给定一个由N个二进制字符串组成的数组arr [] ，任务是计算不包含任何Arc Intersection的字符串的数量。
Connecting consecutive pairs of identical letters using Arcs, if their is an intersection obtained, then it is known as Arc Intersection. Below is the illustration of the same.
例子：
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.
天真的方法：最简单的方法是遍历数组并检查每个字符串，无论相似字符是否在连续索引处组合在一起。如果发现是真实的，不断递增，这样的字符串的计数。最后，打印获得的计数值。
时间复杂度： O(N * M 2 )，其中M是给定数组中字符串的最大长度。
辅助空间： O(1)
高效的方法：为了优化上述方法，我们的想法是使用Stack。请按照以下步骤解决问题：
- 初始化count ，以存储不包含任何弧形交点的字符串的计数。
- 初始化堆栈以将字符串的每个字符存储到其中。
- 迭代给定的字符串并执行以下操作：
- 将当前字符推入堆栈。
- 如果堆栈大小大于2 ，则检查堆栈顶部的两个元素是否相同。如果发现为真，则同时弹出两个字符以删除最近的弧形交点。
- 完成上述步骤后，如果堆栈为空，则不包含任何弧形交点。
- 对数组中的每个字符串执行步骤2到步骤4 ，以检查字符串是否包含弧形交点。如果不包含则计算这个字符串。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to check if there is arc
// intersection or not
int arcIntersection(string S, int len)
{
stack 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
// 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 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
# 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 29C#
// 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 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[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. 输出：
2时间复杂度： O(N * M)，其中M是给定数组中字符串的最大长度。
辅助空间： O(M)，其中M是给定数组中字符串的最大长度。