Given an array arr[] consisting of N binary strings, the task is to count the number of strings that does not contain any Arc Intersection.
Connecting consecutive pairs of identical letters using Arcs, if there is an intersection obtained, then it is known as Arc Intersection. Below is the illustration of the same.
Examples:
Input: arr[] = {“0101”, “0011”, “0110”}
Output: 2
Explanation: The string “0101” have Arc Intersection. Therefore, the count of strings that doesn’t have any Arc Intersection is 2.Input: arr[] = {“0011”, “0110”, “00011000”}
Output: 3
Explanation: All the given strings doesn’t have any Arc Intersection. Therefore, the count is 3.
Naive Approach: The simplest approach is to traverse the array and check for each string, if similar characters are grouped together at consecutive indices or not. If found to be true, keep incrementing count of such strings. Finally, print the value of count obtained.
Time Complexity: O(N*M2), where M is the maximum length of string in the given array.
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use Stack. Follow the steps below to solve the problem:
- Initialize count, to store the count of strings that doesn’t contain any arc intersection.
- Initialize a stack to store every character of the string into it.
-
Iterate the given string and perform the following operations:
- Push the current character into the stack.
- If the stack size is greater than 2, then check if the two elements at the top of the stack are same or not. If found to be true, then pop both the characters as to remove the nearest arc intersection.
- After completing the above steps, if stack is empty, then it doesn’t contain any arc intersection.
- Follow the Step 2 to Step 4 for each string in the array to check whether string contains arc intersection or not. If it doesn’t contain then count this string.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to check if there is arc // intersection or not int arcIntersection(string S, int len)
{ stack< char > stk;
// Traverse the string S
for ( int i = 0; i < len; i++) {
// Insert all the elements in
// the stack one by one
stk.push(S[i]);
if (stk.size() >= 2) {
// Extract the top element
char temp = stk.top();
// Pop out the top element
stk.pop();
// Check if the top element
// is same as the popped element
if (stk.top() == temp) {
stk.pop();
}
// Otherwise
else {
stk.push(temp);
}
}
}
// If the stack is empty
if (stk.empty())
return 1;
return 0;
} // Function to check if there is arc // intersection or not for the given // array of strings void countString(string arr[], int N)
{ // Stores count of string not
// having arc intersection
int count = 0;
// Iterate through array
for ( int i = 0; i < N; i++) {
// Length of every string
int len = arr[i].length();
// Function Call
count += arcIntersection(
arr[i], len);
}
// Print the desired count
cout << count << endl;
} // Driver Code int main()
{ string arr[] = { "0101" , "0011" , "0110" };
int N = sizeof (arr) / sizeof (arr[0]);
// Function Call
countString(arr, N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG
{ // Function to check if there is arc // intersection or not static int arcIntersection(String S, int len)
{ Stack<Character> stk = new Stack<>();
// Traverse the String S
for ( int i = 0 ; i < len; i++)
{
// Insert all the elements in
// the stack one by one
stk.push(S.charAt(i));
if (stk.size() >= 2 )
{
// Extract the top element
char temp = stk.peek();
// Pop out the top element
stk.pop();
// Check if the top element
// is same as the popped element
if (stk.peek() == temp)
{
stk.pop();
}
// Otherwise
else
{
stk.add(temp);
}
}
}
// If the stack is empty
if (stk.isEmpty())
return 1 ;
return 0 ;
} // Function to check if there is arc // intersection or not for the given // array of Strings static void countString(String arr[], int N)
{ // Stores count of String not
// having arc intersection
int count = 0 ;
// Iterate through array
for ( int i = 0 ; i < N; i++)
{
// Length of every String
int len = arr[i].length();
// Function Call
count += arcIntersection(
arr[i], len);
}
// Print the desired count
System.out.print(count + "\n" );
} // Driver Code public static void main(String[] args)
{ String arr[] = { "0101" , "0011" , "0110" };
int N = arr.length;
// Function Call
countString(arr, N);
} } // This code is contributed by 29AjayKumar |
# Python3 program for the above approach # Function to check if there is arc # intersection or not def arcIntersection(S, lenn):
stk = []
# Traverse the string S
for i in range (lenn):
# Insert all the elements in
# the stack one by one
stk.append(S[i])
if ( len (stk) > = 2 ):
# Extract the top element
temp = stk[ - 1 ]
# Pop out the top element
del stk[ - 1 ]
# Check if the top element
# is same as the popped element
if (stk[ - 1 ] = = temp):
del stk[ - 1 ]
# Otherwise
else :
stk.append(temp)
# If the stack is empty
if ( len (stk) = = 0 ):
return 1
return 0
# Function to check if there is arc # intersection or not for the given # array of strings def countString(arr, N):
# Stores count of string not
# having arc intersection
count = 0
# Iterate through array
for i in range (N):
# Length of every string
lenn = len (arr[i])
# Function Call
count + = arcIntersection(arr[i], lenn)
# Print the desired count
print (count)
# Driver Code if __name__ = = '__main__' :
arr = [ "0101" , "0011" , "0110" ]
N = len (arr)
# Function Call
countString(arr, N)
# This code is contributed by mohit kumar 29 |
// C# program for // the above approach using System;
using System.Collections.Generic;
class GFG
{ // Function to check if there is arc // intersection or not static int arcIntersection(String S, int len)
{ Stack< char > stk = new Stack< char >();
// Traverse the String S
for ( int i = 0; i < len; i++)
{
// Insert all the elements in
// the stack one by one
stk.Push(S[i]);
if (stk.Count >= 2)
{
// Extract the top element
char temp = stk.Peek();
// Pop out the top element
stk.Pop();
// Check if the top element
// is same as the popped element
if (stk.Peek() == temp)
{
stk.Pop();
}
// Otherwise
else
{
stk.Push(temp);
}
}
}
// If the stack is empty
if (stk.Count == 0)
return 1;
return 0;
} // Function to check if there is arc // intersection or not for the given // array of Strings static void countString(String []arr, int N)
{ // Stores count of String not
// having arc intersection
int count = 0;
// Iterate through array
for ( int i = 0; i < N; i++)
{
// Length of every String
int len = arr[i].Length;
// Function Call
count += arcIntersection(
arr[i], len);
}
// Print the desired count
Console.Write(count + "\n" );
} // Driver Code public static void Main(String[] args)
{ String [] arr = { "0101" , "0011" , "0110" };
int N = arr.Length;
// Function Call
countString(arr, N);
} } // This code is contributed by jana_sayantan. |
<script> // JavaScript program for the above approach // Function to check if there is arc // intersection or not function arcIntersection(S, len)
{ var stk = [];
// Traverse the string S
for ( var i = 0; i < len; i++) {
// Insert all the elements in
// the stack one by one
stk.push(S[i]);
if (stk.length >= 2) {
// Extract the top element
var temp = stk[stk.length-1];
// Pop out the top element
stk.pop();
// Check if the top element
// is same as the popped element
if (stk[stk.length-1] == temp) {
stk.pop();
}
// Otherwise
else {
stk.push(temp);
}
}
}
// If the stack is empty
if (stk.length==0)
return 1;
return 0;
} // Function to check if there is arc // intersection or not for the given // array of strings function countString(arr, N)
{ // Stores count of string not
// having arc intersection
var count = 0;
// Iterate through array
for ( var i = 0; i < N; i++) {
// Length of every string
var len = arr[i].length;
// Function Call
count += arcIntersection(
arr[i], len);
}
// Print the desired count
document.write( count + "<br>" );
} // Driver Code var arr = [ "0101" , "0011" , "0110" ];
var N = arr.length;
// Function Call countString(arr, N); </script> |
2
Time Complexity: O(N*M), where M is the maximum length of string in the given array.
Auxiliary Space: O(M), where M is the maximum length of string in the given array.