🌵840. Magic Squares In Grid
https://leetcode.com/problems/magic-squares-in-grid/
js
/**
* @param {number[][]} grid
* @return {number}
*/
var numMagicSquaresInside = function(grid) {
const rows = grid.length
const cols = grid[0].length
if (rows < 3 || cols < 3) {
return 0
}
let count = 0
for (let i = 0; i < rows - 2; i++) {
for (let j = 0; j < cols - 2; j++) {
const subgrid = grid.slice(i, i + 3).map(row => row.slice(j, j + 3))
if (subgrid[1][1] !== 5) {
continue
}
const nums = subgrid[0].concat(subgrid[1]).concat(subgrid[2])
const numMatch = new Set(nums).size === 9 && nums.every(n => n >= 1 && n <= 9)
if (!numMatch) {
continue
}
const rowSums = subgrid.map(r => r[0] + r[1] + r[2])
const colSums = [0, 1, 2].map(c => subgrid[0][c] + subgrid[1][c] + subgrid[2][c])
const sumD1 = subgrid[0][0] + subgrid[1][1] + subgrid[2][2]
const sumD2 = subgrid[0][2] + subgrid[1][1] + subgrid[2][0]
if (new Set(rowSums.concat(colSums).concat([sumD1, sumD2])).size === 1) {
count += 1
}
}
}
return count
}py
class Solution:
def numMagicSquaresInside(self, grid):
"""
:type grid: List[List[int]]
:rtype: int
"""
rows = len(grid)
cols = len(grid[0])
if rows < 3 or cols < 3:
return 0
count = 0
for i in range(rows - 2):
for j in range(cols - 2):
subgrid = [row[j:j + 3] for row in grid[i:i + 3]]
if subgrid[1][1] != 5:
continue
num_match = set([x for row in subgrid for x in row]) == set(
range(1, 10))
if not num_match:
continue
row_sums = [sum(row) for row in subgrid]
col_sums = [sum([subgrid[r][c] for r in range(3)])
for c in range(3)]
sum_d1 = subgrid[0][0] + subgrid[1][1] + subgrid[2][2]
sum_d2 = subgrid[0][2] + subgrid[1][1] + subgrid[2][0]
if len(set(row_sums + col_sums + [sum_d1, sum_d2])) == 1:
count += 1
return count