Problem Description

You are given a 2D matrix grid of size m x n. Your task is to check whether the entire grid satisfies two specific conditions for each cell:

1. Vertical Condition: Each cell must be equal to the cell directly below it (if such a cell exists). In other words, grid[i][j] should equal grid[i + 1][j].

2. Horizontal Condition: Each cell must be different from the cell directly to its right (if such a cell exists). In other words, grid[i][j] should not equal grid[i][j + 1].

The problem asks you to return true if all cells in the grid satisfy both conditions, and false if any cell violates either condition.

For example, consider a grid where:

• All values in the same column are identical (satisfying the vertical condition)
• Adjacent columns have different values (satisfying the horizontal condition)

Such a grid would return true.

The solution iterates through each cell in the grid and checks:

• If there's a cell below (i + 1 < m), it verifies that the current cell equals the cell below
• If there's a cell to the right (j + 1 < n), it verifies that the current cell differs from the cell to the right

If any cell fails either check, the function immediately returns false. If all cells pass both checks, it returns true.

Intuition

The key insight is that this is a straightforward validation problem - we need to verify that every cell in the grid follows two simple rules. There's no need for complex algorithms or data structures; we just need to check each cell systematically.

Think about what the conditions actually mean for the grid structure:

• The first condition (each cell equals the cell below it) means that all values in a column must be the same
• The second condition (each cell differs from the cell to its right) means that adjacent columns must have different values

This suggests the grid should look like vertical stripes, where each stripe (column) has a uniform value, and neighboring stripes have different values.

The most direct approach is to simply iterate through each cell and verify both conditions. We don't need to check every relationship twice - when we're at cell grid[i][j], we only need to:

• Check if it matches the cell below (if one exists)
• Check if it differs from the cell to the right (if one exists)

By checking "forward" relationships (down and right) rather than all four directions, we avoid redundant comparisons. Each relationship between adjacent cells is checked exactly once.

The early return strategy is efficient here - as soon as we find any cell that violates either condition, we know the entire grid fails the test, so we can immediately return false. Only if we complete the entire traversal without finding violations can we confidently return true.

Solution Approach

The solution uses a simple simulation approach, iterating through each cell in the grid and checking the two required conditions.

Implementation Steps:

1. Get grid dimensions: First, we extract the dimensions of the grid - m rows and n columns using len(grid) and len(grid[0]).

2. Iterate through each cell: We use nested loops with enumerate() to traverse the grid. The outer loop iterates through rows (index i), and the inner loop iterates through columns (index j). The enumerate() function gives us both the index and the value x at each position.

3. Check vertical condition: For each cell at position (i, j), we check if there's a cell below it by verifying i + 1 < m. If such a cell exists, we compare the current cell's value x with grid[i + 1][j]. If they're not equal, we immediately return False since the condition is violated.

4. Check horizontal condition: Similarly, we check if there's a cell to the right by verifying j + 1 < n. If such a cell exists, we compare the current cell's value x with grid[i][j + 1]. If they're equal, we return False since the condition requires them to be different.

5. Return result: If we complete the entire traversal without finding any violations, all cells satisfy both conditions, so we return True.

Key Implementation Details:

• The boundary checks (i + 1 < m and j + 1 < n) ensure we don't access indices outside the grid boundaries
• The use of early return (return False) as soon as a violation is found makes the solution efficient
• The conditions are checked in the exact form specified: equality for vertical neighbors and inequality for horizontal neighbors

Time Complexity: O(m × n) where m is the number of rows and n is the number of columns, as we visit each cell exactly once.

Space Complexity: O(1) as we only use a constant amount of extra space for variables.

Example Walkthrough

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

Consider this 3×4 grid:

grid = [[1, 2, 1, 2],
        [1, 2, 1, 2],
        [1, 2, 1, 2]]

We'll iterate through each cell and check both conditions:

Step 1: Cell (0,0) with value 1

• Check below: grid[0][0] = 1, grid[1][0] = 1 ✓ (equal)
• Check right: grid[0][0] = 1, grid[0][1] = 2 ✓ (different)

Step 2: Cell (0,1) with value 2

• Check below: grid[0][1] = 2, grid[1][1] = 2 ✓ (equal)
• Check right: grid[0][1] = 2, grid[0][2] = 1 ✓ (different)

Step 3: Cell (0,2) with value 1

• Check below: grid[0][2] = 1, grid[1][2] = 1 ✓ (equal)
• Check right: grid[0][2] = 1, grid[0][3] = 2 ✓ (different)

Step 4: Cell (0,3) with value 2

• Check below: grid[0][3] = 2, grid[1][3] = 2 ✓ (equal)
• Check right: No cell to the right (j+1 = 4 >= n) - skip this check

Step 5: Cell (1,0) with value 1

• Check below: grid[1][0] = 1, grid[2][0] = 1 ✓ (equal)
• Check right: grid[1][0] = 1, grid[1][1] = 2 ✓ (different)

We continue this process for all remaining cells. Notice that:

• Cells in the last row (row 2) don't need vertical checks since there's no row below
• Cells in the last column (column 3) don't need horizontal checks since there's no column to the right

In this example, all cells satisfy both conditions:

• Each column has the same value throughout (column 0 has all 1s, column 1 has all 2s, etc.)
• Adjacent columns have different values (1→2→1→2)

The function returns True.

Counter-example: If we had a grid like:

grid = [[1, 1],
        [2, 1]]

At cell (0,0) with value 1:

• Check below: grid[0][0] = 1, grid[1][0] = 2 ✗ (not equal)

The function would immediately return False without checking other cells.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m × n)

The algorithm iterates through every element in the grid exactly once using nested loops. The outer loop runs m times (number of rows), and for each row, the inner loop runs n times (number of columns). For each element at position (i, j), the algorithm performs constant-time operations:

• Checking if there exists a cell below: i + 1 < m and comparing values x != grid[i + 1][j]
• Checking if there exists a cell to the right: j + 1 < n and comparing values x == grid[i][j + 1]

Since we visit each of the m × n cells once and perform O(1) operations per cell, the total time complexity is O(m × n).

Space Complexity: O(1)

The algorithm uses only a constant amount of extra space:

• Variables m and n to store the dimensions of the grid
• Loop variables i, j, and x from the enumerate functions
• The row reference variable

These variables don't scale with the input size, as no additional data structures like arrays, lists, or recursive call stacks are created. The algorithm operates directly on the input grid without requiring auxiliary space proportional to the input size, resulting in O(1) space complexity.

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

Common Pitfalls

1. Incorrect Boundary Checking Order

A common mistake is checking the cell values before verifying if the neighboring cell exists within bounds. This leads to IndexError exceptions.

Incorrect approach:

# Wrong - checking value before boundary
if grid[i][j] != grid[i + 1][j] and i + 1 < rows:
    return False

Correct approach:

# Right - checking boundary first
if i + 1 < rows and grid[i][j] != grid[i + 1][j]:
    return False

The short-circuit evaluation in Python ensures that if i + 1 < rows is False, the second condition won't be evaluated, preventing index errors.

2. Confusing the Conditions

Another frequent error is mixing up which condition applies to which direction:

• Vertical (same column): Values must be equal
• Horizontal (same row): Values must be different

Incorrect logic:

# Wrong - reversed conditions
if i + 1 < rows and grid[i][j] == grid[i + 1][j]:  # Should be !=
    return False
if j + 1 < cols and grid[i][j] != grid[i][j + 1]:  # Should be ==
    return False

3. Edge Case: Single Cell Grid

While the provided solution handles this correctly, it's worth noting that a grid with dimensions 1x1 should return True since there are no neighbors to violate any conditions. The boundary checks naturally handle this case, but some might overthink and add unnecessary special case handling.

4. Using Wrong Indices After enumerate()

When using enumerate(), confusion can arise about which variable represents what:

Incorrect usage:

for i, row in enumerate(grid):
    for j, val in enumerate(row):
        # Wrong - using i and j incorrectly
        if j + 1 < rows and val != grid[j + 1][i]:  # Mixed up i and j
            return False

Solution: Use descriptive variable names like row_idx and col_idx to avoid confusion, and remember that the first index from enumerate is always the position, not the value.