2498. Frog Jump II


Problem Description

You have an integer array stones where each element represents the position of a stone in a river, and the array is sorted in strictly increasing order. A frog starts on the first stone and wants to make it to the last stone before returning to the first stone. In doing so, the frog is allowed to jump to any stone, but only once per stone.

The length of a jump is calculated as the absolute difference between the positions of the current stone and the stone the frog jumps to. Thus, if the frog jumps from stones[i] to stones[j], the length of the jump is |stones[i] - stones[j]|.

A path's cost is determined by the longest (maximum length) jump the frog makes on its way from the first stone to the last stone and back. The goal is to determine the minimum cost of a path, meaning, to find out the smallest possible maximum jump length that the frog can achieve while still reaching the last stone and coming back to the first one.

Intuition

The intuition behind the solution lies in understanding how the cost of a path is defined and making strategic jumps that minimize the maximum jump length. The fact that the area we cover is the river between the first and last stone, and we eventually need to return to the start, it is reasonable to consider that the costliest jumps will be at the beginning or end of the route. This is because the stones are sorted and there are no gaps, so the largest jumps will be between stones that are the farthest apart.

We notice that the frog has two critical long jumps: the first jump into the river and the last jump out of the river. The first jump can only be between stones[0] and stones[1], and the last jump can be between any two consecutive stones since the frog must return to stones[0].

Hence, we arrive at the idea of checking every pair of stones that could represent the last jump (every two consecutive stones), and for each such pair, calculate the longest jump that occurs if the frog were to make that particular pair its last jump sequence. We can then extract which of these calculated longest jumps represents the smallest maximum jump, yielding the minimum cost of a path.

The implementation of the solution iterates through the stones array, starting from the second index (since the first jump is predefined), and compares the length of the jump two stones apart, updating the answer to the maximum jump length seen so far. This effectively accounts for the costliest jump the frog makes if it decided to make that pair of stones the last jump of its return trip. By the end of the iteration, since all potential last jump pairs are accounted for, the ans variable holds the value of the minimum cost path.

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Solution Approach

The solution uses a simple for-loop to iterate through the array of stones from the second stone onwards, and it performs a comparison operation for each element to find the maximum jump length.

The core algorithm doesn't utilize advanced data structures or complex patterns but relies on understanding the problem and employing a simple iterative approach to solve it.

Here's a breakdown of the implementation steps with a focus on understanding the algorithm and patterns used:

  1. Initialize ans with the difference between the first two stones, since the frog's first jump from stones[0] to stones[1] is predetermined and represents the initial cost.

  2. Iterate through the array of stones starting from index 2 (the third stone). For each stone at index i, consider the previous stone at index i-1 and the one before it at index i-2.

  3. Calculate the length of the jump from stones[i - 2] to stones[i] by taking the absolute difference: stones[i] - stones[i - 2]. This represents the potential maximum jump length if stones[i - 1] and stones[i] are considered as the last jump on the way back.

  4. Update the ans if the current jump length is greater than the previously recorded maximum jump length. This is done using the max function.

  5. Repeat this process until all possible jump lengths (for when each pair of consecutive stones could be the frog's last jump back to the start) have been considered.

  6. At the end of the loop, ans holds the minimum cost of the path that the frog can take.

The reference solution doesn't apply complex algorithms or use additional data structures; it applies a direct approach that uses only simple array indexing and built-in functions. This illustrates an important pattern in many coding challenges: sometimes the straightforward solution is the most effective and efficient one.

Here's the essence of the code snippet that achieves this:

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

Note that the code above assumes that the input will always have more than two elements as per the constraints implied in the problem description.

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1int sqrt(int n) {
2  for (int guess = 1; guess * guess <= n; guess++) {
3    if (guess * guess == n) {
4      return guess;
5    }
6  }
7  return -1;
8}

Example Walkthrough

Let us consider a small example to illustrate the solution approach. Suppose we have the following integer array stones: [2, 4, 7, 10], where each element signifies the stone's position in a river.

Following the logic described above:

  1. Initialize ans with the difference between the first two stones.

    • ans = stones[1] - stones[0]
    • ans = 4 - 2
    • ans = 2
  2. Proceed through the stones to find the maximum jump required if the last jump back were between the second stone and third stone, then the third stone and fourth stone.

  3. We make a jump from stones[0] to stones[2] (from position 2 to position 7):

    • Current jump length = stones[2] - stones[0]
    • Current jump length = 7 - 2
    • Current jump length = 5
  4. Since 5 (current jump length) > 2 (ans), we update ans.

    • ans = 5
  5. Next, consider the stones at positions stones[1], stones[2], and stones[3] to simulate if the last jump is between stones[2] and stones[3].

    • Current jump length would be between stones[1] and stones[3]: from position 4 to position 10.
    • Current jump length = stones[3] - stones[1]
    • Current jump length = 10 - 4
    • Current jump length = 6
  6. Since 6 (current jump length) > 5 (ans), we update ans again.

    • ans = 6
  7. We have now considered all possible last jump pairs, and thus, the ans value is the minimum possible maximum jump length the frog can achieve when completing its path.

    • The final ans = 6

So, the minimum cost of a path which is the smallest maximum jump the frog has to make in this case is 6.

Solution Implementation

1from typing import List
2
3class Solution:
4    def maxJump(self, stones: List[int]) -> int:
5        # Initialize the maximum jump distance as the distance between the first two stones.
6        max_jump = stones[1] - stones[0]
7
8        # Iterate over the stones starting from the third stone.
9        for i in range(2, len(stones)):
10            # For each stone, calculate the jump distance from the stone two positions before it.
11            # Update the maximum jump distance if the current distance is larger.
12            max_jump = max(max_jump, stones[i] - stones[i - 2])
13
14        # Return the maximum jump distance found.
15        return max_jump
16
1class Solution {
2
3    // Method to calculate the maximum jump distance between consecutive stones
4    public int maxJump(int[] stones) {
5        // Initialize the maximum jump to the distance between the first two stones
6        int maxJumpDistance = stones[1] - stones[0];
7      
8        // Loop through the array starting from the third stone
9        for (int i = 2; i < stones.length; ++i) {
10            // Calculate the jump distance between the current stone and the stone two steps back
11            int jumpDistance = stones[i] - stones[i - 2];
12          
13            // Update the maximum jump distance if the current jump is greater
14            maxJumpDistance = Math.max(maxJumpDistance, jumpDistance);
15        }
16      
17        // Return the maximum jump distance found
18        return maxJumpDistance;
19    }
20}
21
1#include <vector>
2#include <algorithm> // Include the algorithm header for the 'max' function
3
4class Solution {
5public:
6    // Function to determine the maximum jump between adjacent stones
7    int maxJump(vector<int>& stones) {
8        // Check if there are less than two stones, return 0 if true,
9        // as no jump can be made with only one stone or no stones
10        if (stones.size() < 2) {
11            return 0;
12        }
13
14        // Initialize 'max_jump' to the jump between the first two stones
15        int max_jump = stones[1] - stones[0];
16
17        // Iterate from the third stone to the end of the vector
18        for (int i = 2; i < stones.size(); ++i) {
19            // Update 'max_jump' to be the maximum between its current value and
20            // the difference between the current stone and the stone two places before
21            max_jump = max(max_jump, stones[i] - stones[i - 2]);
22        }
23
24        // Return the maximum jump found
25        return max_jump;
26    }
27};
28
1// Import the algorithm's max function equivalent in TypeScript
2import { max } from 'lodash';
3
4// Function to determine the maximum jump between adjacent stones
5function maxJump(stones: number[]): number {
6    // Check if there are fewer than two stones, return 0 if true,
7    // as no jump can be made with only one stone or no stones
8    if (stones.length < 2) {
9        return 0;
10    }
11
12    // Initialize 'maxJump' to the jump between the first two stones
13    let maxJump = stones[1] - stones[0];
14
15    // Iterate from the third stone to the end of the array
16    for (let i = 2; i < stones.length; i++) {
17        // Update 'maxJump' to be the maximum between its current value and
18        // the difference between the current stone and the stone two places before
19        maxJump = max([maxJump, stones[i] - stones[i - 2]]);
20    }
21
22    // Return the maximum jump found
23    return maxJump;
24}
25
26// Example on how to use the function
27// let stones = [0, 3, 5, 9, 10];
28// let result = maxJump(stones);
29// console.log(result); // This would print the result of the maximum jump.
30
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Time and Space Complexity

The given Python code snippet defines a method maxJump that determines the maximum jump between consecutive or alternate stones in a list of stones represented by their positions in stones.

Time Complexity

The time complexity of the code is O(n), where n is the length of the stones list. This is because there is a single for loop iterating through the stones array, starting at the second index and performing a constant time operation (calculating the maximum jump and updating ans) during each iteration.

Space Complexity

The space complexity of the code is O(1), which means it is constant space complexity. Apart from the input list itself, the only additional storage used is a single variable ans to keep track of the current maximum jump, which does not rely on the size of the input.

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