How do you solve Maximum Number of Jumps to Reach the Last Index on LeetCode?

Maximum Number of Jumps to Reach the Last Index (LeetCode #2770) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n) space complexity. Key insight: Dynamic programming helps avoid redundant calculations.

What is the brute force solution for Maximum Number of Jumps to Reach the Last Index?

The brute force approach for Maximum Number of Jumps to Reach the Last Index is Brute Force, with O(n²) time complexity and O(n) space complexity. The brute force approach involves exploring all possible jumps from the current index to find the maximum number of jumps to reach the last index. This method checks every possible jump, which can be inefficient. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Maximum Number of Jumps to Reach the Last Index?

Maximum Number of Jumps to Reach the Last Index uses the following concepts: Array, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Maximum Number of Jumps to Reach the Last Index?

Clarify the problem constraints and edge cases. Explain your thought process while coding. Test your solution with multiple examples.

What are common mistakes when solving Maximum Number of Jumps to Reach the Last Index?

Not considering all valid jumps from an index. Overlooking the base case when reaching the last index.

#2770

Maximum Number of Jumps to Reach the Last Index

Medium

ArrayDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n²)Space O(n)

The optimal approach uses dynamic programming to store the maximum number of jumps to each index. This way, we avoid redundant calculations and efficiently find the solution.

⚙️

Algorithm

3 steps

1Step 1: Initialize a dp array of size n with all values set to -1, except dp[0] which is set to 0.
2Step 2: Iterate through each index i from 0 to n-1.
3Step 3: For each index i, check all possible jumps to index j (i < j < n) that satisfy the target condition and update dp[j] accordingly.

solution.py13 lines

1# Full working Python code
2
3def maxJumps(nums, target):
4    n = len(nums)
5    dp = [-1] * n
6    dp[0] = 0
7    for i in range(n):
8        if dp[i] == -1:
9            continue
10        for j in range(i + 1, n):
11            if nums[j] - nums[i] <= target:
12                dp[j] = max(dp[j], dp[i] + 1)
13    return dp[-1]

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we significantly reduce redundant calculations by storing results. The space complexity is O(n) for the dp array.

1Dynamic programming helps avoid redundant calculations.
2Understanding the jump conditions is crucial for optimizing the solution.

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