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Jump Game IV

Hard

Asked by:

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Topics:

ArraysGraphsDynamic Programming

Given an array of integers arr, you are initially positioned at the first index of the array.

In one step you can jump from index i to index:

i + 1 where: i + 1 < arr.length.
i - 1 where: i - 1 >= 0.
j where: arr[i] == arr[j] and i != j.

Return the minimum number of steps to reach the last index of the array.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.

Example 2:

Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You do not need to jump.

Example 3:

Input: arr = [7,6,9,6,9,6,9,7]
Output: 1
Explanation: You can jump directly from index 0 to index 7 which is last index of the array.

Constraints:

1 <= arr.length <= 5 * 10⁴
-10⁸ <= arr[i] <= 10⁸

Try coding it in LeetCode

Jump Game IV

Hard

Asked by:

20 views

Topics:

ArraysGraphsDynamic Programming

Given an array of integers arr, you are initially positioned at the first index of the array.

In one step you can jump from index i to index:

i + 1 where: i + 1 < arr.length.
i - 1 where: i - 1 >= 0.
j where: arr[i] == arr[j] and i != j.

Return the minimum number of steps to reach the last index of the array.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.

Example 2:

Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You do not need to jump.

Example 3:

Input: arr = [7,6,9,6,9,6,9,7]
Output: 1
Explanation: You can jump directly from index 0 to index 7 which is last index of the array.

Constraints:

1 <= arr.length <= 5 * 10⁴
-10⁸ <= arr[i] <= 10⁸

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 are the constraints on the size of the input array? What is the range of values within the array?
Can the array contain duplicate values, and if so, how should the algorithm handle them?
If it's impossible to reach the last element of the array, what should the function return?
Is the goal to find *any* shortest path, or is there a specific shortest path that needs to be returned if multiple exist?
Is the array guaranteed to have at least one element?

Brute Force Solution

Approach

The brute force strategy for this game involves exploring every conceivable path to reach the end. We will repeatedly jump to every possible location we can jump to, until we hopefully find the end. We will keep track of the number of jumps for each path.

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

Start at the beginning.
Consider all possible jumps we can make from our current position: to the next position, the previous position, and any other location with the same value.
For each of those jumps, imagine taking it and then repeating the process from that new location, considering all possible jumps from there.
Keep doing this until we either reach the end or we realize we're stuck in a loop or have gone too far.
If we reach the end, remember the number of jumps it took to get there.
Repeat this whole process, trying every single possible combination of jumps from the start.
Once we've explored every path, compare the number of jumps for all the paths that reached the end, and choose the path with the fewest jumps.

Code Implementation

def jump_game_iv_brute_force(arr):
    def find_all_indices(value):
        indices = []
        for index, element in enumerate(arr):
            if element == value:
                indices.append(index)
        return indices

    def solve(current_index, current_jumps, visited):
        # Reached the end, update the min jumps if necessary
        if current_index == len(arr) - 1:
            return current_jumps

        if current_index in visited:
            return float('inf')

        visited.add(current_index)

        min_jumps = float('inf')

        # Try jumping to the next index
        if current_index + 1 < len(arr):
            min_jumps = min(min_jumps, solve(current_index + 1, current_jumps + 1, visited.copy()))

        # Try jumping to the previous index
        if current_index - 1 >= 0:
            min_jumps = min(min_jumps, solve(current_index - 1, current_jumps + 1, visited.copy()))

        # Try jumping to all other indices with the same value
        same_value_indices = find_all_indices(arr[current_index])

        for next_index in same_value_indices:
            # Avoid jumping to the same index
            if next_index != current_index:
                min_jumps = min(min_jumps, solve(next_index, current_jumps + 1, visited.copy()))

        return min_jumps

    # Visited is used to avoid infinite recursion
    visited_set = set()
    result = solve(0, 0, visited_set)

    # If no solution is found, return -1
    if result == float('inf'):
        return -1
    else:
        return result

Big(O) Analysis

Time Complexity

O(n!) – In the brute force approach, we explore every possible jump sequence. At each index, we consider up to three possible moves: forward, backward, and jumps to other indices with the same value. In the worst case, especially if many elements have the same value, we explore almost all permutations of indices. Therefore, the number of paths can grow factorially with the input size n, leading to O(n!) time complexity.

Space Complexity

O(N) – The brute force approach uses recursion to explore all possible jump combinations. The maximum depth of the recursion can be proportional to N, where N is the length of the input array. Each recursive call adds a new frame to the call stack, resulting in auxiliary space used by the recursion stack. In the worst-case, the recursion depth could be N, leading to O(N) space complexity due to the recursion stack. The algorithm also implicitly stores the number of jumps for each path being explored, although this does not change the O(N) complexity in the worst case.

Optimal Solution

Approach

This problem is about finding the fewest jumps needed to reach the end of a list. The trick is to explore outwards in layers, figuring out all the places you can reach with a certain number of jumps before trying more jumps.

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

Start at the beginning of the list and consider this jump zero.
Find all the positions you can reach from your current position with one jump. This includes jumping to the next spot, the previous spot, and any other spot in the list that has the same value as your current spot. Remember to mark these spots as reachable in one jump.
Now, for each of those reachable spots, find all the new places you can reach with another jump. Again, this includes the next spot, the previous spot, and all other spots with the same value. Mark these as reachable in two jumps.
Keep doing this outward exploration, increasing the jump count each time, until you reach the end of the list. The number of jumps it took to reach the end is your answer.
A crucial optimization: once you've explored all possible jumps from positions with the same value, don't bother looking at those positions again. Remove them so you don't waste time rechecking them. This will dramatically speed up your solution.

Code Implementation

def jump_game_iv(arr):
    array_length = len(arr)
    if array_length <= 1:
        return 0

    # Store indices with the same value
    value_indices = {}
    for index, value in enumerate(arr):
        if value not in value_indices:
            value_indices[value] = []
        value_indices[value].append(index)

    queue = [0]
    visited = {0}
    jump_count = 0

    while queue:
        next_level = []

        for current_index in queue:
            if current_index == array_length - 1:
                return jump_count

            # Add neighbors to the queue.
            neighbors = []
            neighbors.append(current_index + 1)
            neighbors.append(current_index - 1)

            current_value = arr[current_index]
            if current_value in value_indices:
                neighbors.extend(value_indices[current_value])

                # Optimization:
                # remove the list of indices after visiting
                del value_indices[current_value]

            for neighbor_index in neighbors:
                if 0 <= neighbor_index < array_length and neighbor_index not in visited:
                    next_level.append(neighbor_index)
                    visited.add(neighbor_index)

        queue = next_level
        jump_count += 1

    return -1

Big(O) Analysis

Time Complexity

O(n) – The algorithm uses a breadth-first search (BFS) approach. In the worst case, each element of the array is visited and enqueued into the queue once. When processing each element, the algorithm checks at most three neighbors (next, previous, and same value indices). The optimization to remove same-value indices ensures that each index is considered for same-value jumps only once. Therefore, the total number of operations is proportional to the number of elements in the array, resulting in a time complexity of O(n).

Space Complexity

O(N) – The algorithm uses a queue to store the indices to visit during the breadth-first search. In the worst-case scenario, where almost all indices are reachable with the same number of jumps, the queue might contain close to N indices, where N is the length of the input array. Additionally, a hash map is utilized to store indices grouped by their values to avoid revisiting the indices that have the same value. In the worst case, this hash map might also store all the indices, contributing to O(N) space. Therefore, the auxiliary space complexity is O(N).

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately, as no jumps are possible from an empty array.
Array with only one element	Return 0, because the starting position is also the ending position and no jumps are needed.
Array with all identical values	The algorithm should still work correctly, jumping between all the identical values using the hashmap and array traversal.
Large input array causing memory issues	Consider using iterative BFS instead of recursive DFS to avoid stack overflow and optimize memory usage.
No path to the end of the array (unreachable end)	Return -1 if BFS finishes exploring all reachable nodes without finding the last element index.
Integer overflow when calculating indices (though not highly likely)	Ensure the jumps within the bounds of the array using comparison.
Input array with large number of duplicate values.	The algorithm's performance is impacted, but the algorithm will still correctly compute the minimum jumps. Consider clearing the list of indices after processing similar values.
Array where only the first and last elements are identical.	The standard BFS approach should naturally find a path (if one exists) and this case does not require special consideration.

Null or empty input array

How to Handle:

Return 0 immediately, as no jumps are possible from an empty array.

Array with only one element

How to Handle:

Return 0, because the starting position is also the ending position and no jumps are needed.

Array with all identical values

How to Handle:

The algorithm should still work correctly, jumping between all the identical values using the hashmap and array traversal.

Large input array causing memory issues

How to Handle:

Consider using iterative BFS instead of recursive DFS to avoid stack overflow and optimize memory usage.

No path to the end of the array (unreachable end)

How to Handle:

Return -1 if BFS finishes exploring all reachable nodes without finding the last element index.

Integer overflow when calculating indices (though not highly likely)

How to Handle:

Ensure the jumps within the bounds of the array using comparison.

Input array with large number of duplicate values.

How to Handle:

The algorithm's performance is impacted, but the algorithm will still correctly compute the minimum jumps. Consider clearing the list of indices after processing similar values.

Array where only the first and last elements are identical.

How to Handle:

The standard BFS approach should naturally find a path (if one exists) and this case does not require special consideration.

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 are the constraints on the size of the input array? What is the range of values within the array?
Can the array contain duplicate values, and if so, how should the algorithm handle them?
If it's impossible to reach the last element of the array, what should the function return?
Is the goal to find *any* shortest path, or is there a specific shortest path that needs to be returned if multiple exist?
Is the array guaranteed to have at least one element?

Brute Force Solution

Approach

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

Start at the beginning.
Consider all possible jumps we can make from our current position: to the next position, the previous position, and any other location with the same value.
For each of those jumps, imagine taking it and then repeating the process from that new location, considering all possible jumps from there.
Keep doing this until we either reach the end or we realize we're stuck in a loop or have gone too far.
If we reach the end, remember the number of jumps it took to get there.
Repeat this whole process, trying every single possible combination of jumps from the start.
Once we've explored every path, compare the number of jumps for all the paths that reached the end, and choose the path with the fewest jumps.

Code Implementation

def jump_game_iv_brute_force(arr):
    def find_all_indices(value):
        indices = []
        for index, element in enumerate(arr):
            if element == value:
                indices.append(index)
        return indices

    def solve(current_index, current_jumps, visited):
        # Reached the end, update the min jumps if necessary
        if current_index == len(arr) - 1:
            return current_jumps

        if current_index in visited:
            return float('inf')

        visited.add(current_index)

        min_jumps = float('inf')

        # Try jumping to the next index
        if current_index + 1 < len(arr):
            min_jumps = min(min_jumps, solve(current_index + 1, current_jumps + 1, visited.copy()))

        # Try jumping to the previous index
        if current_index - 1 >= 0:
            min_jumps = min(min_jumps, solve(current_index - 1, current_jumps + 1, visited.copy()))

        # Try jumping to all other indices with the same value
        same_value_indices = find_all_indices(arr[current_index])

        for next_index in same_value_indices:
            # Avoid jumping to the same index
            if next_index != current_index:
                min_jumps = min(min_jumps, solve(next_index, current_jumps + 1, visited.copy()))

        return min_jumps

    # Visited is used to avoid infinite recursion
    visited_set = set()
    result = solve(0, 0, visited_set)

    # If no solution is found, return -1
    if result == float('inf'):
        return -1
    else:
        return result

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

Start at the beginning of the list and consider this jump zero.
Find all the positions you can reach from your current position with one jump. This includes jumping to the next spot, the previous spot, and any other spot in the list that has the same value as your current spot. Remember to mark these spots as reachable in one jump.
Now, for each of those reachable spots, find all the new places you can reach with another jump. Again, this includes the next spot, the previous spot, and all other spots with the same value. Mark these as reachable in two jumps.
Keep doing this outward exploration, increasing the jump count each time, until you reach the end of the list. The number of jumps it took to reach the end is your answer.
A crucial optimization: once you've explored all possible jumps from positions with the same value, don't bother looking at those positions again. Remove them so you don't waste time rechecking them. This will dramatically speed up your solution.

Code Implementation

def jump_game_iv(arr):
    array_length = len(arr)
    if array_length <= 1:
        return 0

    # Store indices with the same value
    value_indices = {}
    for index, value in enumerate(arr):
        if value not in value_indices:
            value_indices[value] = []
        value_indices[value].append(index)

    queue = [0]
    visited = {0}
    jump_count = 0

    while queue:
        next_level = []

        for current_index in queue:
            if current_index == array_length - 1:
                return jump_count

            # Add neighbors to the queue.
            neighbors = []
            neighbors.append(current_index + 1)
            neighbors.append(current_index - 1)

            current_value = arr[current_index]
            if current_value in value_indices:
                neighbors.extend(value_indices[current_value])

                # Optimization:
                # remove the list of indices after visiting
                del value_indices[current_value]

            for neighbor_index in neighbors:
                if 0 <= neighbor_index < array_length and neighbor_index not in visited:
                    next_level.append(neighbor_index)
                    visited.add(neighbor_index)

        queue = next_level
        jump_count += 1

    return -1

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Null or empty input array	Return 0 immediately, as no jumps are possible from an empty array.
Array with only one element	Return 0, because the starting position is also the ending position and no jumps are needed.
Array with all identical values	The algorithm should still work correctly, jumping between all the identical values using the hashmap and array traversal.
Large input array causing memory issues	Consider using iterative BFS instead of recursive DFS to avoid stack overflow and optimize memory usage.
No path to the end of the array (unreachable end)	Return -1 if BFS finishes exploring all reachable nodes without finding the last element index.
Integer overflow when calculating indices (though not highly likely)	Ensure the jumps within the bounds of the array using comparison.
Input array with large number of duplicate values.	The algorithm's performance is impacted, but the algorithm will still correctly compute the minimum jumps. Consider clearing the list of indices after processing similar values.
Array where only the first and last elements are identical.	The standard BFS approach should naturally find a path (if one exists) and this case does not require special consideration.

Null or empty input array

How to Handle:

Return 0 immediately, as no jumps are possible from an empty array.

Array with only one element

How to Handle:

Return 0, because the starting position is also the ending position and no jumps are needed.

Array with all identical values

How to Handle:

The algorithm should still work correctly, jumping between all the identical values using the hashmap and array traversal.

Large input array causing memory issues

How to Handle:

Consider using iterative BFS instead of recursive DFS to avoid stack overflow and optimize memory usage.

No path to the end of the array (unreachable end)

How to Handle:

Return -1 if BFS finishes exploring all reachable nodes without finding the last element index.

Integer overflow when calculating indices (though not highly likely)

How to Handle:

Ensure the jumps within the bounds of the array using comparison.

Input array with large number of duplicate values.

How to Handle:

The algorithm's performance is impacted, but the algorithm will still correctly compute the minimum jumps. Consider clearing the list of indices after processing similar values.

Array where only the first and last elements are identical.

How to Handle:

The standard BFS approach should naturally find a path (if one exists) and this case does not require special consideration.