Problem Description

You have a frog that needs to make a round trip across a river using stepping stones. The stones are represented by an array stones where each element indicates the position of a stone, and the array is sorted in strictly increasing order.

The frog starts at the first stone (index 0) and must:

1. Travel to the last stone
2. Return back to the first stone

Key constraints:

• The frog can use each stone at most once during the entire round trip
• This means if the frog jumps on a stone while going forward, it cannot use that same stone on the way back

Jump length and cost:

• The length of a jump between two stones is the absolute difference between their positions: |stones[i] - stones[j]|
• The cost of the entire path is defined as the maximum jump length among all jumps in the complete round trip

Your task is to find the minimum possible cost for the frog to complete this round trip.

Example visualization: If stones = [0, 2, 5, 6, 7], the frog might:

• Go forward: 0 → 2 → 5 → 7 (jumps of length 2, 3, 2)
• Return: 7 → 6 → 0 (jumps of length 1, 6)
• Maximum jump length = 6, so the cost = 6

The goal is to find a path strategy that minimizes this maximum jump length across all possible valid round-trip paths.

Intuition

The key insight is to think about what happens when we try to minimize the maximum jump length. Since each stone can only be used once in the entire round trip, we need to strategically decide which stones to use on the forward path and which to save for the return path.

Consider this: if we visit every single stone on the way forward, we'd have no stones left for the return journey except to jump directly from the last stone back to the first stone. This would result in a very large jump of stones[n-1] - stones[0].

Instead, we should alternate which stones we use. A clever strategy is to:

• On the forward path: visit stones at even indices (0, 2, 4, ...)
• On the return path: visit stones at odd indices (..., 5, 3, 1)

This alternating pattern ensures that:

1. We have stones available for both directions
2. The jumps are relatively balanced in length

With this pattern, the forward jumps would be from stone i to stone i+2 (skipping one stone each time). The return jumps would also skip one stone each time but use the previously skipped stones.

The largest jump in this strategy would be:

• Either a forward jump: stones[i+2] - stones[i] for some even i
• Or the initial jump if we start directly to stone 1: stones[1] - stones[0]
• Or the final return jump to stone 0

By checking all consecutive pairs of stones with distance 2 apart (stones[i] - stones[i-2]), along with the first jump stones[1] - stones[0], we can find the maximum jump length in this optimal alternating pattern. This gives us the minimum possible cost since any other pattern would either create larger jumps or leave us with no valid return path.

Learn more about Greedy and Binary Search patterns.

Solution Approach

The implementation is remarkably simple once we understand the alternating stone pattern strategy:

def maxJump(self, stones: List[int]) -> int:
    ans = stones[1] - stones[0]
    for i in range(2, len(stones)):
        ans = max(ans, stones[i] - stones[i - 2])
    return ans

Step-by-step breakdown:

1. Initialize with the first jump: ans = stones[1] - stones[0]

   • This handles the edge case where we might jump directly from stone 0 to stone 1
   • This is also the minimum jump we must consider since we need to leave the first stone

2. Check all jumps that skip one stone: Loop through indices starting from 2

   • For each stone at index i, calculate the jump distance from stone i-2 to stone i
   • This represents jumps like: stone 0→2, stone 1→3, stone 2→4, etc.
   • Update ans to keep track of the maximum jump distance encountered

3. Return the maximum jump distance found

   • This maximum represents the minimum possible cost for the round trip

Why this works:

The algorithm implicitly constructs the optimal path:

• Forward path: 0 → 2 → 4 → 6 → ... → (last or second-to-last stone)
• Return path: ... → 5 → 3 → 1 → 0

The jumps we're checking (stones[i] - stones[i-2]) cover:

• All forward jumps in the even-indexed path
• All return jumps in the odd-indexed path (since jumping from odd index j to j-2 has the same distance as jumping from j-2 to j)

Time Complexity: O(n) where n is the number of stones - we iterate through the array once

Space Complexity: O(1) - we only use a constant amount of extra space to store the answer

This elegant solution leverages the sorted nature of the stones array and the symmetry of the round-trip requirement to find the optimal jumping pattern with minimal code.

Example Walkthrough

Let's walk through the solution with stones = [0, 3, 9, 10, 12].

Step 1: Initialize with first possible jump

• ans = stones[1] - stones[0] = 3 - 0 = 3
• This represents a potential jump from stone 0 to stone 1

Step 2: Check all jumps that skip one stone

• i = 2: Check jump from stone 0 to stone 2

  • Jump length: stones[2] - stones[0] = 9 - 0 = 9
  • Update: ans = max(3, 9) = 9

• i = 3: Check jump from stone 1 to stone 3

  • Jump length: stones[3] - stones[1] = 10 - 3 = 7
  • Update: ans = max(9, 7) = 9

• i = 4: Check jump from stone 2 to stone 4

  • Jump length: stones[4] - stones[2] = 12 - 9 = 3
  • Update: ans = max(9, 3) = 9

Result: The minimum possible cost is 9

Verifying with the actual path: This algorithm implicitly creates the alternating pattern:

• Forward path: 0 → 2 → 4 (jumps: 9, 3)
• Return path: 4 → 3 → 1 → 0 (jumps: 2, 7, 3)

All jumps in the complete round trip: [9, 3, 2, 7, 3] Maximum jump = 9 ✓

Notice how the algorithm found this optimal cost by checking:

• The jump 0→2 (length 9) - which appears in our forward path
• The jump 1→3 (length 7) - which when reversed (3→1) appears in our return path
• The jump 2→4 (length 3) - which appears in our forward path

The algorithm correctly identifies that the maximum jump of 9 cannot be reduced further while maintaining a valid round trip where each stone is used at most once.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n), where n is the length of the stones array. The algorithm iterates through the stones array once, starting from index 2 to the end. Each iteration performs a constant-time operation (calculating the difference between stones[i] and stones[i-2] and comparing it with the current maximum). Therefore, the overall time complexity is linear.

Space Complexity: O(1). The algorithm uses only a constant amount of extra space. It maintains a single variable ans to track the maximum jump distance throughout the iteration. No additional data structures that scale with the input size are created, resulting in constant space complexity.

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

Common Pitfalls

1. Misunderstanding the Optimal Strategy

Many people initially try complex approaches like dynamic programming or graph algorithms, thinking they need to explicitly track which stones are used in forward vs. return paths. The key insight that's often missed is that the alternating pattern (even indices forward, odd indices back) is always optimal.

Why this happens: The problem statement makes it seem like we need to find two completely separate paths, leading to overthinking.

Solution: Recognize that with sorted stones, the alternating pattern minimizes the maximum jump by distributing stones evenly between forward and return paths.

2. Forgetting Edge Cases with Small Arrays

When stones has only 2 elements [a, b], the frog must jump directly from a to b and back, making the cost b - a. The algorithm handles this correctly by initializing with stones[1] - stones[0], but it's easy to overlook why this initialization is crucial.

Example pitfall code:

def maxJump(self, stones: List[int]) -> int:
    ans = 0  # Wrong initialization!
    for i in range(2, len(stones)):
        ans = max(ans, stones[i] - stones[i - 2])
    return ans  # Returns 0 for 2-element arrays!

Solution: Always initialize with stones[1] - stones[0] to handle the direct jump case.

3. Incorrectly Trying to Handle the Last Stone Separately

Some might think the last stone needs special handling since the frog must reach it. They might add extra logic to ensure the last stone is included in the forward path.

Incorrect approach:

def maxJump(self, stones: List[int]) -> int:
    ans = stones[1] - stones[0]
    for i in range(2, len(stones) - 1):
        ans = max(ans, stones[i] - stones[i - 2])
    # Special case for last stone - UNNECESSARY!
    ans = max(ans, stones[-1] - stones[-2])
    return ans

Solution: The original algorithm already handles this correctly. If there's an even number of stones, the last stone is naturally included in the even-indexed path. If odd, the second-to-last stone jumps to the last stone as part of the pattern.

4. Attempting to Track Used Stones Explicitly

Some might create additional data structures to track which stones have been used:

Overcomplicated approach:

def maxJump(self, stones: List[int]) -> int:
    used = [False] * len(stones)
    # Complex logic to track forward and return paths...

Solution: The beauty of this problem is that the alternating pattern implicitly ensures no stone is used twice. No explicit tracking is needed.