# Check if a string consisting only of a, b, c can be made empty by removing substring “abc” recursively

• Difficulty Level : Medium
• Last Updated : 10 Aug, 2021

Given a string S size of N consisting of characters ‘a‘, ‘b‘, and ‘c‘ only, the task is to check if the given string can be made empty by removing the string “abc” recursively or not. If found to be true, then print “Yes”. Otherwise, print “No”.

Examples:

Input: S = abcabc
Output: Yes
Explanation:
Below are the operations performed to empty the string:

1. Remove the substring S[3, 5] from the string modifies the string S to “abc”.
2. Remove the substring S[0, 2] from the string modifies the string S to “”.

After the above operations, the given string S can be made empty by removing substring “abc”. Therefore, print Yes.

Input: S = abcabcababccc
Output: No

Approach: The given problem can be solved by using a stack. The idea is to minimize the string to an empty string by removing all occurrences of the “abc”. Follow the steps below to solve the given problem:

• Initialize a stack, say Stack to store the characters of the given string S.
• Traverse the given string S and perform the following steps:
• If the current character is ‘a‘ or ‘b‘, then push it to the stack Stack.
• If the current character is ‘c‘ and the last two characters are ‘b‘ and ‘a‘ respectively, then pop twice from the stack.
• If the current character is ‘c‘ and the last two characters are notb‘ and ‘a‘, then the given string S can not be formed.
• After completing the above steps, if the stack is empty, then print “Yes”. Otherwise, print “No”.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach` `#include ``using` `namespace` `std;` `// Function to check if the given``// string S can be made empty``string canMadeEmpty(string s, ``int` `n)``{``    ``// Stores the characters``    ``// of the string S``    ``stack<``char``> St;` `    ``// Traverse the given string``    ``for` `(``int` `i = 0; i < n; i++) {``        ``// If the character is c``        ``if` `(s[i] == ``'c'``) {` `            ``// If stack size is greater``            ``// than 2``            ``if` `(St.size() >= 2) {` `                ``// Pop from the stack``                ``char` `b = St.top();``                ``St.pop();``                ``char` `a = St.top();``                ``St.pop();` `                ``// Top two characters in``                ``// the stack should be 'b'``                ``// and 'a' respectively``                ``if` `(a != ``'a'` `|| b != ``'b'``)``                    ``return` `"No"``;``            ``}` `            ``// Otherwise, print No``            ``else``                ``return` `"No"``;``        ``}` `        ``// If character is 'a' or 'b'``        ``// push to stack``        ``else``            ``St.push(s[i]);``    ``}` `    ``// If stack is empty, then print``    ``// Yes. Otherwise print No``    ``if` `(St.size() == 0) {``        ``return` `"Yes"``;``    ``}``    ``else` `{``        ``return` `"No"``;``    ``}``}` `// Driver Code``int` `main()``{``    ``string S = ``"aabcbc"``;``    ``int` `N = S.length();``    ``cout << canMadeEmpty(S, N);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.util.*;` `class` `GFG``{` `// Function to check if the given``// String S can be made empty``static` `String canMadeEmpty(String s, ``int` `n)``{``  ` `    ``// Stores the characters``    ``// of the String S``    ``Stack St = ``new` `Stack();` `    ``// Traverse the given String``    ``for` `(``int` `i = ``0``; i < n; i++)``    ``{``      ` `        ``// If the character is c``        ``if` `(s.charAt(i) == ``'c'``) {` `            ``// If stack size is greater``            ``// than 2``            ``if` `(St.size() >= ``2``) {` `                ``// Pop from the stack``                ``char` `b = St.peek();``                ``St.pop();``                ``char` `a = St.peek();``                ``St.pop();` `                ``// Top two characters in``                ``// the stack should be 'b'``                ``// and 'a' respectively``                ``if` `(a != ``'a'` `|| b != ``'b'``)``                    ``return` `"No"``;``            ``}` `            ``// Otherwise, print No``            ``else``                ``return` `"No"``;``        ``}` `        ``// If character is 'a' or 'b'``        ``// push to stack``        ``else``            ``St.add(s.charAt(i));``    ``}` `    ``// If stack is empty, then print``    ``// Yes. Otherwise print No``    ``if` `(St.size() == ``0``) {``        ``return` `"Yes"``;``    ``}``    ``else` `{``        ``return` `"No"``;``    ``}``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``String S = ``"aabcbc"``;``    ``int` `N = S.length();``    ``System.out.print(canMadeEmpty(S, N));``}``}` `// This code is contributed by Princi Singh.`

## Python3

 `# Python program for the above approach` `# Function to check if the given``# string S can be made empty``def` `canMadeEmpty(s, n):``  ` `    ``# Stores the characters``    ``# of the string S``    ``st ``=` `[]``    ` `    ``# Traverse the given string``    ``for` `i ``in` `range``(n):``      ` `        ``# If the character is c``        ``if` `s[i] ``=``=` `'c'``:``          ` `             ``# If stack size is greater``            ``# than 2``            ``if` `len``(st) >``=` `2``:``              ` `                ``# Pop from the stack``                ``b ``=` `st[``-``1``]``                ``st.pop()``                ``a ``=` `st[``-``1``]``                ``st.pop()``                ` `                 ``# Top two characters in``                ``# the stack should be 'b'``                ``# and 'a' respectively``                ``if` `a !``=` `'a'` `or` `b !``=` `'b'``:``                    ``return` `"No"``                  ` `             ``# Otherwise, print No``            ``else``:``                ``return` `"No"``              ` `        ``# If character is 'a' or 'b'``        ``# push to stack``        ``else``:``            ``st.append(s[i])``            ` `    ``# If stack is empty, then print``    ``# Yes. Otherwise print No``    ``if` `len``(st) ``=``=` `0``:``        ``return` `"Yes"``    ``else``:``        ``return` `"No"` `# Driver code``s ``=` `"aabcbc"``n ``=` `len``(s)``print``(canMadeEmpty(s, n))` `# This code is contributed by Parth Manchanda`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `public` `class` `GFG``{` `// Function to check if the given``// String S can be made empty``static` `String canMadeEmpty(String s, ``int` `n)``{``  ` `    ``// Stores the characters``    ``// of the String S``    ``Stack<``char``> St = ``new` `Stack<``char``>();` `    ``// Traverse the given String``    ``for` `(``int` `i = 0; i < n; i++)``    ``{``      ` `        ``// If the character is c``        ``if` `(s[i] == ``'c'``) {` `            ``// If stack size is greater``            ``// than 2``            ``if` `(St.Count >= 2) {` `                ``// Pop from the stack``                ``char` `b = St.Peek();``                ``St.Pop();``                ``char` `a = St.Peek();``                ``St.Pop();` `                ``// Top two characters in``                ``// the stack should be 'b'``                ``// and 'a' respectively``                ``if` `(a != ``'a'` `|| b != ``'b'``)``                    ``return` `"No"``;``            ``}` `            ``// Otherwise, print No``            ``else``                ``return` `"No"``;``        ``}` `        ``// If character is 'a' or 'b'``        ``// push to stack``        ``else``            ``St.Push(s[i]);``    ``}` `    ``// If stack is empty, then print``    ``// Yes. Otherwise print No``    ``if` `(St.Count == 0) {``        ``return` `"Yes"``;``    ``}``    ``else` `{``        ``return` `"No"``;``    ``}``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``String S = ``"aabcbc"``;``    ``int` `N = S.Length;``    ``Console.Write(canMadeEmpty(S, N));``}``}` `// This code is contributed by shikhasingrajput`

## Javascript

 ``

Output:

`Yes`

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

My Personal Notes arrow_drop_up