LeetCode Problem Workspace

Jump Game V

Jump Game V is a hard dynamic programming problem that focuses on maximizing jumps between indices in an array.

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Practice Focus

Hard · State transition dynamic programming

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Answer-first summary

Jump Game V is a hard dynamic programming problem that focuses on maximizing jumps between indices in an array.

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Jump Game V is a problem that asks you to determine the maximum number of indices you can visit in an array, following specific jump rules. The solution involves using dynamic programming where dp[i] represents the maximum number of jumps starting from index i. The final answer is the maximum value from the dp array.

Problem Statement

Given an array of integers arr and an integer d, you can jump from index i to index j if arr[i] > arr[j], and there is no intermediate index k (where min(i, j) < k < max(i, j)) such that arr[i] > arr[k]. Additionally, the maximum jump distance is constrained by d.

The goal is to determine the maximum number of indices you can visit, starting from any index in the array. Return the maximum number of indices you can visit while respecting the jump rules.

Examples

Example 1

Input: arr = [6,4,14,6,8,13,9,7,10,6,12], d = 2

Output: 4

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

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

Start at index 0. You can visit all the indicies.

Constraints

  • 1 <= arr.length <= 1000
  • 1 <= arr[i] <= 105
  • 1 <= d <= arr.length

Solution Approach

Dynamic Programming with State Transitions

Use dynamic programming to keep track of the maximum jumps possible from each index. For each index i, calculate the maximum reachable indices by checking the valid jumps based on the constraints. This ensures you account for both the jump distance d and the requirement that no intermediate index should violate the jump condition.

Bottom-Up Approach to Build dp Array

Start by calculating the possible jumps for each index in reverse order. This bottom-up approach ensures that when calculating the jumps from an index, all jumps from subsequent indices are already known, making the calculation for each index efficient.

Maximizing Jumps with Memoization

To optimize, store the results for each index in a dp array where dp[i] represents the maximum number of jumps from index i. Use memoization to avoid redundant calculations and minimize time complexity.

Complexity Analysis

Metric Value
Time Depends on the final approach
Space Depends on the final approach

The time complexity of this problem depends on the approach you take. If you use dynamic programming with memoization and traverse each index only once, the time complexity will be O(n * d), where n is the length of the array and d is the maximum jump distance. The space complexity is O(n), as we store the dp array for each index.

What Interviewers Usually Probe

  • Look for knowledge of dynamic programming and its application to state transitions.
  • Check if the candidate understands how to optimize the solution with memoization.
  • Assess the ability to handle edge cases like arrays where no jumps are possible.

Common Pitfalls or Variants

Common pitfalls

  • Not considering the valid jump conditions correctly, especially the intermediate indices.
  • Misunderstanding how the jump distance d interacts with the array's constraints.
  • Failing to optimize the solution with memoization, leading to inefficient repeated calculations.

Follow-up variants

  • Change the problem to allow multiple jump ranges and ask how to handle varying distances for each jump.
  • Introduce constraints where the jumps can only occur from certain indices and test how the solution adjusts.
  • Modify the problem to test if jumps can be made based on conditions other than values in the array (e.g., modulo operations).

FAQ

What is the primary approach for solving Jump Game V?

The main approach for solving Jump Game V is dynamic programming, where you maintain a dp array to store the maximum jumps possible from each index.

What is the significance of the jump distance d in this problem?

The jump distance d limits how far you can jump from one index to another. It's an important factor in calculating the possible jumps while adhering to the problem's constraints.

How can I optimize the solution for Jump Game V?

You can optimize the solution by using memoization to store the results of subproblems in a dp array, preventing redundant calculations and improving time complexity.

What is the maximum number of indices I can visit in Jump Game V?

The maximum number of indices you can visit depends on the specific input array and the jump distance d. The optimal solution will calculate the maximum possible jumps using dynamic programming.

What should I consider when using dynamic programming for this problem?

When using dynamic programming, ensure that you're correctly calculating the valid jumps and optimizing your solution by avoiding redundant calculations, especially by using memoization.

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Solution

Solution 1

#### Python3

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class Solution:
    def maxJumps(self, arr: List[int], d: int) -> int:
        @cache
        def dfs(i):
            ans = 1
            for j in range(i - 1, -1, -1):
                if i - j > d or arr[j] >= arr[i]:
                    break
                ans = max(ans, 1 + dfs(j))
            for j in range(i + 1, n):
                if j - i > d or arr[j] >= arr[i]:
                    break
                ans = max(ans, 1 + dfs(j))
            return ans

        n = len(arr)
        return max(dfs(i) for i in range(n))

Solution 2

#### Python3

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class Solution:
    def maxJumps(self, arr: List[int], d: int) -> int:
        @cache
        def dfs(i):
            ans = 1
            for j in range(i - 1, -1, -1):
                if i - j > d or arr[j] >= arr[i]:
                    break
                ans = max(ans, 1 + dfs(j))
            for j in range(i + 1, n):
                if j - i > d or arr[j] >= arr[i]:
                    break
                ans = max(ans, 1 + dfs(j))
            return ans

        n = len(arr)
        return max(dfs(i) for i in range(n))
Jump Game V Solution: State transition dynamic programming | LeetCode #1340 Hard