Introduction

In this blog post, we will explore how to solve the LeetCode problem "1654" 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

A certain bug's home is on the x-axis at position x. Help them get there from position 0. The bug jumps according to the following rules: It can jump exactly a positions forward (to the right). It can jump exactly b positions backward (to the left). It cannot jump backward twice in a row. It cannot jump to any forbidden positions. The bug may jump forward beyond its home, but it cannot jump to positions numbered with negative integers. Given an array of integers forbidden, where forbidden[i] means that the bug cannot jump to the position forbidden[i], and integers a, b, and x, return the minimum number of jumps needed for the bug to reach its home. If there is no possible sequence of jumps that lands the bug on position x, return -1. Example 1: Input: forbidden = [14,4,18,1,15], a = 3, b = 15, x = 9 Output: 3 Explanation: 3 jumps forward (0 -> 3 -> 6 -> 9) will get the bug home. Example 2: Input: forbidden = [8,3,16,6,12,20], a = 15, b = 13, x = 11 Output: -1 Example 3: Input: forbidden = [1,6,2,14,5,17,4], a = 16, b = 9, x = 7 Output: 2 Explanation: One jump forward (0 -> 16) then one jump backward (16 -> 7) will get the bug home. Constraints: 1 <= forbidden.length <= 1000 1 <= a, b, forbidden[i] <= 2000 0 <= x <= 2000 All the elements in forbidden are distinct. Position x is not forbidden.

Explanation

• BFS: Use Breadth-First Search to explore possible jump sequences. Each state in the BFS includes the current position and a flag indicating whether the last jump was backward.
  • Avoid Cycles and Forbidden Positions: Keep track of visited states to avoid cycles and check if a position is forbidden before jumping.
  • Limit Search Space: The bug might jump beyond x, but we need to limit how far it can jump to avoid infinite loops. A reasonable upper bound can be calculated based on a, b, and x.

• Runtime Complexity: O(a + b + x), Space Complexity: O(a + b + x)

Code

    from collections import deque

def minimumJumps(forbidden, a, b, x):
    forbidden_set = set(forbidden)
    queue = deque([(0, 0, False)])  # (position, steps, last_jump_backward)
    visited = set()
    visited.add((0, False))
    max_reach = max(x + a, max(forbidden) + a + b)

    while queue:
        position, steps, last_jump_backward = queue.popleft()

        if position == x:
            return steps

        # Jump forward
        next_pos = position + a
        if 0 <= next_pos <= max_reach and next_pos not in forbidden_set and (next_pos, False) not in visited:
            queue.append((next_pos, steps + 1, False))
            visited.add((next_pos, False))

        # Jump backward
        next_pos = position - b
        if not last_jump_backward and 0 <= next_pos <= max_reach and next_pos not in forbidden_set and (next_pos, True) not in visited:
            queue.append((next_pos, steps + 1, True))
            visited.add((next_pos, True))

    return -1