Check if end of given Binary string can be reached by choosing jump value in between given range
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.
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
Input: S = “011010”, L = 2, R = 3
Following are the series of moves having characters at that indices as 0:
S(= 0) -> S(= 0) -> S(= 0) and S is the end of the string S.
Therefore, print Yes.
Input: S = “01101110”, L = 2, R = 3
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 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:
Time Complexity: O(N)
Auxiliary Space: O(N)