Introduction

In this blog post, we will explore how to solve the LeetCode problem "1345" 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

Given an array of integers arr, you are initially positioned at the first index of the array. In one step you can jump from index i to index: i + 1 where: i + 1 < arr.length. i - 1 where: i - 1 >= 0. j where: arr[i] == arr[j] and i != j. Return the minimum number of steps to reach the last index of the array. Notice that you can not jump outside of the array at any time. Example 1: Input: arr = [100,-23,-23,404,100,23,23,23,3,404] Output: 3 Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array. Example 2: Input: arr = [7] Output: 0 Explanation: Start index is the last index. You do not need to jump. Example 3: Input: arr = [7,6,9,6,9,6,9,7] Output: 1 Explanation: You can jump directly from index 0 to index 7 which is last index of the array. Constraints: 1 <= arr.length <= 5 * 104 -108 <= arr[i] <= 108

Explanation

Here's the breakdown of the solution:

• BFS Approach: Use Breadth-First Search (BFS) to explore possible jump paths from the start index to the end index. BFS guarantees finding the shortest path.
• Optimized Jump Logic: Pre-compute a dictionary mapping each number in the array to the indices where it appears. This allows us to efficiently find all possible jumps to indices with the same value. Crucially, clear the entries in the dictionary as we process each number to avoid revisiting already explored nodes and prevent cycles.

• Visited Set: Maintain a visited set to keep track of visited indices, preventing cycles and redundant exploration.

• Runtime Complexity: O(N), where N is the length of the input array arr.

• Storage Complexity: O(N), primarily due to the indices dictionary, visited set, and queue used in BFS.

Code

    from collections import deque

def minJumps(arr):
    n = len(arr)
    if n == 1:
        return 0

    indices = {}
    for i, num in enumerate(arr):
        if num not in indices:
            indices[num] = []
        indices[num].append(i)

    visited = {0}
    queue = deque([(0, 0)])  # (index, steps)

    while queue:
        index, steps = queue.popleft()

        if index == n - 1:
            return steps

        # Jump to same value indices
        value = arr[index]
        if value in indices:
            for j in indices[value]:
                if j not in visited:
                    visited.add(j)
                    queue.append((j, steps + 1))
            indices.pop(value)  # Clear to prevent revisiting

        # Jump to adjacent indices
        if index + 1 < n and index + 1 not in visited:
            visited.add(index + 1)
            queue.append((index + 1, steps + 1))
        if index - 1 >= 0 and index - 1 not in visited:
            visited.add(index - 1)
            queue.append((index - 1, steps + 1))

    return -1  # Should never happen if a path exists