LeetCode #2498 — MEDIUM

Frog Jump II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a 0-indexed integer array stones sorted in strictly increasing order representing the positions of stones in a river.

A frog, initially on the first stone, wants to travel to the last stone and then return to the first stone. However, it can jump to any stone at most once.

The length of a jump is the absolute difference between the position of the stone the frog is currently on and the position of the stone to which the frog jumps.

More formally, if the frog is at stones[i] and is jumping to stones[j], the length of the jump is |stones[i] - stones[j]|.

The cost of a path is the maximum length of a jump among all jumps in the path.

Return the minimum cost of a path for the frog.

Example 1:

Input: stones = [0,2,5,6,7]
Output: 5
Explanation: The above figure represents one of the optimal paths the frog can take.
The cost of this path is 5, which is the maximum length of a jump.
Since it is not possible to achieve a cost of less than 5, we return it.

Example 2:

Input: stones = [0,3,9]
Output: 9
Explanation: 
The frog can jump directly to the last stone and come back to the first stone. 
In this case, the length of each jump will be 9. The cost for the path will be max(9, 9) = 9.
It can be shown that this is the minimum achievable cost.

Constraints:

2 <= stones.length <= 10⁵
0 <= stones[i] <= 10⁹
stones[0] == 0
stones is sorted in a strictly increasing order.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed integer array stones sorted in strictly increasing order representing the positions of stones in a river. A frog, initially on the first stone, wants to travel to the last stone and then return to the first stone. However, it can jump to any stone at most once. The length of a jump is the absolute difference between the position of the stone the frog is currently on and the position of the stone to which the frog jumps. More formally, if the frog is at stones[i] and is jumping to stones[j], the length of the jump is |stones[i] - stones[j]|. The cost of a path is the maximum length of a jump among all jumps in the path. Return the minimum cost of a path for the frog.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search · Greedy

Example 1

[0,2,5,6,7]

Example 2

[0,3,9]

Core Insight

What unlocks the optimal approach

One of the optimal strategies will be to jump to every stone.
Skipping just one stone in every forward jump and jumping to those skipped stones in backward jump can minimize the maximum jump.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2498: Frog Jump II
class Solution {
    public int maxJump(int[] stones) {
        int ans = stones[1] - stones[0];
        for (int i = 2; i < stones.length; ++i) {
            ans = Math.max(ans, stones[i] - stones[i - 2]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2498: Frog Jump II
func maxJump(stones []int) int {
	ans := stones[1] - stones[0]
	for i := 2; i < len(stones); i++ {
		ans = max(ans, stones[i]-stones[i-2])
	}
	return ans
}

# Accepted solution for LeetCode #2498: Frog Jump II
class Solution:
    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

// Accepted solution for LeetCode #2498: Frog Jump II
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2498: Frog Jump II
// class Solution {
//     public int maxJump(int[] stones) {
//         int ans = stones[1] - stones[0];
//         for (int i = 2; i < stones.length; ++i) {
//             ans = Math.max(ans, stones[i] - stones[i - 2]);
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2498: Frog Jump II
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #2498: Frog Jump II
// class Solution {
//     public int maxJump(int[] stones) {
//         int ans = stones[1] - stones[0];
//         for (int i = 2; i < stones.length; ++i) {
//             ans = Math.max(ans, stones[i] - stones[i - 2]);
//         }
//         return ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.