Shift 2D Grid - Practice Coding Problems

Shift 2D Grid - Problem

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:

Element at grid[i][j] moves to grid[i][j + 1]
Element at grid[i][n - 1] moves to grid[i + 1][0]
Element at grid[m - 1][n - 1] moves to grid[0][0]

Return the 2D grid after applying shift operation k times.

Input & Output

Example 1 — Basic Shift

$ Input: grid = [[1,2,3],[4,5,6]], k = 1

› Output: [[6,1,2],[3,4,5]]

💡 Note: After 1 shift: element 6 moves to [0][0], 1 moves to [0][1], 2 moves to [0][2], 3 moves to [1][0], etc.

Example 2 — Multiple Shifts

$ 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]]

💡 Note: After 4 shifts, the last row moves to the top and all other rows shift down by one position.

Example 3 — Full Cycle

$ Input: grid = [[1,2,3],[4,5,6]], k = 6

› Output: [[1,2,3],[4,5,6]]

💡 Note: After 6 shifts (m×n = 2×3 = 6), all elements return to their original positions.

Constraints

m == grid.length
n == grid[i].length
1 ≤ m ≤ 50
1 ≤ n ≤ 50
-1000 ≤ grid[i][j] ≤ 1000
0 ≤ k ≤ 100

Visualization

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Understanding the Visualization

Input Grid

2×3 grid with elements [1,2,3,4,5,6]

Shift Operation

Each element moves right → next row → wraps to [0][0]

Output Grid

Grid after k=1 shifts

Key Takeaway

🎯 Key Insight: Use modular arithmetic to calculate final positions directly instead of simulating k shifts

Asked in

A Amazon 15 M Microsoft 12

The key insight is to use modular arithmetic instead of simulating shifts. Map 2D grid to 1D positions, calculate new positions with (pos + k) % (m×n), then map back to 2D. Best approach is Mathematical Position Mapping. Time: O(m×n), Space: O(m×n)

Common Approaches

Approach	Time	Space	Notes
✓ Simulate K Shifts	O(k × m × n)	O(1)	Perform the shift operation k times, moving elements one position at a time
Mathematical Position Mapping	O(m × n)	O(m × n)	Calculate final position of each element directly using modular arithmetic

Simulate K Shifts — Algorithm Steps

Repeat k times:
Save the last element (bottom-right)
Shift each element to next position
Place saved element at top-left

Visualization

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Step-by-Step Walkthrough

Original Grid

Start with input grid

Shift 1

Move each element one position right/down

Final Result

Grid after k shifts

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int** solution(int** grid, int gridSize, int* gridColSize, int k, int* returnSize, int** returnColumnSizes) {
    int m = gridSize, n = gridColSize[0];
    *returnSize = m;
    *returnColumnSizes = (int*)malloc(m * sizeof(int));
    
    // Allocate result array
    int** result = (int**)malloc(m * sizeof(int*));
    for (int i = 0; i < m; i++) {
        result[i] = (int*)malloc(n * sizeof(int));
        (*returnColumnSizes)[i] = n;
        for (int j = 0; j < n; j++) {
            result[i][j] = grid[i][j];
        }
    }
    
    for (int shift = 0; shift < k; shift++) {
        // Save the last element
        int last = result[m-1][n-1];
        
        // Shift elements within each row
        for (int i = 0; i < m; i++) {
            for (int j = n-1; j > 0; j--) {
                result[i][j] = result[i][j-1];
            }
        }
        
        // Move last element of each row to first element of next row
        for (int i = m-1; i > 0; i--) {
            result[i][0] = result[i-1][n-1];
        }
        
        // Place saved element at top-left
        result[0][0] = last;
    }
    
    return result;
}

int main() {
    // Simple implementation for demonstration
    int grid[2][3] = {{1,2,3},{4,5,6}};
    int* gridArray[2] = {grid[0], grid[1]};
    int gridColSize[2] = {3, 3};
    int k = 1;
    
    int returnSize;
    int* returnColumnSizes;
    
    int** result = solution(gridArray, 2, gridColSize, k, &returnSize, &returnColumnSizes);
    
    printf("[");
    for (int i = 0; i < returnSize; i++) {
        printf("[");
        for (int j = 0; j < returnColumnSizes[i]; j++) {
            printf("%d", result[i][j]);
            if (j < returnColumnSizes[i] - 1) printf(",");
        }
        printf("]");
        if (i < returnSize - 1) printf(",");
    }
    printf("]\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(k × m × n)

For each of k shifts, we process all m×n elements

✓ Linear Growth

Space Complexity

O(1)

Only using temporary variables for shifting

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Simulate K Shifts — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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