Problem Description

You are a hiker planning a route through a terrain represented as a 2D grid. The grid heights has dimensions rows x columns, where heights[row][col] indicates the elevation at position (row, col).

Your journey starts at the top-left corner (0, 0) and your destination is the bottom-right corner (rows-1, columns-1). From any cell, you can move in four directions: up, down, left, or right (no diagonal moves).

The effort of a route is defined as the maximum absolute difference in heights between any two consecutive cells along that route. For example, if you travel through cells with heights [2, 5, 3, 8], the effort would be max(|2-5|, |5-3|, |3-8|) = 5.

Your task is to find the path from start to finish that minimizes this effort value. Return the minimum effort required.

Example visualization:

• If you have a 3x3 grid and travel from (0,0) to (2,2), you might pass through several cells
• The effort for your entire path is determined by the single largest height difference between adjacent cells on your route
• Among all possible paths, you need to find the one where this maximum difference is as small as possible

The problem essentially asks you to find the "smoothest" path through the terrain, where smoothness is measured by avoiding large jumps in elevation between consecutive steps.

Flowchart Walkthrough

First, let's pin down the algorithm using the Flowchart. Here's a step-by-step walkthrough:

Is it a graph?

• Yes: The problem involves a 2D grid where each cell can be treated as a node, and movements between adjacent cells represent edges. We need to find a path from one node (top-left) to another (bottom-right).

Is it a tree?

• No: The grid structure forms a graph with potential cycles (you can move in 4 directions), not a tree structure.

Is the problem related to directed acyclic graphs?

• No: The movement between cells is bidirectional (you can move up/down/left/right), and the graph can contain cycles.

Is the problem related to shortest paths?

• Yes: We're looking for the path with minimum "effort" (though not traditional shortest distance). The effort is defined as the maximum edge weight along the path.

Is the graph weighted?

• Yes: Each edge between adjacent cells has a weight equal to the absolute difference in heights between those cells.

Dijkstra's Algorithm

• The flowchart leads us to Dijkstra's Algorithm for weighted shortest path problems.

Additional Consideration - Connectivity: While the flowchart suggests Dijkstra's, the solution actually uses Union-Find (DSU). This is because the problem can also be viewed as a connectivity problem - finding when the start and end become connected when adding edges in order of increasing weight. The Union-Find approach sorts edges by weight and adds them until start and end are connected, which efficiently solves this "minimum bottleneck path" problem.

Conclusion: The flowchart initially suggests Dijkstra's Algorithm for this weighted shortest path problem. However, the actual solution cleverly reformulates it as a connectivity problem using Union-Find, which is particularly efficient for finding the minimum maximum edge weight path.

Intuition

The key insight is recognizing that we're looking for a path where the maximum edge weight is minimized. This is different from finding the shortest total distance - we care only about the single largest "jump" we have to make along our path.

Think of it this way: if you're hiking and want to minimize effort, you'd prefer a path with many small elevation changes over a path with even one huge cliff to climb, regardless of total distance.

This transforms our problem into finding the "bottleneck" - what's the smallest possible value for the maximum edge we must traverse to connect start to finish?

Here's where the Union-Find approach becomes clever:

1. Sort all edges by weight: We list every possible connection between adjacent cells and sort them by their height difference (effort).

2. Gradually connect cells: Starting with the easiest connections (smallest height differences), we begin connecting cells together into groups.

3. Check connectivity: After adding each edge, we check if the start cell (0,0) and end cell (rows-1, cols-1) belong to the same connected component.

4. The answer emerges: The moment start and end become connected, the weight of the edge we just added is our answer - it's the minimum possible "maximum effort" needed.

Why does this work? By adding edges in order of increasing weight, we ensure that when start and end first become connected, we've used the smallest possible set of edges needed. The largest edge in this set (which is the one we just added) represents the minimum bottleneck for any path between start and end.

This approach is like gradually raising a "water level" - edges with height difference below the water level are traversable. We raise the level until a path from start to finish becomes possible. The final water level is our minimum effort.

Learn more about Depth-First Search, Breadth-First Search, Union Find, Binary Search and Heap (Priority Queue) patterns.

Solution Approach

The implementation uses the Union-Find (Disjoint Set Union) data structure to efficiently track and merge connected components. Let's break down the solution step by step:

1. Union-Find Data Structure Setup

The UnionFind class maintains:

• p: parent array where p[x] points to the parent of node x
• size: array tracking the size of each component for union by rank optimization

Key operations:

• find(x): Returns the root of the component containing x, with path compression for efficiency
• union(a, b): Merges the components containing a and b, using union by rank
• connected(a, b): Checks if a and b are in the same component

2. Graph Construction

For an m × n grid, we:

• Treat each cell (i, j) as a node with index i * n + j (flattening 2D to 1D)
• Create edges between adjacent cells (right and down directions only to avoid duplicates)

dirs = (0, 1, 0)  # Represents (0,1) and (1,0) movements
for i in range(m):
    for j in range(n):
        for a, b in pairwise(dirs):  # Gets pairs: (0,1) and (1,0)
            x, y = i + a, j + b
            if 0 <= x < m and 0 <= y < n:
                # Create edge with weight = height difference
                e.append((abs(heights[i][j] - heights[x][y]), i * n + j, x * n + y))

3. Kruskal's Algorithm Variant

The solution applies a modified Kruskal's algorithm:

1. Sort edges by weight (height difference) in ascending order: e.sort()
2. Process edges one by one, merging components
3. Early termination: Stop as soon as start (0) and end (m * n - 1) are connected

for h, a, b in e:
    uf.union(a, b)  # Merge components containing cells a and b
    if uf.connected(0, m * n - 1):  # Check if start and end are connected
        return h  # Return current edge weight as the answer

4. Why This Works

By processing edges in order of increasing weight, we guarantee that:

• When start and end first become connected, we've used the minimum possible edges
• The current edge weight h is the largest edge needed for connectivity
• This represents the minimum possible "maximum effort" for any path

Time and Space Complexity

• Time: O(E log E) where E = 2mn (number of edges), dominated by sorting
• Space: O(mn) for the Union-Find structure and edge list

The solution elegantly transforms a path-finding problem into a connectivity problem, leveraging Union-Find's efficiency in tracking when two nodes become reachable from each other.

Example Walkthrough

Let's walk through a small example to illustrate the solution approach.

Consider this 3×3 grid:

heights = [[1, 2, 2],
           [3, 8, 2],
           [5, 3, 5]]

We need to find a path from (0,0) to (2,2) that minimizes the maximum height difference between consecutive cells.

Step 1: Create all edges with their weights (height differences)

We'll create edges between adjacent cells. Each cell (i,j) gets an index: i*3 + j.

• Cell (0,0) = index 0
• Cell (0,1) = index 1
• Cell (2,2) = index 8

Edges created:

• (0,0)→(0,1): |1-2| = 1, connects nodes 0→1
• (0,1)→(0,2): |2-2| = 0, connects nodes 1→2
• (0,0)→(1,0): |1-3| = 2, connects nodes 0→3
• (0,1)→(1,1): |2-8| = 6, connects nodes 1→4
• (0,2)→(1,2): |2-2| = 0, connects nodes 2→5
• (1,0)→(1,1): |3-8| = 5, connects nodes 3→4
• (1,1)→(1,2): |8-2| = 6, connects nodes 4→5
• (1,0)→(2,0): |3-5| = 2, connects nodes 3→6
• (1,1)→(2,1): |8-3| = 5, connects nodes 4→7
• (1,2)→(2,2): |2-5| = 3, connects nodes 5→8
• (2,0)→(2,1): |5-3| = 2, connects nodes 6→7
• (2,1)→(2,2): |3-5| = 2, connects nodes 7→8

Step 2: Sort edges by weight

Weight 0: (1→2), (2→5)
Weight 1: (0→1)
Weight 2: (0→3), (3→6), (6→7), (7→8)
Weight 3: (5→8)
Weight 5: (3→4), (4→7)
Weight 6: (1→4), (4→5)

Step 3: Process edges using Union-Find

Initially, each cell is its own component: {0}, {1}, {2}, {3}, {4}, {5}, {6}, {7}, {8}

Processing edges in order:

1. Add edge (1→2) with weight 0: Components become {0}, {1,2}, {3}, {4}, {5}, {6}, {7}, {8}
2. Add edge (2→5) with weight 0: Components become {0}, {1,2,5}, {3}, {4}, {6}, {7}, {8}
3. Add edge (0→1) with weight 1: Components become {0,1,2,5}, {3}, {4}, {6}, {7}, {8}
4. Add edge (0→3) with weight 2: Components become {0,1,2,3,5}, {4}, {6}, {7}, {8}
5. Add edge (3→6) with weight 2: Components become {0,1,2,3,5,6}, {4}, {7}, {8}
6. Add edge (6→7) with weight 2: Components become {0,1,2,3,5,6,7}, {4}, {8}
7. Add edge (7→8) with weight 2: Components become {0,1,2,3,5,6,7,8}, {4}

Now nodes 0 and 8 are connected! The algorithm stops here.

Step 4: Return the answer

The weight of the edge that connected nodes 0 and 8 was 2, so the minimum effort is 2.

This means there exists a path from (0,0) to (2,2) where no consecutive cells differ by more than 2 in height. One such path is: (0,0)→(0,1)→(0,2)→(1,2)→(2,2) with differences [1, 0, 0, 3], but we could also take (0,0)→(1,0)→(2,0)→(2,1)→(2,2) with differences [2, 2, 2, 2], achieving the maximum difference of 2.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m × n × log(m × n))

The time complexity is dominated by the following operations:

• Creating edges: We iterate through all m × n cells, and for each cell, we check at most 2 neighbors (right and down due to the dirs = (0, 1, 0) pattern). This creates approximately 2 × m × n edges, taking O(m × n) time.
• Sorting edges: The edge list contains at most 2 × m × n edges, and sorting them takes O(m × n × log(m × n)) time.
• Union-Find operations: For each edge, we perform union() and connected() operations. With path compression and union by size, each operation has an amortized time complexity of nearly O(1) (specifically O(α(m × n)) where α is the inverse Ackermann function). Processing all edges takes O(m × n × α(m × n)) ≈ O(m × n) time.

The sorting step dominates, giving us a total time complexity of O(m × n × log(m × n)).

Space Complexity: O(m × n)

The space complexity consists of:

• UnionFind data structure: The p and size arrays each store m × n elements, using O(m × n) space.
• Edge list e: Stores at most 2 × m × n edges, where each edge is a tuple of 3 integers, using O(m × n) space.
• Recursion stack for find(): In the worst case (before path compression), the recursion depth could be O(m × n), though path compression reduces this significantly in practice.

The total space complexity is O(m × n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrect Edge Generation - Missing or Duplicate Edges

A common mistake is incorrectly generating edges between adjacent cells, either missing connections or creating duplicates. This often happens when trying to check all four directions.

Pitfall Example:

# Incorrect: Creates duplicate edges
for row in range(rows):
    for col in range(cols):
        # Checking all 4 directions creates duplicates
        for dr, dc in [(0,1), (1,0), (0,-1), (-1,0)]:
            new_row, new_col = row + dr, col + dc
            if 0 <= new_row < rows and 0 <= new_col < cols:
                edges.append((effort, current, neighbor))
# This creates edge (A,B) and later (B,A) - duplicates!

Solution: Only check right and down directions from each cell (as shown in the correct code). This ensures each edge is created exactly once since we're systematically moving through the grid.

2. Incorrect 2D to 1D Index Conversion

When flattening the 2D grid coordinates to 1D indices for Union-Find, using the wrong formula is a frequent error.

Pitfall Example:

# Incorrect formulas:
index = row * rows + col  # Wrong! Should use cols, not rows
index = col * cols + row  # Wrong! Reversed row and col
index = row + col * cols  # Wrong! Order matters

Solution: Always use index = row * cols + col for row-major ordering. Remember: you multiply the row by the number of columns (width), not rows (height).

3. Forgetting Edge Cases - Single Cell Grid

When the grid is 1×1, there are no edges to process, and the loop never executes. If not handled, this could lead to incorrect results.

Pitfall Example:

def minimumEffortPath(self, heights):
    # ... edge generation and sorting ...
  
    for effort, cell_a, cell_b in edges:
        union_find.union(cell_a, cell_b)
        if union_find.connected(0, rows * cols - 1):
            return effort
  
    # If edges is empty (1x1 grid), this line is never reached!
    # Missing return statement

Solution: Add a return statement at the end: return 0. For a single cell, the start and end are the same, so the effort is 0.

4. Using BFS/DFS Instead of Union-Find

While BFS/DFS with binary search can solve this problem, implementing it correctly is more complex and error-prone.

Pitfall Example:

# Attempting BFS with priority queue - more complex and slower
def minimumEffortPath(self, heights):
    # Using Dijkstra's algorithm - correct but more complicated
    pq = [(0, 0, 0)]  # (effort, row, col)
    visited = set()
    # ... complex logic with priority queue ...

Solution: The Union-Find approach with Kruskal's algorithm is cleaner and more efficient for this specific problem since we're looking for the minimum bottleneck path.

5. Incorrect Union-Find Implementation

Forgetting path compression or union by rank can lead to poor performance.

Pitfall Example:

def find(self, x):
    # No path compression - O(n) worst case
    while self.parent[x] != x:
        x = self.parent[x]
    return x

def union(self, a, b):
    # No union by rank - can create unbalanced trees
    root_a, root_b = self.find(a), self.find(b)
    self.parent[root_a] = root_b  # Always attach a to b

Solution: Always implement both optimizations:

• Path compression in find(): self.parent[x] = self.find(self.parent[x])
• Union by rank/size in union(): attach smaller tree to larger tree