Introduction

In this blog post, we will explore how to solve the LeetCode problem "1340" 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 and an integer d. In one step you can jump from index i to index: i + x where: i + x < arr.length and 0 < x <= d. i - x where: i - x >= 0 and 0 < x <= d. In addition, you can only jump from index i to index j if arr[i] > arr[j] and arr[i] > arr[k] for all indices k between i and j (More formally min(i, j) < k < max(i, j)). You can choose any index of the array and start jumping. Return the maximum number of indices you can visit. Notice that you can not jump outside of the array at any time. Example 1: Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2 Output: 4 Explanation: You can start at index 10. You can jump 10 --> 8 --> 6 --> 7 as shown. Note that if you start at index 6 you can only jump to index 7. You cannot jump to index 5 because 13 > 9. You cannot jump to index 4 because index 5 is between index 4 and 6 and 13 > 9. Similarly You cannot jump from index 3 to index 2 or index 1. Example 2: Input: arr = [3,3,3,3,3], d = 3 Output: 1 Explanation: You can start at any index. You always cannot jump to any index. Example 3: Input: arr = [7,6,5,4,3,2,1], d = 1 Output: 7 Explanation: Start at index 0. You can visit all the indicies. Constraints: 1 <= arr.length <= 1000 1 <= arr[i] <= 105 1 <= d <= arr.length

Explanation

• Dynamic Programming with Memoization: Use dynamic programming to store the maximum number of indices reachable from each starting index. Memoization avoids redundant calculations by storing and reusing previously computed results.
  • Depth-First Search (DFS) for Jumps: Employ DFS to explore possible jumps from each index, considering the jump conditions (distance d and height restrictions).
  • Iterate and Maximize: Iterate through each index in the array as a potential starting point and find the maximum reachable indices using the DP-DFS approach.

• Runtime Complexity: O(n*d), where n is the length of the array and d is the maximum jump distance.
• Storage Complexity: O(n) due to the memoization array.

Code

    def maxJumps(arr, d):
    n = len(arr)
    dp = {}  # Memoization dictionary to store results

    def dfs(i):
        if i in dp:
            return dp[i]

        max_reachable = 1  # Initialize with the current index itself

        # Check jumps to the right
        for j in range(i + 1, min(i + d + 1, n)):
            valid_jump = True
            for k in range(i + 1, j):
                if arr[k] >= arr[i]:
                    valid_jump = False
                    break
            if valid_jump and arr[i] > arr[j]:
                max_reachable = max(max_reachable, 1 + dfs(j))

        # Check jumps to the left
        for j in range(i - 1, max(i - d - 1, -1), -1):
            valid_jump = True
            for k in range(i - 1, j, -1):
                if arr[k] >= arr[i]:
                    valid_jump = False
                    break
            if valid_jump and arr[i] > arr[j]:
                max_reachable = max(max_reachable, 1 + dfs(j))

        dp[i] = max_reachable
        return max_reachable

    max_indices = 0
    for i in range(n):
        max_indices = max(max_indices, dfs(i))

    return max_indices