检查一个单元格是否可以在字符串中多次访问

给定一个由 ‘.’ 组成的字符串或任何数字。一种 ‘。’在字符串表示单元格为空，如果有数字$x$在任何单元格中，这意味着可以移动$x$在字符串向右或向左步进。
任务是检查字符串的任何单元格是否可以多次访问。如果是，则打印 YES 否则打印 NO。
例子：

Input : str = ".2...2.."
Output: YES
The fourth cell can be visited twice. One way to reach 
the fourth cell is from 2nd cell by moving 2 steps to right
and another way to reach fourth cell is by moving 2 steps 
left from cell 6.

Input : str = ".2...1"
Output: NO
None of the cells in the given string 
can be visited more than once.

这个想法是采用一个数组visited[]来跟踪可以访问字符串的第i个单元格的次数。现在遍历字符串并检查当前字符是否为 ‘.’或一个数字$x$ .如果当前字符是 ‘.’如果它是一个数字，则不执行任何其他操作，然后将 [ix, i+x] 范围内的访问数组中的访问次数增加 1。
最后，遍历visited[] 数组并检查是否有任何单元格被多次访问。
下面是上述方法的实现：

C++

// C++ program to check if any cell of the
// string can be visited more than once
 
#include 
using namespace std;
 
// Function to check if any cell can be
// visited more than once
bool checkIfOverlap(string str)
{
    int len = str.length();
 
    // Array to mark cells
    int visited[len + 1] = { 0 };
 
    // Traverse the string
    for (int i = 0; i < len; i++) {
        if (str[i] == '.')
            continue;
 
        // Increase the visit count of the left and right
        // cells within the array which can be visited
        for (int j = max(0, i - str[i]); j <= min(len, i + str[i]); j++)
            visited[j]++;
    }
 
    for (int i = 0; i < len; i++) {
        // If any cell can be visited more than once
        // Return True
        if (visited[i] > 1) {
            return true;
        }
    }
 
    return false;
}
 
// Driver code
int main()
{
    string str = ".2..2.";
 
    if (checkIfOverlap(str))
        cout << "YES";
    else
        cout << "NO";
 
    return 0;
}

Java

// Java program to check if any cell of the
// string can be visited more than once
 
import java.io.*;
 
class GFG {
 
// Function to check if any cell can be
// visited more than once
static boolean checkIfOverlap(String str)
{
    int len = str.length();
 
    // Array to mark cells
    int []visited =  new int[len +1];
 
    // Traverse the string
    for (int i = 0; i < len; i++) {
        if (str.charAt(i)== '.')
            continue;
 
        // Increase the visit count of the left and right
        // cells within the array which can be visited
        for (int j = Math.max(0, i - str.charAt(i)); j <= Math.min(len, i + str.charAt(i)); j++)
            visited[j]++;
    }
 
    for (int i = 0; i < len; i++) {
        // If any cell can be visited more than once
        // Return True
        if (visited[i] > 1) {
            return true;
        }
    }
 
    return false;
}
 
// Driver code
 
    public static void main (String[] args) {
        String str = ".2..2.";
 
    if (checkIfOverlap(str))
        System.out.println("YES");
    else
        System.out.print("NO");
    }
}
 
// This code is contributed by inder_verma..

Python 3

# Python3 program to check if
# any cell of the string can
# be visited more than once
 
# Function to check if any cell
# can be visited more than once
def checkIfOverlap(str) :
 
    length = len(str)
 
    # Array to mark cells
    visited = [0] * (length + 1)
 
    # Traverse the string
    for i in range(length) :
        if str[i] == "." :
            continue
 
        # Increase the visit count
        # of the left and right cells
        # within the array which can
        # be visited
        for j in range(max(0, i - ord(str[i]),
                       min(length, i +
                       ord(str[i])) + 1)) :
            visited[j] += 1
 
    # If any cell can be visited
    # more than once, Return True
    for i in range(length) :
 
        if visited[i] > 1 :
            return True
 
    return False
 
# Driver code    
if __name__ == "__main__" :
 
    str = ".2..2."
 
    if checkIfOverlap(str) :
        print("YES")
                         
    else :
        print("NO")
                     
# This code is contributed
# by ANKITRAI1

C#

// C# program to check if any
// cell of the string can be
// visited more than once
using System;
 
class GFG
{
 
// Function to check if any cell
// can be visited more than once
static bool checkIfOverlap(String str)
{
    int len = str.Length;
 
    // Array to mark cells
    int[] visited = new int[len + 1];
 
    // Traverse the string
    for (int i = 0; i < len; i++)
    {
        if (str[i]== '.')
            continue;
 
        // Increase the visit count of
        // the left and right cells
        // within the array which can be visited
        for (int j = Math.Max(0, i - str[i]);
                 j <= Math.Min(len, i + str[i]); j++)
            visited[j]++;
    }
 
    for (int i = 0; i < len; i++)
    {
        // If any cell can be visited
        // more than once, Return True
        if (visited[i] > 1)
        {
            return true;
        }
    }
 
    return false;
}
 
// Driver code
public static void Main ()
{
    String str = ".2..2.";
 
    if (checkIfOverlap(str))
        Console.Write("YES");
    else
        Console.Write("NO");
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

PHP

 1)
        {
            return true;
        }
    }
 
    return false;
}
 
// Driver code
$str = ".2..2.";
 
if (checkIfOverlap($str))
    echo "YES";
else
    echo "NO";
 
// This code is contributed
// by ChitraNayal
?>

Javascript

 
// JavaScript program to check if any cell of the
// string can be visited more than once
     
    // Function to check if any cell can be
// visited more than once
    function checkIfOverlap(str)
    {
        let len = str.length;
   
    // Array to mark cells
    let visited =  new Array(len +1);
      for(let i=0;i<visited.length;i++)
    {
        visited[i]=0;
    }
     
    // Traverse the string
    for (let i = 0; i < len; i++) {
        if (str[i]== '.')
            continue;
   
        // Increase the visit count of the left and right
        // cells within the array which can be visited
        for (let j = Math.max(0, i - str[i]);
        j <= Math.min(len, i + str[i]); j++)
            visited[j]++;
    }
   
    for (let i = 0; i < len; i++) {
        // If any cell can be visited more than once
        // Return True
        if (visited[i] > 1) {
            return true;
        }
    }
   
    return false;
    }
     
    // Driver code
    let str = ".2..2.";
   
    if (checkIfOverlap(str))
        document.write("YES");
    else
        document.write("NO");
     
 
// This code is contributed by rag2127
 

输出：

YES