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Frog Jump II

Medium

Google

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Topics:

ArraysBinary SearchDynamic Programming

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:

stones = [0,2,5,6,7] Output: 5

Example 2:

stones = [0,3,9] Output: 9

Explain an efficient algorithm to solve this problem.

Frog Jump II

Medium

Google

5 views

Topics:

ArraysBinary SearchDynamic Programming

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:

stones = [0,2,5,6,7] Output: 5

Example 2:

stones = [0,3,9] Output: 9

Explain an efficient algorithm to solve this problem.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible value for any element in the `stones` array?
Can the input array `stones` ever be empty or contain only one element?
If it is impossible for the frog to reach the last stone and return to the first, what should the function return?
Are the stones guaranteed to be in strictly increasing order?
Could you provide an example input and expected output for a case with more than 3 stones?

Brute Force Solution

Approach

The frog needs to cross some stones, and we want to find the smallest maximum jump distance. The brute force method means we will try every possible combination of jumps forward and backward to see if the frog can make it to the last stone.

Here's how the algorithm would work step-by-step:

Consider the first stone. The frog can jump to any other stone.
For each of those stones, consider where the frog can jump next (forward or backward).
Keep trying all possible jump sequences until the frog either reaches the last stone or gets stuck.
If the frog reaches the last stone, record the longest jump made in that particular sequence.
Repeat this process for every initial jump possibility from the starting stone.
Once we've explored all possible jump sequences that reach the end, we pick the smallest of all the recorded longest jumps.

Code Implementation

def frog_jump_brute_force(stones):    smallest_maximum_jump = float('inf')
    def solve(current_stone_index, visited_stones, current_maximum_jump):
        nonlocal smallest_maximum_jump
        # If we've reached the last stone, update the result.
        if current_stone_index == len(stones) - 1:
            smallest_maximum_jump = min(smallest_maximum_jump, current_maximum_jump)
            return

        # Iterate through all possible next stones.
        for next_stone_index in range(len(stones)):            if next_stone_index != current_stone_index and next_stone_index not in visited_stones:
                jump_distance = abs(stones[next_stone_index] - stones[current_stone_index])
                # Recursively explore the path.
                solve(
                    next_stone_index,
                    visited_stones | {next_stone_index},
                    max(current_maximum_jump, jump_distance)
                )

    # Iterate through all possible first jumps.
    for first_jump_index in range(1, len(stones)):        jump_distance_first = abs(stones[first_jump_index] - stones[0])
        # Initialize the recursive calls.
        solve(first_jump_index, {0, first_jump_index}, jump_distance_first)
    return smallest_maximum_jump

Big(O) Analysis

Time Complexity

O(2^n) – The given solution uses a brute-force approach, exploring every possible combination of forward and backward jumps. At each stone, the frog has multiple choices of where to jump next. In the worst-case scenario, the frog might explore almost all possible paths. Since the number of stones is 'n', and at each stone the frog has a choice of going to several other stones (both forward and backward), the number of possible paths grows exponentially with 'n'. Hence, the time complexity is O(2^n) because it resembles exploring a binary tree with a depth related to 'n', where each node represents a stone and each branch represents a jump decision.

Space Complexity

O(2^N) – The brute force approach explores all possible jump sequences using recursion. In the worst-case scenario, the frog can jump to any of the N stones from each stone, leading to a branching factor of approximately N at each level of the recursion. Since the maximum depth of the recursion is also proportional to N (the number of stones), the total number of possible jump sequences explored grows exponentially, resulting in a recursion tree with up to roughly N^N nodes. Each recursive call stores its state (return address, parameters), so the space used by the recursion stack can grow up to O(2^N) due to the number of paths that may be explored, since it's a full traversal of all permutations.

Optimal Solution

Approach

The key to solving this problem efficiently is to realize that the biggest possible jump is always between the first and last stones. The optimal strategy then involves only considering jumps between the first and last stones and the jumps between adjacent stones.

Here's how the algorithm would work step-by-step:

Calculate the distance between the first and last stone.
Go through the other stones and find the largest distance between any two stones that are next to each other.
The biggest of the two distances found in the previous steps is the smallest amount of energy the frog needs to complete the course.

Code Implementation

def frog_jump_ii(stones):
    # Calculate the distance between the first and last stone.
    max_energy = stones[-1] - stones[0]

    # Find the largest distance between adjacent stones.
    for i in range(1, len(stones)):
        distance_between_stones = stones[i] - stones[i-1]
        if distance_between_stones > max_energy:
            max_energy = distance_between_stones

    return max_energy

Big(O) Analysis

Time Complexity

O(n) – The algorithm first calculates the distance between the first and last stone, which takes constant time O(1). Then, it iterates through the stones array once to find the largest distance between adjacent stones. This iteration depends on the number of stones, n. Therefore, the dominant operation is the single loop with n iterations, resulting in a linear time complexity. The total operations approximate to 'n', which simplifies to O(n).

Space Complexity

O(1) – The algorithm calculates the distance between the first and last stone and iterates through the other stones to find the largest distance between adjacent stones. This involves storing a constant number of variables to hold distances and track the maximum distance found so far. The space used does not depend on the number of stones (N), and no auxiliary data structures like arrays or hash maps are created. Therefore, the space complexity is constant.

Edge Cases

Case	How to Handle
Empty or null stones array	Return 0 since no jump is possible.
Stones array with only one stone	Return 0 since no jump is needed.
Stones array with only two stones	Return the difference between the two stones as the minimal cost.
Stones array with stones not in strictly increasing order	Sort the stones array and proceed with cost calculation (or return error, based on constraints).
Large difference between consecutive stones, leading to large cost	The algorithm naturally handles it by calculating and maximizing the differences.
Maximum sized array of stones to test for performance	Ensure the solution has O(n) complexity to process this efficiently.
Stones array with potential Integer overflow when calculating cost.	Use long data type to store distances between stones to prevent overflow.
Negative stone positions (if problem definition allows)	Handle negative stone positions by shifting the origin or using absolute differences.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible value for any element in the `stones` array?
Can the input array `stones` ever be empty or contain only one element?
If it is impossible for the frog to reach the last stone and return to the first, what should the function return?
Are the stones guaranteed to be in strictly increasing order?
Could you provide an example input and expected output for a case with more than 3 stones?

Brute Force Solution

Approach

Here's how the algorithm would work step-by-step:

Consider the first stone. The frog can jump to any other stone.
For each of those stones, consider where the frog can jump next (forward or backward).
Keep trying all possible jump sequences until the frog either reaches the last stone or gets stuck.
If the frog reaches the last stone, record the longest jump made in that particular sequence.
Repeat this process for every initial jump possibility from the starting stone.
Once we've explored all possible jump sequences that reach the end, we pick the smallest of all the recorded longest jumps.

Code Implementation

def frog_jump_brute_force(stones):    smallest_maximum_jump = float('inf')
    def solve(current_stone_index, visited_stones, current_maximum_jump):
        nonlocal smallest_maximum_jump
        # If we've reached the last stone, update the result.
        if current_stone_index == len(stones) - 1:
            smallest_maximum_jump = min(smallest_maximum_jump, current_maximum_jump)
            return

        # Iterate through all possible next stones.
        for next_stone_index in range(len(stones)):            if next_stone_index != current_stone_index and next_stone_index not in visited_stones:
                jump_distance = abs(stones[next_stone_index] - stones[current_stone_index])
                # Recursively explore the path.
                solve(
                    next_stone_index,
                    visited_stones | {next_stone_index},
                    max(current_maximum_jump, jump_distance)
                )

    # Iterate through all possible first jumps.
    for first_jump_index in range(1, len(stones)):        jump_distance_first = abs(stones[first_jump_index] - stones[0])
        # Initialize the recursive calls.
        solve(first_jump_index, {0, first_jump_index}, jump_distance_first)
    return smallest_maximum_jump

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

Calculate the distance between the first and last stone.
Go through the other stones and find the largest distance between any two stones that are next to each other.
The biggest of the two distances found in the previous steps is the smallest amount of energy the frog needs to complete the course.

Code Implementation

def frog_jump_ii(stones):
    # Calculate the distance between the first and last stone.
    max_energy = stones[-1] - stones[0]

    # Find the largest distance between adjacent stones.
    for i in range(1, len(stones)):
        distance_between_stones = stones[i] - stones[i-1]
        if distance_between_stones > max_energy:
            max_energy = distance_between_stones

    return max_energy

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty or null stones array	Return 0 since no jump is possible.
Stones array with only one stone	Return 0 since no jump is needed.
Stones array with only two stones	Return the difference between the two stones as the minimal cost.
Stones array with stones not in strictly increasing order	Sort the stones array and proceed with cost calculation (or return error, based on constraints).
Large difference between consecutive stones, leading to large cost	The algorithm naturally handles it by calculating and maximizing the differences.
Maximum sized array of stones to test for performance	Ensure the solution has O(n) complexity to process this efficiently.
Stones array with potential Integer overflow when calculating cost.	Use long data type to store distances between stones to prevent overflow.
Negative stone positions (if problem definition allows)	Handle negative stone positions by shifting the origin or using absolute differences.