Problem Description

You have an array nums with n integers (0-indexed) and an integer value target. You start at index 0 and want to reach the last index (n-1) by making jumps.

From any index i, you can jump to any index j if these conditions are met:

• j must be ahead of i (meaning 0 <= i < j < n)
• The difference between the values at these positions must be within the target range: -target <= nums[j] - nums[i] <= target

Your goal is to find the maximum number of jumps you can make to reach the last index. If it's impossible to reach the last index, return -1.

For example, if you're at index i, you can only jump to index j if j is ahead of you and the absolute difference |nums[j] - nums[i]| is at most target. You want to maximize the number of jumps taken while still being able to reach the end.

Intuition

The key insight is that we want to maximize the number of jumps, not minimize them. This means we should explore all possible valid paths from the starting position to the end and choose the one with the most jumps.

Think of this problem backwards: from any position i, we need to determine what's the maximum number of jumps we can make to reach the end. If we're already at the last index, we need 0 jumps. Otherwise, we look at all positions we can jump to from our current position and pick the path that gives us the most jumps.

This naturally leads to a recursive approach. For each position i, we:

1. Check all positions j where j > i (positions ahead of us)
2. Verify if we can jump there: |nums[j] - nums[i]| <= target
3. If we can jump to j, calculate how many jumps we can make from j to the end
4. Take the maximum among all valid jumps and add 1 (for the current jump)

The recursive nature of this problem makes it a perfect candidate for dynamic programming with memoization. Since we might visit the same position multiple times through different paths, we can cache the result for each position to avoid recalculating it.

The function dfs(i) represents "what's the maximum number of jumps from position i to reach the end?" If we can't reach the end from position i, we return negative infinity to indicate it's impossible. Starting from dfs(0) gives us the answer for the entire problem.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution implements a top-down dynamic programming approach using memoization with Python's @cache decorator.

Implementation Details:

1. Main Function Setup: We define n = len(nums) to get the array length and call dfs(0) to start the recursion from index 0.

2. Recursive Function dfs(i): This function calculates the maximum jumps from position i to reach index n-1.

   • Base Case: If i == n - 1, we've reached the destination, so return 0 (no more jumps needed)
   • Recursive Case: Initialize ans = -inf to track the maximum jumps. This negative infinity helps identify when no valid path exists.

3. Exploring Valid Jumps: For each position i, we iterate through all possible jump destinations:

   for j in range(i + 1, n):
       if abs(nums[i] - nums[j]) <= target:
           ans = max(ans, 1 + dfs(j))

   • We check positions j from i+1 to n-1
   • For each valid jump (where |nums[i] - nums[j]| <= target), we calculate 1 + dfs(j)
   • The 1 accounts for the current jump from i to j
   • dfs(j) gives us the maximum jumps from j to the end
   • We keep the maximum value among all possible paths

4. Memoization: The @cache decorator automatically stores results of dfs(i) for each unique i, preventing redundant calculations when the same position is reached through different paths.

5. Final Result: After dfs(0) completes:

   • If ans < 0 (still negative infinity or negative), it means no valid path exists, so return -1
   • Otherwise, return ans as the maximum number of jumps

Time Complexity: O(n²) where each position is computed once (due to memoization) and for each position, we check up to n-1 possible jumps.

Space Complexity: O(n) for the memoization cache and recursion stack depth.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Example: nums = [1, 3, 6, 4, 1], target = 2

We want to find the maximum number of jumps from index 0 to index 4.

Step-by-step execution of dfs(0):

Starting at index 0 (value = 1):

• Check index 1: |3 - 1| = 2 ≤ 2 ✓ Valid jump → Calculate 1 + dfs(1)
• Check index 2: |6 - 1| = 5 > 2 ✗ Invalid jump
• Check index 3: |4 - 1| = 3 > 2 ✗ Invalid jump
• Check index 4: |1 - 1| = 0 ≤ 2 ✓ Valid jump → Calculate 1 + dfs(4)

Computing dfs(1) (value = 3):

• Check index 2: |6 - 3| = 3 > 2 ✗ Invalid jump
• Check index 3: |4 - 3| = 1 ≤ 2 ✓ Valid jump → Calculate 1 + dfs(3)
• Check index 4: |1 - 3| = 2 ≤ 2 ✓ Valid jump → Calculate 1 + dfs(4)

Computing dfs(3) (value = 4):

• Check index 4: |1 - 4| = 3 > 2 ✗ Invalid jump
• No valid jumps → Return -inf

Computing dfs(4) (value = 1):

• Base case: We're at the last index → Return 0

Backtracking with results:

• dfs(4) = 0 (base case)
• dfs(3) = -inf (no valid jumps to end)
• dfs(1) chooses max between:
  • Path through index 3: 1 + (-inf) = -inf
  • Path through index 4: 1 + 0 = 1
  • Result: dfs(1) = 1
• dfs(0) chooses max between:
  • Path through index 1: 1 + 1 = 2
  • Path through index 4: 1 + 0 = 1
  • Result: dfs(0) = 2

Final answer: 2 jumps (path: 0 → 1 → 4)

The algorithm explores all valid paths and selects the one with maximum jumps. The memoization ensures each position is computed only once, even if reached through multiple paths.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n^2)

The solution uses dynamic programming with memoization. The dfs function can be called at most n times (once for each index from 0 to n-1). For each call to dfs(i), we iterate through all indices from i+1 to n-1 in the worst case, which takes O(n) time. Since we use memoization with @cache, each state i is computed only once. Therefore, the total time complexity is O(n) * O(n) = O(n^2).

Space Complexity: O(n)

The space complexity consists of two components:

1. The recursion call stack can go up to depth n in the worst case (when we jump one index at a time from 0 to n-1), contributing O(n) space.
2. The memoization cache stores at most n different states (one for each index), contributing O(n) space.

The total space complexity is O(n) + O(n) = O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Base Case Handling

A common mistake is forgetting to handle the edge case where the input array has only one element. When nums has length 1, we're already at the last index (index 0), so no jumps are needed.

Incorrect approach:

def maximumJumps(self, nums: List[int], target: int) -> int:
    @cache
    def dfs(i):
        if i == n - 1:
            return 0
        # ... rest of code
  
    n = len(nums)
    result = dfs(0)
    return -1 if result < 0 else result

Issue: When len(nums) == 1, the function returns 0 instead of 0 jumps, which is actually correct but might be confusing in implementation.

Solution: Add explicit handling or ensure the base case is clear:

def maximumJumps(self, nums: List[int], target: int) -> int:
    n = len(nums)
    # Handle single element array explicitly
    if n == 1:
        return 0
  
    @cache
    def dfs(i):
        if i == n - 1:
            return 0
        # ... rest of code

2. Using Wrong Initial Value for Maximum Tracking

Another pitfall is initializing the maximum jumps variable incorrectly, such as using 0 or -1 instead of negative infinity.

Incorrect approach:

@cache
def dfs(i):
    if i == n - 1:
        return 0
  
    ans = -1  # Wrong! Should be -inf
    for j in range(i + 1, n):
        if abs(nums[i] - nums[j]) <= target:
            ans = max(ans, 1 + dfs(j))
    return ans

Issue: If no valid jumps exist from position i, returning -1 will propagate through the recursion. When we compute 1 + dfs(j) where dfs(j) = -1, we get 0, which incorrectly suggests a valid path exists.

Solution: Always use negative infinity for impossible cases:

from math import inf

@cache
def dfs(i):
    if i == n - 1:
        return 0
  
    ans = -inf  # Correct initialization
    for j in range(i + 1, n):
        if abs(nums[i] - nums[j]) <= target:
            ans = max(ans, 1 + dfs(j))
    return ans

3. Misunderstanding the Jump Condition

Some might incorrectly interpret the condition as requiring exact equality or using the wrong comparison.

Incorrect approaches:

# Wrong: Using wrong range check
if nums[j] - nums[i] <= target:  # Missing absolute value
    ans = max(ans, 1 + dfs(j))

# Wrong: Checking only positive difference
if 0 <= nums[j] - nums[i] <= target:  # Ignores negative differences
    ans = max(ans, 1 + dfs(j))

Solution: Always check the absolute difference:

if abs(nums[i] - nums[j]) <= target:
    ans = max(ans, 1 + dfs(j))

4. Memory Optimization Oversight

While @cache is convenient, in competitive programming or memory-constrained environments, it might cause issues with large inputs.

Alternative Solution using Bottom-Up DP:

def maximumJumps(self, nums: List[int], target: int) -> int:
    n = len(nums)
    dp = [-inf] * n
    dp[n-1] = 0
  
    for i in range(n-2, -1, -1):
        for j in range(i+1, n):
            if abs(nums[i] - nums[j]) <= target:
                dp[i] = max(dp[i], 1 + dp[j])
  
    return -1 if dp[0] < 0 else dp[0]

This approach uses O(n) space without recursion stack overhead and provides clearer memory usage patterns.