# Check if given String can be split only into subsequences ABC

• Last Updated : 12 Oct, 2021

Given a string S of length N where each character of the string is either ‘A‘, ‘B‘ or ‘C‘. The task is to find if is it possible to split the string into subsequences “ABC“. If it is possible to split then print “Yes“. Otherwise, print “No“.

Examples:

Input: S = “ABABCC”
Output: Yes
Explanation:
One of the possible way of splitting is to split the string in 2 subsequences of “ABC” is following:

• First form a subsequence “ABC” by taking the character at indices 0, 1, and 4.
• Again form a subsequence “ABC” by taking the character at indices 2, 3, and 5.

Therefore, the string can be split in 2 subsequences of “ABC”.

Input: S = “AABBCC”
Output: Yes
Explanation:
One of the possible way of splitting is to split the string in 2 subsequences of “ABC” is following:

• First form a subsequence “ABC” by taking the character at indices 0, 2, and 4
• Again form a subsequence “ABC” by taking the character at indices 1, 3, and 5.

Therefore, the string can be split in 2 subsequences of “ABC”.

Input: S = “BAC”
Output: No

Approach: The given problem can be solved based on the following observations:

• It can be observed that if N is not multiple of 3 or count of ‘A‘, ‘B‘, and ‘C‘ are not equal then it will be impossible to split the string satisfying the conditions.
• Also, for every ‘B‘ there must be at least one ‘A‘ to its left and one ‘C’ to its right.

Follow the steps below to solve the problem:

• Initialize 3 deques of integers say A, B, and C to store the indices of characters ‘A‘, ‘B‘ and ‘C‘ respectively.
• Iterate over characters of the string S and in each iteration, if the current character is ‘A‘ then push the index i into A. Else if the current character is ‘B‘ then push the index i into B. Otherwise, push the index i into C.
• If N is not multiple of 3 and count of characters ‘A‘, ‘B‘ and ‘C‘ are not equal then print “No”.
• Otherwise, Iterate over the deque B using the variable i, and in each iteration, if B[i] is greater than the front element of deque A then Pop the front element of deque A. Otherwise print “No” and return.
• Now again iterate over the deque B using the variable i in reverse and in each iteration if B[i] is lesser than the last element of deque C then Pop the last element of the deque C. Otherwise print “No” and return.
• Finally, if none of the above cases satisfy then print “Yes” as the answer.

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 can be splitted into` `// subsequences "ABC"` `string check(string S)` `{` `    ``// Stores the length` `    ``// of the string` `    ``int` `N = S.length();`   `    ``// Stores the indices of 'A'` `    ``deque<``int``> A;` `    ``// Stores the indices of 'B'` `    ``deque<``int``> B;` `    ``// Stores the indices of 'C'` `    ``deque<``int``> C;`   `    ``// Traverse the string S` `    ``for` `(``int` `i = 0; i < N; i++) {`   `        ``// If S[i] is equal to 'A'` `        ``if` `(S[i] == ``'A'``) {` `            ``// Push the index i in A` `            ``A.push_back(i);` `        ``}` `        ``// Else if S[i] is equal` `        ``// to 'B'` `        ``else` `if` `(S[i] == ``'B'``) {` `            ``// Push the index i in B` `            ``B.push_back(i);` `        ``}`   `        ``// Else if S[i] is equal` `        ``// to 'C'` `        ``else` `{` `            ``// Push the index i in C` `            ``C.push_back(i);` `        ``}` `    ``}`   `    ``// If N is not multiple of 3 or` `    ``// count of characters 'A', 'B'` `    ``// and 'C' are not equal` `    ``if` `(N % 3 || A.size() != B.size()` `        ``|| A.size() != C.size()) {`   `        ``// Return "No"` `        ``return` `"No"``;` `    ``}` `    ``// Iterate over the deque B` `    ``for` `(``int` `i = 0; i < B.size(); i++) {`   `        ``// If A is not empty and` `        ``// B[i] is greater than` `        ``// A[0]` `        ``if` `(!A.empty() && B[i] > A[0]) {` `            ``// Removes the front` `            ``// element of A` `            ``A.pop_front();` `        ``}` `        ``// Else return "No"` `        ``else` `{` `            ``return` `"No"``;` `        ``}` `    ``}` `    ``// Iterate over the deque` `    ``// B in reverse` `    ``for` `(``int` `i = B.size() - 1; i >= 0; i--) {` `        ``// If C is not empty and` `        ``// last element of C is` `        ``// greater thab B[i]` `        ``if` `(!C.empty() && B[i] < C.back()) {` `            ``// Removes the last` `            ``// element of C` `            ``C.pop_back();` `        ``}` `        ``// Else return "No"` `        ``else` `{` `            ``return` `"No"``;` `        ``}` `    ``}`   `    ``// If none of the above` `    ``// cases satisfy return` `    ``//"Yes'` `    ``return` `"Yes"``;` `}`   `// Driver Code` `int` `main()` `{` `    ``// Input` `    ``string S = ``"ABABCC"``;`   `    ``// Function call` `    ``cout << check(S) << endl;` `    ``return` `0;` `}`

## Java

 `// Java program for the above approach` `import` `java.io.*;` `import` `java.util.*;`   `class` `GFG{` `    `  `// Function to check if the given` `// string can be splitted into` `// subsequences "ABC"` `public` `static` `String check(String S)` `{` `    `  `    ``// Stores the length` `    ``// of the string` `    ``int` `N = S.length();`   `    ``// Stores the indices of 'A'` `    ``Deque A = ``new` `LinkedList();` `    `  `    ``// Stores the indices of 'B'` `    ``Deque B = ``new` `LinkedList();` `    `  `    ``// Stores the indices of 'C'` `    ``Deque C = ``new` `LinkedList();` `    `  `    ``// Traverse the string S` `    ``for``(``int` `i = ``0``; i < N; i++)` `    ``{` `        `  `        ``// If S[i] is equal to 'A'` `        ``if` `(S.charAt(i) == ``'A'``) ` `        ``{` `            `  `            ``// Push the index i in A` `            ``A.addLast(i);` `        ``}` `        `  `        ``// Else if S[i] is equal` `        ``// to 'B'` `        ``else` `if` `(S.charAt(i) == ``'B'``) ` `        ``{` `            `  `            ``// Push the index i in B` `            ``B.addLast(i);` `        ``}` `        `  `        ``// Else if S[i] is equal` `        ``// to 'C'` `        ``else` `        ``{` `            `  `            ``// Push the index i in C` `            ``C.addLast(i);` `        ``}` `    ``}`   `    ``// If N is not multiple of 3 or` `    ``// count of characters 'A', 'B'` `    ``// and 'C' are not equal` `    ``if` `(N % ``3` `> ``0` `|| A.size() != B.size() ||` `        ``A.size() != C.size())` `    ``{` `        `  `        ``// Return "No"` `        ``return` `"No"``;` `    ``}` `    `  `    ``// Iterate over the deque B` `    ``for``(Iterator itr = B.iterator(); itr.hasNext();) ` `    ``{` `        ``Integer b = (Integer)itr.next();` `        `  `        ``// If A is not empty and` `        ``// curr B is greater than` `        ``// A[0]` `        ``if` `(A.size() > ``0` `&& b > A.getFirst())` `        ``{` `            `  `            ``// Removes the front` `            ``// element of A` `            ``A.pop();` `        ``}` `        `  `        ``// Else return "No"` `        ``else` `        ``{` `            ``return` `"No"``;` `        ``}` `    ``}`   `    ``// Iterate over the deque` `    ``// B in reverse` `    ``for``(Iterator itr = B.descendingIterator();` `        ``itr.hasNext();)` `    ``{` `        ``Integer b = (Integer)itr.next();` `        `  `        ``// If C is not empty and` `        ``// last element of C is` `        ``// greater thab B[i]` `        ``if` `(C.size() > ``0` `&& b < C.getLast()) ` `        ``{` `            `  `            ``// Removes the last` `            ``// element of C` `            ``C.pollLast();` `        ``}` `        `  `        ``// Else return "No"` `        ``else` `        ``{` `            ``return` `"No"``;` `        ``}` `    ``}` `    `  `    ``// If none of the above` `    ``// cases satisfy return` `    ``//"Yes'` `    ``return` `"Yes"``;` `}`   `// Driver code` `public` `static` `void` `main(String[] args) ` `{` `    `  `    ``// Input` `    ``String S = ``"ABABCC"``;`   `    ``// Function call` `    ``System.out.println(check(S));` `}` `}`   `// This code is contributed by Manu Pathria`

## Python3

 `# Python3 program for the above approach` `from` `collections ``import` `deque`   `# Function to check if the given` `# can be splitted into` `# subsequences "ABC"` `def` `check(S):` `    `  `    ``# Stores the length` `    ``# of the string` `    ``N ``=` `len``(S)`   `    ``# Stores the indices of 'A'` `    ``A ``=` `deque()` `    `  `    ``# Stores the indices of 'B'` `    ``B ``=` `deque()` `    `  `    ``# Stores the indices of 'C'` `    ``C ``=` `deque()`   `    ``# Traverse the S` `    ``for` `i ``in` `range``(N):` `        `  `        ``# If S[i] is equal to 'A'` `        ``if` `(S[i] ``=``=` `'A'``):` `            `  `            ``# Push the index i in A` `            ``A.append(i)` `            `  `        ``# Else if S[i] is equal` `        ``# to 'B'` `        ``elif` `(S[i] ``=``=` `'B'``):` `            `  `            ``# Push the index i in B` `            ``B.append(i)` `            `  `        ``# Else if S[i] is equal` `        ``# to 'C'` `        ``else``:` `            `  `            ``# Push the index i in C` `            ``C.append(i)`   `    ``# If N is not multiple of 3 or` `    ``# count of characters 'A', 'B'` `    ``# and 'C' are not equal` `    ``if` `(N ``%` `3` `or` `len``(A) !``=` `len``(B) ``or` `                 ``len``(A) !``=` `len``(C)):` `                     `  `        ``# Return "No"` `        ``return` `"No"` `        `  `    ``# Iterate over the deque B` `    ``for` `i ``in` `range``(``len``(B)):` `        `  `        ``# If A is not empty and` `        ``# B[i] is greater than` `        ``# A[0]` `        ``if` `(``len``(A) > ``0` `and` `B[i] > A[``0``]):` `            `  `            ``# Removes the front` `            ``# element of A` `            ``A.popleft()` `            `  `        ``# Else return "No"` `        ``else``:` `            ``return` `"No"`   `    ``# Iterate over the deque` `    ``# B in reverse` `    ``for` `i ``in` `range``(``len``(B) ``-` `1``, ``-``1``, ``-``1``):` `        `  `        ``# If C is not empty and` `        ``# last element of C is` `        ``# greater thab B[i]` `        ``if` `(``len``(C) > ``0` `and` `B[i] < C[``-``1``]):` `            `  `            ``# Removes the last` `            ``# element of C` `            ``C.popleft()` `            `  `        ``# Else return "No"` `        ``else``:` `            ``return` `"No"`   `    ``# If none of the above` `    ``# cases satisfy return` `    ``# "Yes'` `    ``return` `"Yes"`   `# Driver Code` `if` `__name__ ``=``=` `'__main__'``:` `    `  `    ``# Input` `    ``S ``=` `"ABABCC"`   `    ``# Function call` `    ``print``(check(S))`   `# This code is contributed by mohit kumar 29`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;`   `public` `static` `class` `GFG {` `    `  `    ``public` `static` `IEnumerable Reverse(``this` `LinkedList list) {` `        ``var` `el = list.Last;` `        ``while` `(el != ``null``) {` `            ``yield` `return` `el.Value;` `            ``el = el.Previous;` `        ``}` `    ``}` `    `  `    ``// Function to check if the given` `    ``// string can be splitted into` `    ``// subsequences "ABC"` `    ``public` `static` `string` `check(``string` `S)` `    ``{` `        `  `        ``// Stores the length` `        ``// of the string` `        ``int` `N = S.Length;` `    `  `        ``// Stores the indices of 'A'` `        ``LinkedList<``int``> A = ``new` `LinkedList<``int``>();` `        `  `        ``// Stores the indices of 'B'` `        ``LinkedList<``int``> B = ``new` `LinkedList<``int``>();` `        `  `        ``// Stores the indices of 'C'` `        ``LinkedList<``int``> C = ``new` `LinkedList<``int``>();` `        `  `        ``// Traverse the string S` `        ``for``(``int` `i = 0; i < N; i++)` `        ``{` `            `  `            ``// If S[i] is equal to 'A'` `            ``if` `(S[i] == ``'A'``) ` `            ``{` `                `  `                ``// Push the index i in A` `                ``A.AddLast(i);` `            ``}` `            `  `            ``// Else if S[i] is equal` `            ``// to 'B'` `            ``else` `if` `(S[i] == ``'B'``) ` `            ``{` `                `  `                ``// Push the index i in B` `                ``B.AddLast(i);` `            ``}` `            `  `            ``// Else if S[i] is equal` `            ``// to 'C'` `            ``else` `            ``{` `                `  `                ``// Push the index i in C` `                ``C.AddLast(i);` `            ``}` `        ``}` `    `  `        ``// If N is not multiple of 3 or` `        ``// count of characters 'A', 'B'` `        ``// and 'C' are not equal` `        ``if` `(N % 3 > 0 || A.Count != B.Count ||` `            ``A.Count != C.Count)` `        ``{` `            `  `            ``// Return "No"` `            ``return` `"No"``;` `        ``}` `        `  `        ``// Iterate over the deque B` `        ``foreach``(``var` `itr ``in` `B)` `        ``{` `            ``int` `b = itr;` `            `  `            ``// If A is not empty and` `            ``// curr B is greater than` `            ``// A[0]` `            ``if` `(A.Count > 0 && b > A.First.Value)` `            ``{` `                `  `                ``// Removes the front` `                ``// element of A` `                ``A.RemoveFirst();` `            ``}` `            `  `            ``// Else return "No"` `            ``else` `            ``{` `                ``return` `"No"``;` `            ``}` `        ``}` `    `  `        ``// Iterate over the deque` `        ``// B in reverse` `        ``foreach``(``var` `itr ``in` `B.Reverse()) ` `        ``{` `            ``int` `b = itr;` `            `  `            ``// If C is not empty and` `            ``// last element of C is` `            ``// greater thab B[i]` `            ``if` `(C.Count > 0 && b < C.Last.Value) ` `            ``{` `                `  `                ``// Removes the last` `                ``// element of C` `                ``C.RemoveLast();` `            ``}` `            `  `            ``// Else return "No"` `            ``else` `            ``{` `                ``return` `"No"``;` `            ``}` `        ``}` `        `  `        ``// If none of the above` `        ``// cases satisfy return` `        ``//"Yes'` `        ``return` `"Yes"``;` `    ``}` `    `  `    ``// Driver code` `    ``static` `public` `void` `Main (){` `        `  `        ``// Input` `        ``string` `S = ``"ABABCC"``;` `    `  `        ``// Function call` `        ``Console.Write(check(S));` `    ``}` `}`   `// This Code is contributed by ShubhamSingh10`

Output

`Yes`

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

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