Given two positive integers L and R and a binary string S of size N, the task is to check if the end of the string is reached from the index 0 by a series of jumps of indices, say i such that S[i] is 0 jumps are allowed in the range [L, R]. If it is possible to reach, then print Yes. Otherwise, print No.
Examples:
Input: S = “011010”, L = 2, R = 3
Output: Yes
Explanation:
Following are the series of moves having characters at that indices as 0:
S[0](= 0) -> S[3](= 0) -> S[5](= 0) and S[5] is the end of the string S.
Therefore, print Yes.Input: S = “01101110”, L = 2, R = 3
Output: No
Approach: The above problem can be solved with the help of Dynamic Programming, the idea is to maintain 1D array, say dp[] where dp[i] will store the possibility of reaching the ith position and update each index accordingly. Below are the steps:
- Initialize an array, dp[] such that dp[i] stores whether any index i can be reached from index 0 or not. Update the value of dp[0] as 1 as it is the current standing index.
- Initialize a variable, pre as 0 that stores the number of indices from which the current index is reachable.
- Iterate over the range [1, N) and update the value of pre variable as follows:
- If the value of (i >= minJump), then increment the value of pre by dp[i – minJump].
- If the value of (i > maxJump), then decrement the value of pre by dp[i – maxJump – 1].
- If the value of pre is positive, then there is at least 1 index from which the current index is reachable. Therefore, update the value of dp[i] = 1, if the value of S[i] = 0.
- After completing the above steps, if the value of dp[N – 1] is positive, then print Yes. Otherwise, print No.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if it is possible // to reach the end of the binary string // using the given jumps bool canReach(string s, int L, int R) { // Stores the DP states vector< int > dp(s.length()); // Initial state dp[0] = 1; // Stores count of indices from which // it is possible to reach index i int pre = 0; // Traverse the given string for ( int i = 1; i < s.length(); i++) { // Update the values of pre // accordingly if (i >= L) { pre += dp[i - L]; } // If the jump size is out of // the range [L, R] if (i > R) { pre -= dp[i - R - 1]; } dp[i] = (pre > 0) and (s[i] == '0' ); } // Return answer return dp[s.length() - 1]; } // Driver Code int main() { string S = "01101110" ; int L = 2, R = 3; cout << (canReach(S, L, R) ? "Yes" : "No" ); return 0; } |
Java
// Java program for the above approach public class GFG { // Function to check if it is possible // to reach the end of the binary string // using the given jumps static int canReach(String s, int L, int R) { // Stores the DP states int dp[] = new int [s.length()]; // Initial state dp[ 0 ] = 1 ; // Stores count of indices from which // it is possible to reach index i int pre = 0 ; // Traverse the given string for ( int i = 1 ; i < s.length(); i++) { // Update the values of pre // accordingly if (i >= L) { pre += dp[i - L]; } // If the jump size is out of // the range [L, R] if (i > R) { pre -= dp[i - R - 1 ]; } if (pre > 0 && s.charAt(i) == '0' ) dp[i] = 1 ; else dp[i] = 0 ; } // Return answer return dp[s.length() - 1 ]; } // Driver Code public static void main (String[] args) { String S = "01101110" ; int L = 2 , R = 3 ; if (canReach(S, L, R) == 1 ) System.out.println( "Yes" ); else System.out.println( "No" ); } } // This code is contributed by AnkThon |
Python3
# python program for the above approach # Function to check if it is possible # to reach the end of the binary string # using the given jumps def canReach(s, L, R): # Stores the DP states dp = [ 0 for _ in range ( len (s))] # Initial state dp[ 0 ] = 1 # Stores count of indices from which # it is possible to reach index i pre = 0 # Traverse the given string for i in range ( 1 , len (s)): # Update the values of pre # accordingly if (i > = L): pre + = dp[i - L] # If the jump size is out of # the range [L, R] if (i > R): pre - = dp[i - R - 1 ] dp[i] = (pre > 0 ) and (s[i] = = '0' ) # Return answer return dp[ len (s) - 1 ] # Driver Code if __name__ = = "__main__" : S = "01101110" L = 2 R = 3 if canReach(S, L, R): print ( "Yes" ) else : print ( "No" ) # This code is contributed by rakeshsahni |
C#
// C# program for the above approach using System; public class GFG { // Function to check if it is possible // to reach the end of the binary string // using the given jumps static int canReach(String s, int L, int R) { // Stores the DP states int [] dp = new int [s.Length]; // Initial state dp[0] = 1; // Stores count of indices from which // it is possible to reach index i int pre = 0; // Traverse the given string for ( int i = 1; i < s.Length; i++) { // Update the values of pre // accordingly if (i >= L) { pre += dp[i - L]; } // If the jump size is out of // the range [L, R] if (i > R) { pre -= dp[i - R - 1]; } if (pre > 0 && s[i] == '0' ) dp[i] = 1; else dp[i] = 0; } // Return answer return dp[s.Length - 1]; } // Driver Code public static void Main() { String S = "01101110" ; int L = 2, R = 3; if (canReach(S, L, R) == 1) Console.WriteLine( "Yes" ); else Console.WriteLine( "No" ); } } // This code is contributed by Saurabh |
Javascript
<script> // JavaScript program for the above approach // Function to check if it is possible // to reach the end of the binary string // using the given jumps const canReach = (s, L, R) => { // Stores the DP states let dp = new Array(s.length).fill(1); // Stores count of indices from which // it is possible to reach index i let pre = 0; // Traverse the given string for (let i = 1; i < s.length; i++) { // Update the values of pre // accordingly if (i >= L) { pre += dp[i - L]; } // If the jump size is out of // the range [L, R] if (i > R) { pre -= dp[i - R - 1]; } dp[i] = (pre > 0) && (s[i] == '0' ); } // Return answer return dp[s.length - 1]; } // Driver Code let S = "01101110" ; let L = 2, R = 3; if (canReach(S, L, R)) document.write( "Yes" ); else document.write( "No" ); // This code is contributed by rakeshsahni </script> |
No
Time Complexity: O(N)
Auxiliary Space: O(N)