Given a string S consisting of characters '(', ')', '[', ']', '{', '}', the task is to remove all balanced bracket subsequences from the string and print the remaining characters.
Examples:
Input: S = "((){()({})"
Output: "({"
Explanation:
- S[1] and S[2] forms a regular bracket sequence. Therefore, remove them from the string. S = "({()({})"
- S[2] and S[3] are regular bracket sequence. Therefore, remove them from the string. S = "({({})"
- String S[2...4] is regular bracket sequence. Therefore, remove them from the string. S = "({"
Remaining string does not contain any regular bracket sequence. Therefore, print all remaining characters.
Input: S = "{[}])("
Output: ")("
Approach: The idea is to use Stack data structure to solve this problem. Follow the steps below to solve the problem:
- Initialize three stacks, say A, B and C, for storing each type of parenthesis.
- Initialize a boolean array, say vis[], to mark the already visited characters.
- Store indexes of char '(' in stack A. Similarly, stacks B and C stores the positions of '{' and '[' in the string.
- Traverse through the characters of string str and perform the following:
- If the current character is ')':
- If the stack A is not empty, mark the current character in the string and vis[A.top()] as false.
- Pop the top element of the stack A.
- If the current character is '}':
- If the stack B is not empty, mark the current character in the string and vis[B.top()] as false.
- Pop-out the top element of the stack B.
- If the current character is ']':
- If the stack C is not empty, mark the current character in the string and vis[C.top()] as false.
- Pop-out the top element of stack C.
- If the current character is ')':
- After completing all the operations, print the characters of the string whose index is marked true in the vis[] array.
Below is the implementation of the above approach:
// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to remove all possible valid
// bracket subsequences
void removeValidBracketSequences(string& str,
int N)
{
// Stores indexes of '(' in
// valid subsequences
stack<int> A;
// Stores indexes of '{' in
// valid subsequences
stack<int> B;
// Stores indexes of '[' in
// valid subsequences
stack<int> C;
// vis[i]: Check if character at
// i-th index is removed or not
bool vis[N];
// Mark vis[i] as not removed
memset(vis, true, sizeof(vis));
// Iterate over the characters of string
for (int i = 0; i < N; i++) {
// If current character is '('
if (str[i] == '(') {
A.push(i);
}
// If current character is '{'
else if (str[i] == '{') {
B.push(i);
}
// If current character is '['
else if (str[i] == '[') {
C.push(i);
}
// If current character is ')' and
// top element of A is '('
else if (str[i] == ')' && !A.empty()) {
// Mark the top element
// of A as removed
vis[A.top()] = false;
A.pop();
// Mark current character
// as removed
vis[i] = false;
}
// If current character is '}' and
// top element of B is '{'
else if (str[i] == '}' && !B.empty()) {
// Mark the top element
// of B as removed
vis[B.top()] = false;
B.pop();
// Mark current character
// as removed
vis[i] = false;
}
// If current character is ']' and
// top element of B is '['
else if (str[i] == ']' && !C.empty()) {
// Mark the top element
// of C as removed
vis[C.top()] = false;
C.pop();
// Mark current character
// as removed
vis[i] = false;
}
}
// Print the remaining characters
// which is not removed from S
for (int i = 0; i < N; ++i) {
if (vis[i])
cout << str[i];
}
}
// Driver Code
int main()
{
// Given string
string str = "((){()({})";
// Size of the string
int N = str.length();
// Function Call
removeValidBracketSequences(str, N);
return 0;
}
// Java program of the above approach
import java.util.*;
public class GFG
{
// Function to remove all possible valid
// bracket subsequences
static void removeValidBracketSequences(String str, int N)
{
// Stores indexes of '(' in
// valid subsequences
Vector<Integer> A = new Vector<Integer>();
// Stores indexes of '{' in
// valid subsequences
Vector<Integer> B = new Vector<Integer>();
// Stores indexes of '[' in
// valid subsequences
Vector<Integer> C = new Vector<Integer>();
// vis[i]: Check if character at
// i-th index is removed or not
boolean[] vis = new boolean[N];
// Mark vis[i] as not removed
for(int i = 0; i < N; i++)
{
vis[i] = true;
}
// Iterate over the characters of string
for (int i = 0; i < N; i++) {
// If current character is '('
if (str.charAt(i) == '(') {
A.add(i);
}
// If current character is '{'
else if (str.charAt(i) == '{') {
B.add(i);
}
// If current character is '['
else if (str.charAt(i) == '[') {
C.add(i);
}
// If current character is ')' and
// top element of A is '('
else if (str.charAt(i) == ')' && (A.size() > 0)) {
// Mark the top element
// of A as removed
vis[A.get(A.size() - 1)] = false;
A.remove(A.size() - 1);
// Mark current character
// as removed
vis[i] = false;
}
// If current character is '}' and
// top element of B is '{'
else if (str.charAt(i) == '}' && (B.size() > 0)) {
// Mark the top element
// of B as removed
vis[B.get(B.size() - 1)] = false;
B.remove(B.size() - 1);
// Mark current character
// as removed
vis[i] = false;
}
// If current character is ']' and
// top element of B is '['
else if (str.charAt(i) == ']' && (C.size() > 0)) {
// Mark the top element
// of C as removed
vis[C.get(C.size() - 1)] = false;
C.remove(C.size() - 1);
// Mark current character
// as removed
vis[i] = false;
}
}
// Print the remaining characters
// which is not removed from S
for (int i = 0; i < N; ++i)
{
if (vis[i])
System.out.print(str.charAt(i));
}
}
// Driver code
public static void main(String[] args)
{
// Given string
String str = "((){()({})";
// Size of the string
int N = str.length();
// Function Call
removeValidBracketSequences(str, N);
}
}
// This code is contributed by divyeshrabadiya07.
# Python3 program of the above approach
# Function to remove all possible valid
# bracket subsequences
def removeValidBracketSequences(str, N):
# Stores indexes of '(' in
# valid subsequences
A = []
# Stores indexes of '{' in
# valid subsequences
B = []
# Stores indexes of '[' in
# valid subsequences
C = []
# vis[i]: Check if character at
# i-th index is removed or not
vis = [True for i in range(N)]
# Iterate over the characters of string
for i in range(N):
# If current character is '('
if (str[i] == '('):
A.append(i)
# If current character is '{'
elif (str[i] == '{'):
B.append(i)
# If current character is '['
elif (str[i] == '['):
C.append(i)
# If current character is ')' and
# top element of A is '('
elif(str[i] == ')' and len(A) != 0):
# Mark the top element
# of A as removed
vis[A[-1]] = False
A.pop()
# Mark current character
# as removed
vis[i] = False
# If current character is '}' and
# top element of B is '{'
elif (str[i] == '}' and len(B) != 0):
# Mark the top element
# of B as removed
vis[B[-1]] = False
B.pop()
# Mark current character
# as removed
vis[i] = False
# If current character is ']' and
# top element of B is '['
elif(str[i] == ']' and len(C) != 0):
# Mark the top element
# of C as removed
vis[C[-1]] = False
C.pop()
# Mark current character
# as removed
vis[i] = False
# Print the remaining characters
# which is not removed from S
for i in range(N):
if (vis[i]):
print(str[i], end = '')
# Driver Code
if __name__=='__main__':
# Given string
str = "((){()({})"
# Size of the string
N = len(str)
# Function Call
removeValidBracketSequences(str, N)
# This code is contributed by rutvik_56
// C# program of the above approach
using System;
using System.Collections.Generic;
class GFG {
// Function to remove all possible valid
// bracket subsequences
static void removeValidBracketSequences(string str, int N)
{
// Stores indexes of '(' in
// valid subsequences
List<int> A = new List<int>();
// Stores indexes of '{' in
// valid subsequences
List<int> B = new List<int>();
// Stores indexes of '[' in
// valid subsequences
List<int> C = new List<int>();
// vis[i]: Check if character at
// i-th index is removed or not
bool[] vis = new bool[N];
// Mark vis[i] as not removed
for(int i = 0; i < N; i++)
{
vis[i] = true;
}
// Iterate over the characters of string
for (int i = 0; i < N; i++) {
// If current character is '('
if (str[i] == '(') {
A.Add(i);
}
// If current character is '{'
else if (str[i] == '{') {
B.Add(i);
}
// If current character is '['
else if (str[i] == '[') {
C.Add(i);
}
// If current character is ')' and
// top element of A is '('
else if (str[i] == ')' && (A.Count > 0)) {
// Mark the top element
// of A as removed
vis[A[A.Count - 1]] = false;
A.RemoveAt(A.Count - 1);
// Mark current character
// as removed
vis[i] = false;
}
// If current character is '}' and
// top element of B is '{'
else if (str[i] == '}' && (B.Count > 0)) {
// Mark the top element
// of B as removed
vis[B[B.Count - 1]] = false;
B.RemoveAt(B.Count - 1);
// Mark current character
// as removed
vis[i] = false;
}
// If current character is ']' and
// top element of B is '['
else if (str[i] == ']' && (C.Count > 0)) {
// Mark the top element
// of C as removed
vis[C[C.Count - 1]] = false;
C.RemoveAt(C.Count - 1);
// Mark current character
// as removed
vis[i] = false;
}
}
// Print the remaining characters
// which is not removed from S
for (int i = 0; i < N; ++i) {
if (vis[i])
Console.Write(str[i]);
}
}
// Driver code
static void Main()
{
// Given string
string str = "((){()({})";
// Size of the string
int N = str.Length;
// Function Call
removeValidBracketSequences(str, N);
}
}
// This code is contributed by divyesh072019.
<script>
// Javascript program of the above approach
// Function to remove all possible valid
// bracket subsequences
function removeValidBracketSequences(str, N)
{
// Stores indexes of '(' in
// valid subsequences
let A = [];
// Stores indexes of '{' in
// valid subsequences
let B = [];
// Stores indexes of '[' in
// valid subsequences
let C = [];
// vis[i]: Check if character at
// i-th index is removed or not
let vis = new Array(N);
// Mark vis[i] as not removed
for(let i = 0; i < N; i++)
{
vis[i] = true;
}
// Iterate over the characters of string
for (let i = 0; i < N; i++) {
// If current character is '('
if (str[i] == '(') {
A.push(i);
}
// If current character is '{'
else if (str[i] == '{') {
B.push(i);
}
// If current character is '['
else if (str[i] == '[') {
C.push(i);
}
// If current character is ')' and
// top element of A is '('
else if (str[i] == ')' && (A.length > 0)) {
// Mark the top element
// of A as removed
vis[A[A.length - 1]] = false;
A.pop();
// Mark current character
// as removed
vis[i] = false;
}
// If current character is '}' and
// top element of B is '{'
else if (str[i] == '}' && (B.length > 0)) {
// Mark the top element
// of B as removed
vis[B[B.length - 1]] = false;
B.pop();
// Mark current character
// as removed
vis[i] = false;
}
// If current character is ']' and
// top element of B is '['
else if (str[i] == ']' && (C.length > 0)) {
// Mark the top element
// of C as removed
vis[C[C.length - 1]] = false;
C.pop();
// Mark current character
// as removed
vis[i] = false;
}
}
// Print the remaining characters
// which is not removed from S
for (let i = 0; i < N; ++i) {
if (vis[i])
document.write(str[i]);
}
}
// Given string
let str = "((){()({})";
// Size of the string
let N = str.length;
// Function Call
removeValidBracketSequences(str, N);
// This code is contributed by suresh07
</script>
Output:
({Time Complexity: O(N)
Auxiliary Space: O(N)
