Given a string which consists of either ‘.’ or any number. A ‘.’ in the string means that the cell is empty and if there is a number 
in any cell, it means one can move steps to right or left within the string.
The task is to check if any cell in the string can be visited more than once. If so, print YES otherwise print NO.
Examples:
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.
The idea is to take an array visited[] to keep track of the number of times i-th cell of the string can be visited. Now traverse the string and check if the current character is a ‘.’ or a number . If the current character is a ‘.’ then do nothing otherwise if it is a number then increase the count of visits in the visited array within the range [i-x, i+x] by 1.
Finally, traverse the visited[] array and check if any cell is visited more than once.
Below is the implementation of the above approach:
C++
// C++ program to check if any cell of the // string can be visited more than once #include <bits/stdc++.h> 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; } |
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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.. |
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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 |
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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) |
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PHP
<?php // PHP 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) { $len = strlen($str); // Array to mark cells $visited = array_fill(0, $len + 1, NULL); // Traverse the string for ($i = 0; $i < $len; $i++) { if ($str[$i] == '.') continue; // Increase the visit count of the // left and right cells within the // array which can be visited for ($j = max(0, $i - $str[$i]); $j <= min($len, $i + $str[$i]); $j++) $visited[$j]++; } for ($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 $str = ".2..2."; if (checkIfOverlap($str)) echo "YES"; else echo "NO"; // This code is contributed // by ChitraNayal ?> |
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Output:
YES
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