Description
Given an m x n binary grid grid where each 1 marks the home of one friend, return the minimal total travel distance.
The total travel distance is the sum of the distances between the houses of the friends and the meeting point.
The distance is calculated using Manhattan Distance, where distance(p1, p2) = |p2.x - p1.x| + |p2.y - p1.y|.
Example 1:
Input: grid = [[1,0,0,0,1],[0,0,0,0,0],[0,0,1,0,0]]
Output: 6
Explanation: Given three friends living at (0,0), (0,4), and (2,2).
The point (0,2) is an ideal meeting point, as the total travel distance of 2 + 2 + 2 = 6 is minimal.
So return 6.
Example 2:
Input: grid = [[1,1]]
Output: 1
Constraints:
m == grid.lengthn == grid[i].length1 <= m, n <= 200grid[i][j] is either 0 or 1.- There will be at least two friends in the
grid.
Solutions
Solution 1
PythonJavaC++GoRust
class Solution:
def minTotalDistance(self, grid: List[List[int]]) -> int:
def f(arr, x):
return sum(abs(v - x) for v in arr)
rows, cols = [], []
for i, row in enumerate(grid):
for j, v in enumerate(row):
if v:
rows.append(i)
cols.append(j)
cols.sort()
i = rows[len(rows) >> 1]
j = cols[len(cols) >> 1]
return f(rows, i) + f(cols, j)(code-box)
class Solution {
public int minTotalDistance(int[][] grid) {
int m = grid.length, n = grid[0].length;
List<Integer> rows = new ArrayList<>();
List<Integer> cols = new ArrayList<>();
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == 1) {
rows.add(i);
cols.add(j);
}
}
}
Collections.sort(cols);
int i = rows.get(rows.size() >> 1);
int j = cols.get(cols.size() >> 1);
return f(rows, i) + f(cols, j);
}
private int f(List<Integer> arr, int x) {
int s = 0;
for (int v : arr) {
s += Math.abs(v - x);
}
return s;
}
}(code-box)
class Solution {
public:
int minTotalDistance(vector<vector<int>>& grid) {
int m = grid.size(), n = grid[0].size();
vector<int> rows;
vector<int> cols;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j]) {
rows.emplace_back(i);
cols.emplace_back(j);
}
}
}
sort(cols.begin(), cols.end());
int i = rows[rows.size() / 2];
int j = cols[cols.size() / 2];
auto f = [](vector<int>& arr, int x) {
int s = 0;
for (int v : arr) {
s += abs(v - x);
}
return s;
};
return f(rows, i) + f(cols, j);
}
};(code-box)
func minTotalDistance(grid [][]int) int {
rows, cols := []int{}, []int{}
for i, row := range grid {
for j, v := range row {
if v == 1 {
rows = append(rows, i)
cols = append(cols, j)
}
}
}
sort.Ints(cols)
i := rows[len(rows)>>1]
j := cols[len(cols)>>1]
f := func(arr []int, x int) int {
s := 0
for _, v := range arr {
s += abs(v - x)
}
return s
}
return f(rows, i) + f(cols, j)
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}(code-box)
impl Solution {
#[allow(dead_code)]
pub fn min_total_distance(grid: Vec<Vec<i32>>) -> i32 {
let n = grid.len();
let m = grid[0].len();
let mut row_vec = Vec::new();
let mut col_vec = Vec::new();
// Initialize the two vector
for i in 0..n {
for j in 0..m {
if grid[i][j] == 1 {
row_vec.push(i as i32);
col_vec.push(j as i32);
}
}
}
// Since the row vector is originally sorted, we only need to sort the col vector here
col_vec.sort();
Self::compute_manhattan_dis(&row_vec, row_vec[row_vec.len() / 2])
+ Self::compute_manhattan_dis(&col_vec, col_vec[col_vec.len() / 2])
}
#[allow(dead_code)]
fn compute_manhattan_dis(v: &Vec<i32>, e: i32) -> i32 {
let mut ret = 0;
for num in v {
ret += (num - e).abs();
}
ret
}
}(code-box)
