How do you solve Jump Game V on LeetCode?

Jump Game V (LeetCode #1340) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Dynamic programming allows us to build solutions incrementally, avoiding redundant calculations.

What is the time complexity of Jump Game V?

The optimal solution for Jump Game V runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to sorting the indices, while the space complexity is O(n) for the dp array.

What is the brute force solution for Jump Game V?

The brute force approach for Jump Game V is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will try to jump from every index in the array and explore all possible paths. For each index, we will check all valid jumps and count how many indices we can visit. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Jump Game V?

Jump Game V uses the following concepts: Array, Dynamic Programming, Sorting. The recommended patterns to study are: Dynamic Programming, Sorting.

What are the interview tips for Jump Game V?

Always clarify the problem constraints and edge cases before diving into coding. Explain your thought process and approach clearly to the interviewer. Practice writing clean and efficient code, as readability is crucial in interviews.

What are common mistakes when solving Jump Game V?

Failing to consider all valid jumps when updating the dp array. Not properly handling the boundaries of the array when checking jumps.

#1340

Jump Game V

Hard

ArrayDynamic ProgrammingSortingDynamic ProgrammingSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n log n)Space O(n)

The optimal solution uses dynamic programming to store the maximum jumps possible from each index, allowing us to avoid redundant calculations. We can fill this dp array by checking all possible jumps efficiently.

⚙️

Algorithm

4 steps

1Step 1: Initialize a dp array where dp[i] represents the maximum jumps from index i.
2Step 2: Sort the indices based on the values in arr to ensure we process larger values first.
3Step 3: For each index, check all possible jumps (both left and right) within the distance d and update the dp array accordingly.
4Step 4: The result is the maximum value in the dp array.

solution.py25 lines

1# Full working Python code
2
3def maxJumps(arr, d):
4    n = len(arr)
5    dp = [1] * n  # Each index can at least reach itself
6    indices = sorted(range(n), key=lambda i: arr[i])  # Sort indices based on arr values
7
8    for i in indices:
9        # Check right jumps
10        for x in range(1, d + 1):
11            if i + x < n and arr[i] > arr[i + x]:
12                dp[i] = max(dp[i], 1 + dp[i + x])
13            else:
14                break
15        # Check left jumps
16        for x in range(1, d + 1):
17            if i - x >= 0 and arr[i] > arr[i - x]:
18                dp[i] = max(dp[i], 1 + dp[i - x])
19            else:
20                break
21
22    return max(dp)
23
24# Example usage
25print(maxJumps([6,4,14,6,8,13,9,7,10,6,12], 2))

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the indices, while the space complexity is O(n) for the dp array.

1Dynamic programming allows us to build solutions incrementally, avoiding redundant calculations.
2Sorting indices based on values helps ensure we always process larger values first, maximizing potential jumps.

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