Best Meeting Point - Leetcode Solution

The Best Meeting Point problem on LeetCode is a problem of finding the best place to meet for a group of people in a 2D grid. The objective is to minimize the total distance that the people have to travel to reach the meeting point. In this problem, the grid is represented by a binary matrix, where 1 represents an empty cell and 0 represents an obstacle.

Approach:

One approach to solve this problem is to find the medians of the rows and columns separately. To do this, we can iterate through all the cells of the grid, count the number of people in each row and column, and calculate the total distance to each cell of that row or column. Then, we can find the cell with the minimum distance, which will be the median of that row or column.

Next, we can calculate the total distance of all the people to this cell. We can repeat the same process for the columns, and find the cell which has the minimum total distance.

Implementation:

The following is the stepwise implementation of the Best Meeting Point problem in Python:

1. Find the row medians:

def minTotalDistance(grid):
    rows = len(grid)
    cols = len(grid[0])

    row_sum = [0] * rows
    col_sum = [0] * cols

    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == 1:
                row_sum[i] += 1
                col_sum[j] += 1

    row_median = findMedian(row_sum)
    col_median = findMedian(col_sum)

    return calcDistance(grid, row_median, col_median)

Here, we are finding the medians of the rows and columns separately, and then we are calling the function calcDistance to calculate the minimum distance. The findMedian function is used to find the median of the given array.

2. Find the median of an array:

def findMedian(arr):
    median = 0
    n = len(arr)
    if n % 2 == 0:
        median = (arr[n // 2 - 1] + arr[n // 2]) // 2
    else:
        median = arr[n // 2]
    return median

This function calculates the median of an array by checking whether the length of the array is even or odd.

3. Calculate the distance:

def calcDistance(grid, row_median, col_median):
    total_distance = 0
    rows = len(grid)
    cols = len(grid[0])

    for i in range(rows):
        for j in range(cols):
            if grid[i][j] == 1:
                total_distance += abs(row_median - i) + abs(col_median - j)

    return total_distance

This function calculates the minimum distance by iterating through all the cells of the grid, and calculating the distance of each cell to the median of the rows and columns.

Time Complexity:

The time complexity of this algorithm is O(m * n), where m and n are the dimensions of the grid.

Space Complexity:

The space complexity of this algorithm is O(m + n), where m and n are the dimensions of the grid.

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