Introduction

In this blog post, we will explore how to solve the LeetCode problem "1871" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a 0-indexed binary string s and two integers minJump and maxJump. In the beginning, you are standing at index 0, which is equal to '0'. You can move from index i to index j if the following conditions are fulfilled: i + minJump <= j <= min(i + maxJump, s.length - 1), and s[j] == '0'. Return true if you can reach index s.length - 1 in s, or false otherwise. Example 1: Input: s = "011010", minJump = 2, maxJump = 3 Output: true Explanation: In the first step, move from index 0 to index 3. In the second step, move from index 3 to index 5. Example 2: Input: s = "01101110", minJump = 2, maxJump = 3 Output: false Constraints: 2 <= s.length <= 105 s[i] is either '0' or '1'. s[0] == '0' 1 <= minJump <= maxJump < s.length

Explanation

Here's a breakdown of the solution:

• Breadth-First Search (BFS): We use BFS to explore possible jump destinations. Starting from index 0, we iteratively check all reachable indices within the minJump and maxJump range.
• Visited Tracking: To avoid cycles and redundant computations, we keep track of visited indices.

• Early Exit: If we reach the last index (s.length - 1), we immediately return True.

• Runtime Complexity: O(n), where n is the length of the string s. Storage Complexity: O(n).

Code

    from collections import deque

def canReach(s: str, minJump: int, maxJump: int) -> bool:
    """
    Determines if the last index of a binary string can be reached from the first index
    given jump constraints.

    Args:
        s: The binary string.
        minJump: The minimum jump distance.
        maxJump: The maximum jump distance.

    Returns:
        True if the last index can be reached, False otherwise.
    """

    n = len(s)
    if s[n - 1] == '1':
        return False

    q = deque([0])
    visited = {0}
    reachable = [False] * n
    reachable[0] = True

    while q:
        i = q.popleft()

        if i == n - 1:
            return True

        for j in range(i + minJump, min(i + maxJump + 1, n)):
            if s[j] == '0' and j not in visited:
                q.append(j)
                visited.add(j)
                reachable[j] = True

    return False