# 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 ` `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 mp =  ` `        ``new` `HashMap(); ` ` `  `    ``// 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 `

Output:

```YES
```

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