Path With Minimum Effort

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Binary SearchGraphs

You are a hiker preparing for an upcoming hike. You are given heights, a 2D array of size rows x columns, where heights[row][col] represents the height of cell (row, col). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move up, down, left, or right, and you wish to find a route that requires the minimum effort.

A route's effort is the maximum absolute difference in heights between two consecutive cells of the route.

Return the minimum effort required to travel from the top-left cell to the bottom-right cell.

Example 1:

Input: heights = [[1,2,2],[3,8,2],[5,3,5]]
Output: 2
Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.

Example 2:

Input: heights = [[1,2,3],[3,8,4],[5,3,5]]
Output: 1
Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].

Example 3:

Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
Output: 0
Explanation: This route does not require any effort.

Constraints:

rows == heights.length
columns == heights[i].length
1 <= rows, columns <= 100
1 <= heights[i][j] <= 10^6

Path With Minimum Effort

Medium

Google

4 views

Topics:

Binary SearchGraphs

You are a hiker preparing for an upcoming hike. You are given heights, a 2D array of size rows x columns, where heights[row][col] represents the height of cell (row, col). You are situated in the top-left cell, (0, 0), and you hope to travel to the bottom-right cell, (rows-1, columns-1) (i.e., 0-indexed). You can move up, down, left, or right, and you wish to find a route that requires the minimum effort.

A route's effort is the maximum absolute difference in heights between two consecutive cells of the route.

Return the minimum effort required to travel from the top-left cell to the bottom-right cell.

Example 1:

Input: heights = [[1,2,2],[3,8,2],[5,3,5]]
Output: 2
Explanation: The route of [1,3,5,3,5] has a maximum absolute difference of 2 in consecutive cells.
This is better than the route of [1,2,2,2,5], where the maximum absolute difference is 3.

Example 2:

Input: heights = [[1,2,3],[3,8,4],[5,3,5]]
Output: 1
Explanation: The route of [1,2,3,4,5] has a maximum absolute difference of 1 in consecutive cells, which is better than route [1,3,5,3,5].

Example 3:

Input: heights = [[1,2,1,1,1],[1,2,1,2,1],[1,2,1,2,1],[1,2,1,2,1],[1,1,1,2,1]]
Output: 0
Explanation: This route does not require any effort.

Constraints:

rows == heights.length
columns == heights[i].length
1 <= rows, columns <= 100
1 <= heights[i][j] <= 10^6

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 dimensions of the `heights` matrix? What are the minimum and maximum possible values for `m` and `n`?
What is the range of possible height values within the `heights` matrix? Are the heights non-negative integers?
If there are multiple paths with the same minimum effort, does it matter which path I return the effort for?
If it's impossible to reach the bottom right cell from the top left cell, what value should I return?
Are we only allowed to move up, down, left and right, or can we move diagonally?

Brute Force Solution

Approach

The brute force approach to finding the path with minimum effort involves exploring every possible route from the starting point to the destination. We will consider each path, calculate the effort required for it, and then choose the path with the least effort. It's like trying every single road in a city to find the easiest one to drive.

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

Start at the beginning location.
From your current location, consider all possible neighboring locations you can move to.
For each of these neighboring locations, calculate the 'effort' it would take to move there from your current location (think of 'effort' as how difficult the step is).
Now, for each of those possible next locations, repeat the process: consider all of *their* neighbors and calculate the effort to move to them.
Keep doing this, exploring every possible path, until you reach the destination.
As you explore each path, remember the total effort required for that entire path.
Once you've explored every possible path from start to finish, compare the total effort of each path.
Select the path that had the lowest total effort; that's your path with minimum effort.

Code Implementation

def path_with_minimum_effort_brute_force(heights):
    rows = len(heights)
    columns = len(heights[0])
    minimum_effort = float('inf')

    def calculate_effort(path):
        effort = 0
        for i in range(len(path) - 1):
            effort = max(effort, abs(heights[path[i][0]][path[i][1]] - heights[path[i+1][0]][path[i+1][1]])) 
        return effort

    def find_all_paths(row, column, current_path):
        nonlocal minimum_effort

        #If we've reached the destination
        if row == rows - 1 and column == columns - 1:
            current_path.append((row, column))
            effort = calculate_effort(current_path)
            minimum_effort = min(minimum_effort, effort)
            current_path.pop()
            return

        current_path.append((row, column))
        # Mark current cell as visited to avoid cycles
        
        #Explore all possible paths to neighbors
        neighbors = [(row - 1, column), (row + 1, column), (row, column - 1), (row, column + 1)]
        for next_row, next_column in neighbors:
            if 0 <= next_row < rows and 0 <= next_column < columns and (next_row, next_column) not in current_path:
                find_all_paths(next_row, next_column, current_path)

        current_path.pop()

    find_all_paths(0, 0, [])
    return minimum_effort

Big(O) Analysis

Time Complexity

O(4^(m*n)) – The brute force approach explores every possible path from the top-left cell to the bottom-right cell in the m x n grid. At each cell, we can move in (at most) four directions (up, down, left, right). Since we are exploring every possible path, the number of paths can grow exponentially with the size of the grid. In the worst case, we are effectively exploring a quaternary tree where each node has four children, leading to a time complexity of O(4^(m*n)), where m and n are the dimensions of the grid.

Space Complexity

O(2^(rows*cols)) – The brute force approach explores every possible path from the starting point to the destination, potentially visiting each cell multiple times through different paths. The number of possible paths can grow exponentially with the size of the grid, as at each step, multiple branching decisions are made. Each of these paths needs to be tracked (implicitly or explicitly) leading to a tree-like exploration. Therefore, the auxiliary space used is proportional to the number of possible paths, which can be approximated as O(2^(rows*cols)), where rows and cols are dimensions of the height map.

Optimal Solution

Approach

The goal is to find the path from the start to the end that requires the least amount of effort, where effort is the largest height difference you encounter on that path. We will use a strategy similar to finding the shortest route on a map, but instead of distance, we care about the maximum height difference along the path. The process efficiently explores possible paths, prioritizing those with lower effort, until the destination is reached.

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

Create a table that stores the minimum effort needed to reach each location on the grid, starting with a value of zero for the starting location and a very large value for all other locations, indicating that we haven't found a good path there yet.
We will explore the grid, always choosing to go to the location that currently has the smallest known effort. This is like picking the most promising path so far.
When we move to a new location, we calculate the effort it takes to move there from our current spot. This effort is the absolute difference in height between the two locations.
We then consider the maximum effort it took us to get to our current spot and the effort it takes to move to the new location. The larger of these two becomes the potential effort needed to reach the new location.
If this potential effort is smaller than the current minimum effort recorded for the new location, it means we've found a better way to reach that location. We update the table with this new, smaller effort.
Repeat this process of exploring locations and updating the minimum effort table until we reach the final location. This ensures that we have explored the best possible routes from start to finish.
The value in the table for the final location represents the minimum effort required to traverse from the start to the end of the grid.

Code Implementation

import heapq

def path_with_minimum_effort(heights):
    number_of_rows = len(heights)
    number_of_columns = len(heights[0])

    # Initialize effort table with infinity for all cells except the start.
    effort_to_reach = [[float('inf')] * number_of_columns for _ in range(number_of_rows)]
    effort_to_reach[0][0] = 0

    priority_queue = [(0, 0, 0)]  # (effort, row, col)
    directions = [(0, 1), (0, -1), (1, 0), (-1, 0)]

    while priority_queue:
        current_effort, current_row, current_column = heapq.heappop(priority_queue)

        # If we've reached the destination, return the effort.
        if current_row == number_of_rows - 1 and current_column == number_of_columns - 1:
            return current_effort

        # Skip if current effort is greater than already known minimum effort.
        if current_effort > effort_to_reach[current_row][current_column]:
            continue

        for row_direction, column_direction in directions:
            new_row = current_row + row_direction
            new_column = current_column + column_direction

            if 0 <= new_row < number_of_rows and 0 <= new_column < number_of_columns:
                # Calculate the effort to move to the neighbor cell.
                new_path_effort = max(current_effort, abs(heights[current_row][current_column] - heights[new_row][new_column]))

                # Update the effort table if a better path is found.
                if new_path_effort < effort_to_reach[new_row][new_column]:
                    effort_to_reach[new_row][new_column] = new_path_effort

                    # Add the new cell to the priority queue for further exploration.
                    heapq.heappush(priority_queue, (new_path_effort, new_row, new_column))

    return effort_to_reach[number_of_rows - 1][number_of_columns - 1]

Big(O) Analysis

Time Complexity

O(M * N * log(M * N)) – The algorithm uses a priority queue (or min-heap) to repeatedly select the location with the smallest known effort, which contributes a logarithmic cost. In the worst case, we might add each cell of the M x N grid to the priority queue. Each cell will be processed, and for each cell, we examine at most four neighbors. Adding and removing from the priority queue take O(log(M * N)) time, and we potentially do this for every cell. Therefore, the time complexity is O(M * N * log(M * N)).

Space Complexity

O(M * N) – The algorithm uses a table (or 2D array) to store the minimum effort needed to reach each cell in the grid. The dimensions of this table are the same as the input grid, which is M rows and N columns, where M is the number of rows and N is the number of columns in the input height map. Therefore, the auxiliary space used by the algorithm is proportional to the size of the grid. Thus, the space complexity is O(M * N).

Edge Cases

Case	How to Handle
Empty height matrix	Return -1 or throw an exception as there is no path.
Single-cell height matrix (1x1)	Return 0, as there is no effort required to reach the destination from the start.
Large height matrix, approaching memory limits	Ensure the algorithm's memory usage scales reasonably, potentially using iterative deepening or other memory optimization techniques.
Height matrix with all identical values	The path with minimum effort will have an effort of 0, and the algorithm should correctly find a valid path.
Height matrix with extremely large height differences between adjacent cells	The algorithm should handle large integer differences without overflow, possibly requiring a larger integer type.
A path does not exist to the destination	Return -1 or a specific error value to indicate no path was found after exploring all possibilities.
Integer overflow when calculating effort differences	Use appropriate data types (e.g., long) or modulo operations if applicable, to prevent overflow during effort calculation.
Negative height values in the height matrix	Handle negative height values as per the problem statement if acceptable, otherwise throw exception