Maximize Rook Square Values - Google Top Interview Questions
Problem Statement :
You are given a two-dimensional list of integers board representing a chess board.
Return the maximum sum you can attain by placing two rooks in the board such that they can't attack each other.
The sum is made by adding the two numbers where the rooks are placed.
Constraints
2 ≤ n * m ≤ 200,000 where n and m are the number of rows and columns in board
Example 1
Input
board = [
[1, 9, 3, 1, 9],
[1, 1, 1, 1, 1],
[8, 1, 1, 1, 1]
]
Output
17
Explanation
We can take the 8 square and one of the 9 squares.
Solution :
Solution in C++ :
int solve(vector<vector<int>>& board) {
int n = board.size(), m = board[0].size();
auto go = [&]() {
int t = 0;
vector<int> a(m);
for (int i = 0; i < n; ++i) {
if (i > 0) {
int x = 0;
for (int j = 1; j < m; ++j) {
x = max(x, a[j - 1]);
t = max(t, x + board[i][j]);
}
}
for (int j = 0; j < m; ++j) a[j] = max(a[j], board[i][j]);
}
return t;
};
int ret = 0;
for (int i = 0; i < 2; ++i) {
ret = max(ret, go());
reverse(board.begin(), board.end());
}
return ret;
}
Solution in Java :
import java.util.*;
class Solution {
public int solve(int[][] grid) {
final int R = grid.length, C = grid[0].length;
int[][] topleft = A(grid, R, C);
int[][] topright = B(grid, R, C);
int[][] bottomleft = C(grid, R, C);
int[][] bottomright = D(grid, R, C);
int res = Integer.MIN_VALUE;
for (int r = 0; r != R; r++) {
for (int c = 0; c != C; c++) {
int tmp = Integer.MIN_VALUE;
if (r != 0 && c != 0)
tmp = Math.max(tmp, topleft[r - 1][c - 1]);
if (r != 0 && c != C - 1)
tmp = Math.max(tmp, topright[r - 1][c + 1]);
if (r != R - 1 && c != 0)
tmp = Math.max(tmp, bottomleft[r + 1][c - 1]);
if (r != R - 1 && c != C - 1)
tmp = Math.max(tmp, bottomright[r + 1][c + 1]);
res = Math.max(res, tmp + grid[r][c]);
}
}
return res;
}
private int[][] A(int[][] grid, final int R, final int C) { // top left
int[][] res = new int[R][C];
res[0][0] = grid[0][0];
for (int r = 0; r != R; r++) {
if (r != 0)
res[r][0] = Math.max(res[r - 1][0], grid[r][0]);
for (int c = 1; c != C; c++) {
int tmp = grid[r][c];
if (r != 0)
tmp = Math.max(tmp, res[r - 1][c]);
if (c != 0)
tmp = Math.max(tmp, res[r][c - 1]);
res[r][c] = tmp;
}
}
return res;
}
private int[][] B(int[][] grid, final int R, final int C) { // top right
int[][] res = new int[R][C];
res[0][C - 1] = grid[0][C - 1];
for (int r = 0; r != R; r++) {
if (r != 0)
res[r][C - 1] = Math.max(res[r - 1][C - 1], grid[r][C - 1]);
for (int c = C - 2; c != -1; c--) {
int tmp = grid[r][c];
if (r != 0)
tmp = Math.max(tmp, res[r - 1][c]);
if (c != C - 1)
tmp = Math.max(tmp, res[r][c + 1]);
res[r][c] = tmp;
}
}
return res;
}
private int[][] C(int[][] grid, final int R, final int C) { // bottom left
int[][] res = new int[R][C];
res[R - 1][0] = grid[R - 1][0];
for (int r = R - 1; r != -1; r--) {
if (r != R - 1)
res[r][0] = Math.max(res[r + 1][0], grid[r][0]);
for (int c = 1; c != C; c++) {
int tmp = grid[r][c];
if (r != R - 1)
tmp = Math.max(tmp, res[r + 1][c]);
if (c != 0)
tmp = Math.max(tmp, res[r][c - 1]);
res[r][c] = tmp;
}
}
return res;
}
private int[][] D(int[][] grid, final int R, final int C) { // bottom right
int[][] res = new int[R][C];
res[R - 1][C - 1] = grid[R - 1][C - 1];
for (int r = R - 1; r != -1; r--) {
if (r != R - 1)
res[r][C - 1] = Math.max(res[r + 1][C - 1], grid[r][C - 1]);
for (int c = C - 2; c != -1; c--) {
int tmp = grid[r][c];
if (r != R - 1)
tmp = Math.max(tmp, res[r + 1][c]);
if (c != C - 1)
tmp = Math.max(tmp, res[r][c + 1]);
res[r][c] = tmp;
}
}
return res;
}
}
Solution in Python :
def solve(self, board):
board_vals = [(val, (r, c)) for r, row in enumerate(board) for c, val in enumerate(row)]
mx = max(board_vals)
mx2 = max(v for v in board_vals if v != mx)
mxs = [mx, mx2]
result = 0
for v, (r, c) in mxs:
for nv, (nr, nc) in board_vals:
if nr != r and c != nc:
result = max(result, v + nv)
return result
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