How do you solve Minimum Jumps to Reach Home on LeetCode?

Minimum Jumps to Reach Home (LeetCode #1654) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The bug can jump forward and backward, but cannot jump backward twice in a row, which adds complexity to the problem.

What is the time complexity of Minimum Jumps to Reach Home?

The optimal solution for Minimum Jumps to Reach Home runs in O(n) time and uses O(n) space. The complexity is linear because we explore each position at most once, and the queue operations are efficient.

What is the brute force solution for Minimum Jumps to Reach Home?

The brute force approach for Minimum Jumps to Reach Home is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach explores all possible jumps the bug can make, tracking its position and the number of jumps taken. It uses a recursive method to simulate each jump until it either reaches the target position or exhausts all options. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Jumps to Reach Home?

Minimum Jumps to Reach Home uses the following concepts: Array, Hash Table, Breadth-First Search. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Jumps to Reach Home?

Clearly explain your thought process and the rules governing the bug's jumps. Use a diagram or visual representation if possible to illustrate your approach. Practice BFS problems to get comfortable with queue operations and state management.

What are common mistakes when solving Minimum Jumps to Reach Home?

Not properly handling the condition of not jumping backward twice in a row. Failing to account for forbidden positions when exploring potential jumps.

#1654

Minimum Jumps to Reach Home

Medium

ArrayHash TableBreadth-First SearchHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a breadth-first search (BFS) strategy to explore all possible positions efficiently. By tracking the last jump direction, we ensure that the bug never jumps backward twice in a row, while also avoiding forbidden positions.

⚙️

Algorithm

3 steps

1Step 1: Initialize a queue for BFS starting from position 0 with 0 jumps.
2Step 2: Use a set to track visited positions to avoid cycles.
3Step 3: For each position, enqueue possible forward and backward jumps while respecting the jump rules.

solution.py18 lines

1# Full working Python code
2from collections import deque
3
4def minJumps(forbidden, a, b, x):
5    forbidden_set = set(forbidden)
6    queue = deque([(0, 0, 0)])  # (position, jumps, last_jump_direction)
7    visited = set()
8    while queue:
9        pos, jumps, last_jump = queue.popleft()
10        if pos == x:
11            return jumps
12        if pos < 0 or pos in forbidden_set or pos in visited:
13            continue
14        visited.add(pos)
15        queue.append((pos + a, jumps + 1, 1))  # Jump forward
16        if last_jump != -1:  # Jump backward only if last jump was not backward
17            queue.append((pos - b, jumps + 1, -1))  # Jump backward
18    return -1

ℹ

Complexity note: The complexity is linear because we explore each position at most once, and the queue operations are efficient.

1The bug can jump forward and backward, but cannot jump backward twice in a row, which adds complexity to the problem.
2Using a BFS approach ensures that we explore the shortest path to the target position efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.