Minimum Obstacle Removal To Reach Corner - Leetcode Solution

The problem statement for Minimum Obstacle Removal to Reach Corner on Leetcode is as follows:

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] becomes at grid[i][j + 1]. Element at grid[i][n - 1] becomes at grid[i + 1][0]. Element at grid[m - 1][n - 1] becomes at grid[0][0]. Return the minimum number of operations required to move all the 1's present in the grid to a single cell.

If it is not possible to move all 1's to a single cell, return -1.

The grid is rectangular and consists of only 0's and 1's.

Examples:

Example 1:

Input: grid = [[0,0,0,0,0,1], [0,0,0,0,1,0], [0,0,0,1,0,0], [0,0,1,0,0,0], [0,1,0,0,0,0], [1,0,0,0,0,0]], k = 2 Output: 6 Explanation: Shifting grid 2 times horizontally and 3 times vertically give the following grid:

0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 1

The cell (6,4) is the ideal target because all the 1's are in that cell.

Example 2:

Input: grid = [[0,1,1,0], [0,1,1,0], [0,0,0,1]], k = 1 Output: -1 Explanation: We need to move the 1's towards a single cell which in this case, is impossible.

Approach:

This problem is an implementation problem. We need to simulate the movement of the grid as mentioned in the problem statement k times and count the number of obstacles we have removed in the process. An obstacle is a cell with a value of 1 in the grid. At the end of shifting the grid k times, we need to ensure that all 1's in the grid are present in a single cell.

If at the end of the simulation, all 1's are present in the same cell, then we return the number of obstacles we have removed. Otherwise, we return -1.

Algorithm:

Initialize a variable called obstacle_count to 0. Set a flag called all_obstacles_together to False. Iterate k times. Within the iteration, iterate through each cell in the grid and apply the three steps that are mentioned in the problem statement to move the grid. We update the cell values in the grid in-place. In each iteration, check if all the obstacles are present in the same cell or not. If they are present in the same cell, then set the all_obstacles_together flag to True and break out of the loop. Otherwise, continue the iterations. Finally, if all_obstacles_together is True, then return obstacle_count. If not, return -1.

Code:

Here's the Python code to solve the problem:

class Solution: def minMoves(self, grid: List[List[int]], k: int) -> int: obstacle_count = 0 all_obstacles_together = False n = len(grid) m = len(grid[0]) for move in range(k): tmp_grid = [[0] * m for _ in range(n)] for i in range(n): for j in range(m): if grid[i][j] == 1: if j+m-1 >= k: continue tmp_grid[i][(j+move) % m] = 1 grid = tmp_grid for i in range(n): if all(grid[i]): all_obstacles_together = True break if all_obstacles_together: break if all_obstacles_together: for i in range(n): for j in range(m): if grid[i][j] == 1: obstacle_count += abs(i - (n // 2)) + abs(j - (m // 2)) return obstacle_count else: return -1

Time Complexity:

The time complexity of the above algorithm is O(k * mn), where k is the number of moves and n and m are the dimensions of the grid. We have to iterate through the entire grid in each move to apply the shift operations, which leads to a time complexity of O(k * mn).

Space Complexity:

The space complexity of the above algorithm is O(nm), where n and m are the dimensions of the grid. We need to create a temporary copy of the grid in each move to apply the shift operations, which leads to a space complexity of O(nm).

1