Shift 2D Grid Leetcode Problem 1260 [Python Solution]

If you're new to programming or just getting started with LeetCode, you might come across problems that seem a bit challenging at first glance.

The key to tackling these problems is breaking them down step by step, understanding the problem statement, and then coming up with an efficient solution.

In this blog post, we'll walk you through the Shift 2D Grid problem on LeetCode, provide you with a Python solution, and explain the reasoning behind our approach.

Problem Overview

The Shift 2D Grid problem on LeetCode is categorized under Math and Geometry.

The problem statement is as follows:

Problem Statement: Given a 2D grid of size m x n and an integer k, you need to shift the grid k times.

In one shift operation:
1. Element at grid[i][j] moves to grid[i][j + 1].
2. Element at grid[i][n – 1] moves to grid[i + 1][0].
3. Element at grid[m – 1][n – 1] moves to grid[0][0].

Your task is to return the 2D grid after applying the shift operation k times.

Example 1:

Input:

grid = [[1,2,3],[4,5,6],[7,8,9]]
k = 1


Output:

[[9,1,2],[3,4,5],[6,7,8]]


Example 2:

Input:

grid = [[3,8,1,9],[19,7,2,5],[4,6,11,10],[12,0,21,13]]
k = 4


Output:

[[12,0,21,13],[3,8,1,9],[19,7,2,5],[4,6,11,10]]


Example 3:

Input:

grid = [[1,2,3],[4,5,6],[7,8,9]]
k = 9


Output:

[[1,2,3],[4,5,6],[7,8,9]]


Constraints:
– m == grid.length
– n == grid[i].length
– 1 <= m <= 50
– 1 <= n <= 50
– -1000 <= grid[i][j] <= 1000
– 0 <= k <= 100

Now that we understand the problem statement and constraints, let's move on to the solution.

Efficient Python Code Solution

def shiftGrid(grid, k):
    M, N = len(grid), len(grid[0])

    def posToVal(r, c):
        return r * N + c

    def valToPos(v):
        return [v // N, v % N]

    res = [[0] * N for i in range(M)]
    for r in range(M):
        for c in range(N):
            newVal = (posToVal(r, c) + k) % (M * N)
            newR, newC = valToPos(newVal)
            res[newR][newC] = grid[r][c]
    return res

Now, let's break down the code and understand how it works.

Approach Explanation

The idea behind this solution is to treat the 2D grid as a one-dimensional array for the purpose of shifting.

We will convert the two-dimensional coordinates into one-dimensional indices, perform the shifts, and then convert them back into two-dimensional coordinates to construct the final grid.

Here's how our approach works step by step:

1. Get the dimensions of the grid, where M is the number of rows, and N is the number of columns.

2. Define two helper functions:

   • posToVal(r, c) converts a row and column pair into a one-dimensional index.
   • valToPos(v) converts a one-dimensional index back into row and column coordinates.

3. Create a result grid of the same dimensions as the input grid, initialized with zeros.

4. Iterate through every position in the original grid (row by row and column by column).

5. Calculate the new one-dimensional index for each position after shifting.

We add k to the current position's index and take the modulo of (M * N) to ensure it stays within bounds.

1. Convert the new one-dimensional index back into two-dimensional coordinates using the valToPos function.

2. Update the corresponding position in the result grid with the value from the original grid.

3. Return the result grid as the final answer.

The time complexity of this solution is O(M * N) because we iterate through every position in the grid.

The space complexity is also O(M * N) since we create a new grid to store the result.

Reasoning Behind Our Efficient Approach

This approach simplifies the problem by treating the 2D grid as a one-dimensional array, making the shifting operation straightforward.

We use helper functions to convert between one-dimensional and two-dimensional coordinates, which helps us keep the code clean and understandable.

The use of modular arithmetic ensures that we handle out-of-bounds situations correctly, making the solution robust.

Overall, this approach is both efficient and concise.

Time and Space Complexity

• Time Complexity: O(M * N)
  • We iterate through every position in the grid once, where M and N are the dimensions of the grid.
• Space Complexity: O(M * N)
  • We create a result grid of the same dimensions as the input grid, which requires additional space.

Related Interview Questions By Company:

• Contains Duplicate II LeetCode
• Majority Element LeetCode
• Find The Index Of The First Occurrence In A String LeetCode

Related Interview Questions By Difficulty:

• Roman To Integer LeetCode
• Palindrome Number LeetCode
• Shift 2D Grid LeetCode

Related Interview Questions By Category:

• Combination Sum Iv LeetCode
• Find Critical And Pseudo Critical Edges In Minimum Spanning Tree
• Design Browser History LeetCode

Edge Cases a Valid Solution Must Consider

While solving the Shift 2D Grid problem, it's essential to consider edge cases and constraints to ensure your solution works correctly.

Here are some key points to keep in mind:

1. Handling the case where k is equal to or greater than the total number of elements in the grid.
2. Handling grids with a single row or single column, as these have special shift patterns.
3. Dealing with negative values for k.
4. Ensuring that the code works efficiently for the upper constraints of m, n, and k as mentioned in the problem statement.

By considering these edge cases, you can ensure that your solution is robust and handles all possible inputs correctly.

In summary, the Shift 2D Grid problem on LeetCode can be solved efficiently by treating the 2D grid as a one-dimensional array, using modular arithmetic to handle shifts, and converting between one-dimensional and two-dimensional coordinates.

The provided Python solution offers a clear and effective approach to solving this problem.

If you have any questions, suggestions, or comments, please feel free to share them in the comments section.

We hope you found this guide helpful, and we encourage you to try solving the problem on your own as well.

Learning through practice is an excellent way to improve your coding skills.

Happy coding!

Question Link

We encourage you to comment, ask questions, make suggestions, and share this content to help others in their coding journey.