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Jump Game VIII

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ArraysGreedy Algorithms

You are given a 0-indexed array of integers nums. You are currently at index 0.

In each step, you can jump from index i to any index j such that nums[i] == nums[j].

Let x be the minimum number of steps to reach index n - 1. Return x if it is possible to reach index n - 1, otherwise, return -1.

Example 1:

Input: nums = [1,2,3,4,5,6,7,8,9,10]
Output: -1
Explanation: It is impossible to reach index 9 from index 0 since there are no equal values in the array.

Example 2:

Input: nums = [1,2,3,4,5,6,7,8,9,10,1,2,3]
Output: 2
Explanation: You can reach index 12 from index 0 in the following way:
- Jump from index 0 to index 10.
- Jump from index 10 to index 12.

Example 3:

Input: nums = [1,2,3,4,5,6,7,8,9,10,1,2,3,4,5]
Output: 4

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Try coding it in LeetCode

Jump Game VIII

Medium

Asked by:

10 views

Topics:

ArraysGreedy Algorithms

You are given a 0-indexed array of integers nums. You are currently at index 0.

In each step, you can jump from index i to any index j such that nums[i] == nums[j].

Let x be the minimum number of steps to reach index n - 1. Return x if it is possible to reach index n - 1, otherwise, return -1.

Example 1:

Input: nums = [1,2,3,4,5,6,7,8,9,10]
Output: -1
Explanation: It is impossible to reach index 9 from index 0 since there are no equal values in the array.

Example 2:

Input: nums = [1,2,3,4,5,6,7,8,9,10,1,2,3]
Output: 2
Explanation: You can reach index 12 from index 0 in the following way:
- Jump from index 0 to index 10.
- Jump from index 10 to index 12.

Example 3:

Input: nums = [1,2,3,4,5,6,7,8,9,10,1,2,3,4,5]
Output: 4

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵

Try coding it in LeetCode

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 range of values for elements in the input array arr, and what is the range for the integer d?
Can the input array arr be empty, or can d be zero? If so, what should the output be?
If there are multiple ways to reach the end of the array, do I need to return the minimum number of jumps, or is any valid path sufficient?
Are the values in the input array guaranteed to be integers, or could they be floating-point numbers?
If it is not possible to reach the end of the array, what value should I return?

Brute Force Solution

Approach

The core idea is to explore every conceivable path to reach the end. We start at the beginning and at each step, try all possible jump lengths to see if we can get closer to the goal.

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

Begin at the starting position.
From the current position, explore all possible jump lengths allowed.
For each possible jump, check if we land on a valid position (within the bounds).
If the jump leads to the end, we have found a solution.
If the jump leads to a position that is not the end, repeat the process of exploring possible jump lengths from that new position.
If a path does not lead to the end, simply abandon that path and try another one.
Continue exploring all possible paths until either a path to the end is found or all paths have been exhausted.

Code Implementation

def can_reach_end_brute_force(heights, jump_sizes):
    def solve(current_index):
        # If at the end, we found a solution
        if current_index == len(heights) - 1:
            return True

        # Iterate through all possible jump sizes.
        for jump in jump_sizes:
            next_index = current_index + jump

            # Ensure the new index is within bounds
            if 0 <= next_index < len(heights):

                # Recursively check if the end can be reached from the new index.
                if solve(next_index):
                    return True

        return False

    # Initiate recursion, beginning at the start
    return solve(0)

Big(O) Analysis

Time Complexity

O(k^n) – The provided solution explores all possible paths through the array. At each position, there are 'k' possible jump lengths (where k is the size of the jumps array). Since we can potentially visit almost all 'n' positions in the input array, and from each position we are branching into k paths, the worst-case scenario involves exploring every combination of jumps. This creates a branching factor of k at each step, and can reach a depth of n, leading to a time complexity of O(k^n), where k represents the number of possible jumps at each position and n is the number of elements in the input array.

Space Complexity

O(N) – The algorithm explores all possible paths using a recursive approach. In the worst-case scenario, the recursion depth could reach N, where N is the length of the input array, as each jump explores a new index. Each recursive call consumes stack space. Therefore, the maximum size of the call stack would be proportional to N, leading to an auxiliary space complexity of O(N).

Optimal Solution

Approach

The key to solving this puzzle efficiently is to avoid recomputing information. We want to efficiently keep track of what positions can be reached at each step, using information from previous steps to expand our reach without redundant checks.

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

Start at the beginning position and mark it as reachable.
Look at all the positions we can reach from the start and explore moving to positions with the same value nearby, both forward and backward.
Whenever we move, mark that new position as reachable.
Once we've moved to all positions with the same value, avoid rechecking those positions again.
After exploring reachable positions based on value, explore moving based on jump length.
Repeat the process by looking at the next set of reachable positions and explore new reachable positions based on same value and jump length, being careful to only explore positions we haven't reached before.
Continue until you reach the end position, or there are no new reachable positions to explore.
If you reach the end position, you can jump to the last index; otherwise, you cannot.

Code Implementation

def can_reach_end(array):
    array_length = len(array)
    reachable_positions = [False] * array_length
    reachable_positions[0] = True

    queue = [0]
    visited_values = {}

    while queue:
        current_index = queue.pop(0)
        current_value = array[current_index]

        # Explore positions with the same value
        if current_value not in visited_values:
            for index, value in enumerate(array):
                if value == current_value and not reachable_positions[index]:
                    reachable_positions[index] = True
                    queue.append(index)
            visited_values[current_value] = True

        # Explore positions based on jump length
        jump_length = array[current_index]

        # Move forward
        forward_index = current_index + jump_length
        if 0 <= forward_index < array_length and not reachable_positions[forward_index]:
            reachable_positions[forward_index] = True
            queue.append(forward_index)

            # Key step: Check if the end can be jumped to.
            if forward_index == array_length - 1:
                return True

        # Move backward
        backward_index = current_index - jump_length
        if 0 <= backward_index < array_length and not reachable_positions[backward_index]:
            reachable_positions[backward_index] = True
            queue.append(backward_index)
            # Key step: Check if the end can be jumped to.
            if backward_index == array_length - 1:
                return True

    # Check if the last index is reachable
    return reachable_positions[array_length - 1]

Big(O) Analysis

Time Complexity

O(n) – The algorithm's time complexity is O(n) because each index of the input array arr of size n is visited at most a constant number of times. The first source of operations involves exploring adjacent elements with the same value, which each element is only added and removed from the queue once in the worst case. The second source of operations comes from jumping forward and backward k steps which takes O(1) time per reachable index. Each jump will result in marking the new index as visited meaning each index will have constant operations performed on it.

Space Complexity

O(N) – The algorithm uses a queue to keep track of reachable positions and a boolean array of size N to mark visited positions. The queue, in the worst-case scenario, can contain all N indices of the input array, and the visited array also requires N space. Therefore, the auxiliary space is proportional to the input size N, resulting in O(N) space complexity.

Edge Cases

Case	How to Handle
Null or empty input array	Return false immediately as no jumps are possible.
Single element array with zero value	Return true as we are already at the end.
Single element array with non-zero value	Return true if the array length is 1, regardless of the single element's value since we are at the end.
Array where first element is zero and length > 1	Return false, since we can't start at index 0 and jump with zero length.
Array where it's impossible to reach the end (e.g., [3,2,1,0,4])	The algorithm should eventually return false if it gets stuck at a point where it can't reach the end.
Array with very large numbers that could cause integer overflow if added (though unlikely in this specific problem)	The problem doesn't involve arithmetic operations that could cause integer overflow so no special handling needed, but be aware if multiplying array value and jump length in another version of this problem.
Array with negative numbers (if allowed by the problem description, though unlikely)	If negative numbers represent backward jumps, the algorithm must handle revisiting previous indices, possibly with a visited set to avoid infinite loops.
Maximum-sized input array with minimal jump lengths, requiring many iterations	Ensure the algorithm's time complexity is optimized, likely requiring dynamic programming or a greedy approach to avoid exceeding time limits.

Null or empty input array

How to Handle:

Return false immediately as no jumps are possible.

Single element array with zero value

How to Handle:

Return true as we are already at the end.

Single element array with non-zero value

How to Handle:

Return true if the array length is 1, regardless of the single element's value since we are at the end.

Array where first element is zero and length > 1

How to Handle:

Return false, since we can't start at index 0 and jump with zero length.

Array where it's impossible to reach the end (e.g., [3,2,1,0,4])

How to Handle:

The algorithm should eventually return false if it gets stuck at a point where it can't reach the end.

Array with very large numbers that could cause integer overflow if added (though unlikely in this specific problem)

How to Handle:

The problem doesn't involve arithmetic operations that could cause integer overflow so no special handling needed, but be aware if multiplying array value and jump length in another version of this problem.

Array with negative numbers (if allowed by the problem description, though unlikely)

How to Handle:

If negative numbers represent backward jumps, the algorithm must handle revisiting previous indices, possibly with a visited set to avoid infinite loops.

Maximum-sized input array with minimal jump lengths, requiring many iterations

How to Handle:

Ensure the algorithm's time complexity is optimized, likely requiring dynamic programming or a greedy approach to avoid exceeding time limits.

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 range of values for elements in the input array arr, and what is the range for the integer d?
Can the input array arr be empty, or can d be zero? If so, what should the output be?
If there are multiple ways to reach the end of the array, do I need to return the minimum number of jumps, or is any valid path sufficient?
Are the values in the input array guaranteed to be integers, or could they be floating-point numbers?
If it is not possible to reach the end of the array, what value should I return?

Brute Force Solution

Approach

The core idea is to explore every conceivable path to reach the end. We start at the beginning and at each step, try all possible jump lengths to see if we can get closer to the goal.

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

Begin at the starting position.
From the current position, explore all possible jump lengths allowed.
For each possible jump, check if we land on a valid position (within the bounds).
If the jump leads to the end, we have found a solution.
If the jump leads to a position that is not the end, repeat the process of exploring possible jump lengths from that new position.
If a path does not lead to the end, simply abandon that path and try another one.
Continue exploring all possible paths until either a path to the end is found or all paths have been exhausted.

Code Implementation

def can_reach_end_brute_force(heights, jump_sizes):
    def solve(current_index):
        # If at the end, we found a solution
        if current_index == len(heights) - 1:
            return True

        # Iterate through all possible jump sizes.
        for jump in jump_sizes:
            next_index = current_index + jump

            # Ensure the new index is within bounds
            if 0 <= next_index < len(heights):

                # Recursively check if the end can be reached from the new index.
                if solve(next_index):
                    return True

        return False

    # Initiate recursion, beginning at the start
    return solve(0)

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start at the beginning position and mark it as reachable.
Look at all the positions we can reach from the start and explore moving to positions with the same value nearby, both forward and backward.
Whenever we move, mark that new position as reachable.
Once we've moved to all positions with the same value, avoid rechecking those positions again.
After exploring reachable positions based on value, explore moving based on jump length.
Repeat the process by looking at the next set of reachable positions and explore new reachable positions based on same value and jump length, being careful to only explore positions we haven't reached before.
Continue until you reach the end position, or there are no new reachable positions to explore.
If you reach the end position, you can jump to the last index; otherwise, you cannot.

Code Implementation

def can_reach_end(array):
    array_length = len(array)
    reachable_positions = [False] * array_length
    reachable_positions[0] = True

    queue = [0]
    visited_values = {}

    while queue:
        current_index = queue.pop(0)
        current_value = array[current_index]

        # Explore positions with the same value
        if current_value not in visited_values:
            for index, value in enumerate(array):
                if value == current_value and not reachable_positions[index]:
                    reachable_positions[index] = True
                    queue.append(index)
            visited_values[current_value] = True

        # Explore positions based on jump length
        jump_length = array[current_index]

        # Move forward
        forward_index = current_index + jump_length
        if 0 <= forward_index < array_length and not reachable_positions[forward_index]:
            reachable_positions[forward_index] = True
            queue.append(forward_index)

            # Key step: Check if the end can be jumped to.
            if forward_index == array_length - 1:
                return True

        # Move backward
        backward_index = current_index - jump_length
        if 0 <= backward_index < array_length and not reachable_positions[backward_index]:
            reachable_positions[backward_index] = True
            queue.append(backward_index)
            # Key step: Check if the end can be jumped to.
            if backward_index == array_length - 1:
                return True

    # Check if the last index is reachable
    return reachable_positions[array_length - 1]

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return false immediately as no jumps are possible.
Single element array with zero value	Return true as we are already at the end.
Single element array with non-zero value	Return true if the array length is 1, regardless of the single element's value since we are at the end.
Array where first element is zero and length > 1	Return false, since we can't start at index 0 and jump with zero length.
Array where it's impossible to reach the end (e.g., [3,2,1,0,4])	The algorithm should eventually return false if it gets stuck at a point where it can't reach the end.
Array with very large numbers that could cause integer overflow if added (though unlikely in this specific problem)	The problem doesn't involve arithmetic operations that could cause integer overflow so no special handling needed, but be aware if multiplying array value and jump length in another version of this problem.
Array with negative numbers (if allowed by the problem description, though unlikely)	If negative numbers represent backward jumps, the algorithm must handle revisiting previous indices, possibly with a visited set to avoid infinite loops.
Maximum-sized input array with minimal jump lengths, requiring many iterations	Ensure the algorithm's time complexity is optimized, likely requiring dynamic programming or a greedy approach to avoid exceeding time limits.

Null or empty input array

How to Handle:

Return false immediately as no jumps are possible.

Single element array with zero value

How to Handle:

Return true as we are already at the end.

Single element array with non-zero value

How to Handle:

Return true if the array length is 1, regardless of the single element's value since we are at the end.

Array where first element is zero and length > 1

How to Handle:

Return false, since we can't start at index 0 and jump with zero length.

Array where it's impossible to reach the end (e.g., [3,2,1,0,4])

How to Handle:

The algorithm should eventually return false if it gets stuck at a point where it can't reach the end.

Array with very large numbers that could cause integer overflow if added (though unlikely in this specific problem)

How to Handle:

Array with negative numbers (if allowed by the problem description, though unlikely)

How to Handle:

If negative numbers represent backward jumps, the algorithm must handle revisiting previous indices, possibly with a visited set to avoid infinite loops.

Maximum-sized input array with minimal jump lengths, requiring many iterations

How to Handle:

Ensure the algorithm's time complexity is optimized, likely requiring dynamic programming or a greedy approach to avoid exceeding time limits.