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package g2401_2500.s2428_maximum_sum_of_an_hourglass;
// #Medium #Array #Matrix #Prefix_Sum #2022_12_07_Time_4_ms_(93.51%)_Space_44.1_MB_(86.28%)
public class Solution {
public int maxSum(int[][] grid) {
int m = grid.length;
int n = grid[0].length;
int res = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (isHourGlass(i, j, m, n)) {
res = Math.max(res, calculate(i, j, grid));
} else {
// If we cannot form an hour glass from the current row anymore, just move to
// the next one
break;
}
}
}
return res;
}
// Check if an hour glass can be formed from the current position
private boolean isHourGlass(int r, int c, int m, int n) {
return r + 2 < m && c + 2 < n;
}
// Once we know an hourglass can be formed, just traverse the value
private int calculate(int r, int c, int[][] grid) {
int sum = 0;
// Traverse the bottom and the top row of the hour glass at the same time
for (int i = c; i <= c + 2; i++) {
sum += grid[r][i];
sum += grid[r + 2][i];
}
// Add the middle vlaue
sum += grid[r + 1][c + 1];
return sum;
}
}
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