
        else if (mp.hasOwnProperty(
            String.fromCharCode(S2[i].charCodeAt(0) - 1)) &&
            mp.hasOwnProperty(
            String.fromCharCode(S2[i].charCodeAt(0) - 2)))
        {
            mp[String.fromCharCode(
                S2[i].charCodeAt(0) - 1)] =
            mp[String.fromCharCode(
                S2[i].charCodeAt(0) - 1)] - 1;
            mp[String.fromCharCode(
                S2[i].charCodeAt(0) - 2)] =
            mp[String.fromCharCode(
                S2[i].charCodeAt(0) - 2)] - 1;
        }
        else
        {
            return false;
        }
    }
    return true;
}

// Driver Code
var S1 = "abbat";
var S2 = "cat";

// Calling function to check
if (check(S1, S2))
    document.write("YES");
else
    document.write("NO");

// This code is contributed by rdtank

</script>

Output:

YES

Time Complexity : O(m + n), where m is the length of string s1 and n is the length of string s2.
Auxiliary Space : O(m), where m is the length of string s1.